---
title: "Differential Calculus"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 15
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/15-differential-calculus
---

# Chapter 15 — Differential Calculus

The two-variable calculus of the Year 1 volume matures into the differential calculus of maps between normed spaces: the *[differential](#def-b2-diffcalc-differential)* as the best linear approximation, the chain rule in full generality, Schwarz’s symmetry theorem *proved*, Taylor formulas, and the complete second-order analysis of extrema. The inverse function theorem, crown of the theory, is stated with its proof strategy — a Banach fixed point — made explicit.

Throughout, $U$ is an [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) subset of $\R^n$ (or a normed space; the finite-dimensional case carries all the ideas), $f \colon U \to
\R^m$.

## 15.1 The differential

**Definition 15.1.**

$f$ is *differentiable* at $a$ when there is a ([continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)) linear map $\dd f_a \colon \R^n \to \R^m$ with

$$
f(a + h) = f(a) + \dd f_a(h) + o(\norm h)
\qquad (h \to 0).
$$

The map $\dd f_a$, the *differential* of $f$ at $a$, is unique; its matrix in the canonical bases is the *Jacobian matrix* $J_f(a) =
\bigl(\frac{\partial f_i}{\partial x_j}(a)\bigr)$. Differentiability implies [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and the existence of all directional derivatives $\dd f_a(v) = \lim_{t\to0}\frac{f(a + tv)
- f(a)}{t}$; the converse fails ([Exercise 15.2](#exo-b2-diffcalc-2)). For $m = 1$, $\dd f_a(h) = \langle
\nabla f(a), h\rangle$: the gradient of Year 1, now understood as the vector representing the differential.

**Theorem 15.2 (C1C^1C1 criterion).**

If all partial derivatives of $f$ exist on $U$ and are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $a$, then $f$ is [differentiable](#def-b2-diffcalc-differential) at $a$. “$C^1$ on $U$” — [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) partials — therefore implies [differentiability](#def-b2-diffcalc-differential) everywhere, with [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) [differential](#def-b2-diffcalc-differential).

**Proof.** Component by component ($m = 1$ suffices). The Year 1 proof for two variables — move one coordinate at a time, apply the one-variable mean value theorem on each leg, and use [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of the partials at $a$ — generalizes verbatim to $n$ legs:

$$
f(a + h) - f(a) = \sum_{j=1}^{n} \bigl(f(a + h^{(j)}) - f(a +
h^{(j-1)})\bigr)
= \sum_j h_j\,\frac{\partial f}{\partial x_j}(\xi_j),
$$

where $h^{(j)}$ freezes the first $j$ coordinates of $h$ and $\xi_j$ lies on the $j$-th leg; [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) turns each $\frac{\partial f}{\partial x_j}(\xi_j)$ into $\frac{\partial
f}{\partial x_j}(a) + o(1)$, and the error is $o(\norm h)$. ∎

**Theorem 15.3 (Chain rule).**

If $f$ is [differentiable](#def-b2-diffcalc-differential) at $a$ and $g$ at $f(a)$, then $g \circ f$ is [differentiable](#def-b2-diffcalc-differential) at $a$ with

$$
\dd(g \circ f)_a = \dd g_{f(a)} \circ \dd f_a ,
\qquad
J_{g\circ f}(a) = J_g\bigl(f(a)\bigr)\,J_f(a) :
$$

Jacobians multiply.

**Proof.** Write $f(a + h) = f(a) + \dd f_a(h) + \norm h\,\varepsilon_1(h)$ and $g(b + k) = g(b) + \dd g_b(k) + \norm k\,\varepsilon_2(k)$ with $b = f(a)$, $k = k(h) = \dd f_a(h) + \norm h \varepsilon_1(h)$. Substituting,

$$
g(f(a+h)) = g(b) + \dd g_b\bigl(\dd f_a(h)\bigr)
+ \norm h\,\dd g_b(\varepsilon_1(h)) + \norm{k}\,\varepsilon_2(k),
$$

and both error terms are $o(\norm h)$: the first since $\dd g_b$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and $\varepsilon_1 \to 0$; the second since $\norm k
\leq C\norm h$ (bounded linear map plus small term) and $\varepsilon_2(k) \to 0$ as $h \to 0$. ∎

**Example 15.4 (Radial functions, once and for all).**

Let $r(x) = \norm x_2$ on $\R^n\setminus\{0\}$ and $f = g \circ
r$ with $g$ a $C^1$ function of one variable. First, $r$ is [differentiable](#def-b2-diffcalc-differential) away from $0$: from $r^2 = \sum x_i^2$,

$$
\frac{\partial r}{\partial x_i} = \frac{x_i}{r},
\qquad\text{i.e.}\qquad
\nabla r(x) = \frac{x}{\norm x} ,
$$

the unit radial vector (differentiate $r^2$ and divide — or apply the chain rule to $\sqrt{\cdot}$). Then the chain rule gives, for every radial function,

$$
\nabla f(x) = g'\bigl(\norm x\bigr)\,\frac{x}{\norm x} .
$$

Worked instance: $g(r) = \frac1r$ yields $\nabla\frac{1}{\norm
x} = -\frac{x}{\norm x^3}$, the inverse-square field of gravitation and electrostatics — direction radial, magnitude $\frac{1}{\norm x^2}$. Closing insight: gradients of radial functions are radial because the level sets are spheres and the gradient is orthogonal to level sets; at $x = 0$, by contrast, $r$ is *not* [differentiable](#def-b2-diffcalc-differential) (no candidate linear map matches $\norm h$ from all directions) — smooth radial profiles need $g'(0) = 0$ to cross the origin gracefully.

**Theorem 15.5 (Mean value inequality).**

Let $f$ be [differentiable](#def-b2-diffcalc-differential) on $U$ and let the segment $\intcc{a}{b}
= \{a + t(b-a)\}$ lie in $U$. Then

$$
\norm{f(b) - f(a)} \leq \norm{b - a}\,
\sup_{x \in \intcc{a}{b}} \vertiii{\dd f_x} .
$$

In particular, a [differentiable](#def-b2-diffcalc-differential) map with zero [differential](#def-b2-diffcalc-differential) on a *[connected](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-connected)* [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) set is constant.

**Proof.** The function $\varphi(t) = f(a + t(b-a))$ is [differentiable](#def-b2-diffcalc-differential) on $\intcc{0}{1}$ with $\varphi'(t) = \dd f_{a + t(b-a)}(b - a)$ (chain rule), of [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $\leq M\norm{b-a}$ with $M$ the displayed sup. For $\R$-valued $f$ the one-variable mean value inequality concludes; for vector values apply it to $t \mapsto \langle u,
\varphi(t)\rangle$ with $u$ the unit vector along $f(b) - f(a)$. Constancy: locally constant (segments in balls) plus [connectedness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-connected) (the set where $f$ equals a given value is [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) and closed: [Chapter 4](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#ch-b2-metric)). ∎

**Example 15.6 (A Lipschitz constant from the MVI).**

Is $f(x, y) = \sin x\,\sin y$ [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\R^2$, and with which constant? Its gradient is $\nabla f = (\cos x\sin y,\
\sin x\cos y)$, of squared [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm)

$$
\cos^2x\sin^2y + \sin^2x\cos^2y
\leq \sin^2 y + \cos^2y\cdot 1 = 1
$$

(bound $\cos^2x$ and $\sin^2x$ by $1$ separately), so $\vertiii{\dd f_{(x,y)}} = \norm{\nabla f} \leq 1$ everywhere, and [Theorem 15.5](#thm-b2-diffcalc-mvi) on the segment between any two points gives

$$
\abs{f(b) - f(a)} \leq \norm{b - a}_2 :
$$

$f$ is $1$-Lipschitz, and the constant is sharp (near the origin, $f(x, \tfrac\pi2) = \sin x$ has slope $1$). Closing insight: the mean value inequality converts a [pointwise](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) bound on the [differential](#def-b2-diffcalc-differential) into a global modulus of [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) — the standard route to [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) estimates in every dimension, and the engine inside [Exercise 15.12](#exo-b2-diffcalc-12).

## 15.2 Second derivatives

**Theorem 15.7 (Schwarz).**

If $f$ is $C^2$ on $U$ (all second partials exist and are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)), then for all $i, j$:

$$
\frac{\partial^2 f}{\partial x_i\,\partial x_j}
= \frac{\partial^2 f}{\partial x_j\,\partial x_i} .
$$

**Proof.** Two variables suffice ($x = x_i$, $y = x_j$, others frozen). Consider the second difference

$$
\Delta(h) = f(a + h, b + h) - f(a + h, b) - f(a, b + h) + f(a,b) .
$$

Fix $h$ and set $\varphi(x) = f(x, b+h) - f(x, b)$: then $\Delta(h)
= \varphi(a + h) - \varphi(a)$, and two applications of the mean value theorem give

$$
\Delta(h) = h\,\varphi'(\xi)
= h\Bigl(\frac{\partial f}{\partial x}(\xi, b+h) -
\frac{\partial f}{\partial x}(\xi, b)\Bigr)
= h^2\,\frac{\partial^2 f}{\partial y\,\partial x}(\xi, \eta),
$$

with $\xi \in \intoo{a}{a+h}$, $\eta \in \intoo{b}{b+h}$. By [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), $\frac{\Delta(h)}{h^2} \to \frac{\partial^2 f}{\partial
y\partial x}(a, b)$ as $h \to 0$. The same computation with the roles of the variables exchanged (freeze the second variable first) gives $\frac{\Delta(h)}{h^2} \to \frac{\partial^2 f}{\partial x
\partial y}(a,b)$: the two limits of the same quantity coincide. ∎

**Example 15.8 (Why C2C^2C2 is needed: Peano’s counterexample).**

Let $f(x, y) = \dfrac{xy(x^2 - y^2)}{x^2 + y^2}$, $f(0,0) = 0$. Away from the origin $f$ is $C^\infty$; at the origin all first and second partials exist, but the mixed ones disagree. Compute along the axes: $f(x, 0) = f(0, y) = 0$, and for $y \neq 0$,

$$
\frac{\partial f}{\partial x}(0, y)
= \lim_{x\to0}\frac{f(x,y)}{x}
= \frac{y(0 - y^2)}{y^2} = -y ,
\qquad\text{symmetrically}\qquad
\frac{\partial f}{\partial y}(x, 0) = x .
$$

Hence

$$
\frac{\partial^2 f}{\partial y\,\partial x}(0,0)
= \frac{\dd}{\dd y}\Bigl[\frac{\partial f}{\partial
x}(0,y)\Bigr]_{y=0} = -1,
\qquad
\frac{\partial^2 f}{\partial x\,\partial y}(0,0) = +1 :
$$

the two mixed partials exist and differ. No contradiction with [Theorem 15.7](#thm-b2-diffcalc-schwarz): the second partials of $f$ are not [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$ (test along $y = tx$). Closing insight: Schwarz’s theorem is a genuine theorem about *[continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)*, not a formal identity — and the hypothesis “$C^2$” in Taylor–Young below is doing real work.

**Theorem 15.9 (Taylor–Young at order 2).**

Let $f \colon U \to \R$ be $C^2$ and $a \in U$. Then, as $h \to 0$,

$$
f(a + h) = f(a) + \langle\nabla f(a), h\rangle
+ \frac12\, \langle H_a h,\, h\rangle + o\bigl(\norm h^2\bigr),
$$

where $H_a = \bigl(\frac{\partial^2 f}{\partial x_i\partial
x_j}(a)\bigr)$ is the ([symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint), by Schwarz) *Hessian matrix*.

**Proof.** Apply the one-variable Taylor–Young theorem (Year 1 volume) to $\varphi(t) = f(a + th)$ on $\intcc{0}{1}$: by the chain rule, $\varphi'(t) = \langle \nabla f(a + th), h\rangle$ and $\varphi''(t)
= \langle H_{a+th}h, h\rangle$, both [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $t$. Then $\varphi(1) = \varphi(0) + \varphi'(0) + \frac12\varphi''(\theta)$ (Taylor–Lagrange) with $\theta \in \intoo{0}{1}$, and [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of the second partials converts $\varphi''(\theta) =
\varphi''(0) + o(1)\cdot\norm h^2$-uniformly: the displayed expansion. ∎

**Example 15.10 (An expansion two ways).**

Expand $f(x, y) = \eu^x\cos y$ at the origin to order $2$. *By composition of one-variable expansions:*

$$
\eu^x\cos y = \Bigl(1 + x + \frac{x^2}{2} +
o(x^2)\Bigr)\Bigl(1 - \frac{y^2}{2} + o(y^2)\Bigr)
= 1 + x + \frac{x^2 - y^2}{2} + o\bigl(\norm{(x,y)}^2\bigr) .
$$

*By partial derivatives:* $f_x = \eu^x\cos y$, $f_y =
-\eu^x\sin y$, so $\nabla f(0) = (1, 0)$; and $f_{xx} = f$, $f_{yy} = -f$, $f_{xy} = -\eu^x\sin y$ give $H_0 =
\operatorname{diag}(1, -1)$: [Theorem 15.9](#thm-b2-diffcalc-taylor) reproduces $1 + x + \frac12(x^2 - y^2)$. The two computations agree, and the composition route was faster — no second partials at all. Closing insight: the origin is *not* a critical point ($\nabla f \neq 0$), so despite the indefinite Hessian there is no saddle to declare: the linear term rules, and the second-order test only ever speaks at critical points.

**Theorem 15.11 (Second-order extremum test, proved).**

Let $f$ be $C^2$ near a critical point $a$ ($\nabla f(a) = 0$), with Hessian $H = H_a$.

1. If $H$ is positive definite, $a$ is a strict local minimum (negative definite: maximum).
2. If $H$ has [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of both signs, $a$ is a saddle: no extremum.
3. If $H$ is singular (and semidefinite), no conclusion.

The Year 1 “$rt - s^2$” test is the case $n = 2$: $\det H = rt -
s^2$, $\operatorname{tr}$-sign read off $r$.

**Proof.** By the spectral theorem ([Theorem 12.13](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#thm-b2-quadratic-spectral)), $H$’s [quadratic form](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def) is squeezed between its extreme [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen): $\lambda_{\min}\norm h^2 \leq \langle Hh, h\rangle \leq
\lambda_{\max}\norm h^2$.

(1) If $\lambda_{\min} > 0$: Taylor–Young gives

$$
f(a + h) - f(a) \geq \frac{\lambda_{\min}}{2}\norm h^2 -
o(\norm h^2) > 0
$$

for small $h \neq 0$: strict local minimum.

(2) Along an [eigenvector](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $v_+$ with $\lambda_+ > 0$: $f(a + tv_+) -
f(a) = \frac{\lambda_+}{2}t^2 + o(t^2) > 0$ for small $t$; along $v_-$ with $\lambda_- < 0$ the difference is negative: both signs occur in every neighborhood.

(3) $f(x,y) = x^2 + y^4$, $x^2 - y^4$, $x^2 + y^3$ share the same singular semidefinite Hessian at $0$ with three different behaviors. ∎

**Example 15.12 (A complete classification, global included).**

Classify all extrema of $f(x, y) = x^4 + y^4 - 4xy$ on $\R^2$. *Critical points:* $\nabla f = (4x^3 - 4y,\ 4y^3 - 4x) =
0$ gives $y = x^3$ and $x = y^3 = x^9$, so $x(x^8 - 1) = 0$: the real solutions are $(0,0)$, $(1,1)$, $(-1,-1)$. *Hessians:* $H = \begin{pmatrix} 12x^2 & -4\\ -4 &
12y^2\end{pmatrix}$. At $(\pm1, \pm1)$: $\begin{pmatrix} 12 &
-4\\ -4 & 12\end{pmatrix}$, [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $8$ and $16$: positive definite, strict local minima with $f = -2$. At $(0,0)$: $\begin{pmatrix} 0 & -4\\ -4 & 0\end{pmatrix}$, [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\pm4$: a saddle. *Globality:* from $2\abs{xy} \leq x^2 +
y^2$,

$$
f(x, y) \geq x^4 + y^4 - 2(x^2 + y^2)
= (x^2 - 1)^2 + (y^2 - 1)^2 + x^2 + y^2 - 2
\xrightarrow[\norm{(x,y)}\to\infty]{} +\infty :
$$

$f$ is coercive, so it attains a global minimum ([compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) of sublevel sets), necessarily at a critical point: the value $-2$, at both $(1,1)$ and $(-1,-1)$, is the global minimum; there is no maximum ($f$ unbounded above). Closing insight: the local test classifies candidates, but only a growth argument turns “local” into “global” — the two-step pattern of every optimization proof in this book.

**Method 15.13 (Classifying the extrema of f ⁣:Rn→Rf \colon \R^n \to \Rf:Rn→R).**

1. Solve $\nabla f = 0$ (all critical points; on a domain with boundary, treat the boundary separately as in [Exercise 15.7](#exo-b2-diffcalc-7) ).
2. At each critical point, compute the Hessian and its [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) signs — in dimension $2$ , just $\det H$ and $\operatorname{tr} H$ : $\det < 0$ saddle; $\det >  0$ extremum, of the type given by the sign of the trace; $\det = 0$ : the test is silent, study $f$ along curves.
3. For global statements, add a [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) or coercivity argument ( $f \to +\infty$ at infinity, or a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) constraint set), then compare the critical values.

## 15.3 The inverse function theorem

**Theorem 15.14 (Inverse function theorem).**

Let $f \colon U \to \R^n$ be $C^1$ and $a \in U$ with $\dd f_a$ *invertible*. Then there are [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) neighborhoods $V \ni a$, $W
\ni f(a)$ such that $f \colon V \to W$ is a bijection with $C^1$ inverse, and

$$
\dd (f^{-1})_{f(x)} = (\dd f_x)^{-1} \qquad (x \in V).
$$

**Proof.** *Admitted at this level.* ∎

**Remark 15.15 (Why it is true: the fixed-point strategy).**

Solving $f(x) = y$ near $a$ rewrites as the fixed-point equation $x = x + \dd f_a^{-1}\bigl(y - f(x)\bigr) =: \Phi_y(x)$; the map $\Phi_y$ has [differential](#def-b2-diffcalc-differential) $\mathrm{id} - \dd f_a^{-1}\dd f_x$, small near $a$ by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $\dd f$, so $\Phi_y$ is a contraction on a small closed ball and the Banach fixed point theorem ([Theorem 4.12](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-banach)) provides the unique local solution $x = f^{-1}(y)$. [Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and [differentiability](#def-b2-diffcalc-differential) of the inverse then follow from the estimates in the contraction. The full bookkeeping is carried out in Year 3; the strategy — and the statement — are used freely from now on. The companion *implicit function theorem* (solving $F(x, y) = 0$ for $y(x)$ when $\frac{\partial F}{\partial y}$ is invertible) follows by applying the theorem to $(x, y) \mapsto (x, F(x,y))$.

**Example 15.16 (Polar coordinates).**

$\Phi(r, \theta) = (r\cos\theta, r\sin\theta)$ has Jacobian

$$
J_\Phi = \begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta &
r\cos\theta \end{pmatrix},
\qquad \det J_\Phi = r :
$$

invertible for $r \neq 0$, so $\Phi$ is a local $C^1$ diffeomorphism away from the origin — the licence to “change to polar coordinates”, renewed for the multiple integrals of [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint).

**Example 15.17 (Local everywhere, global nowhere).**

Let $f(x, y) = \bigl(\eu^x\cos y,\ \eu^x\sin y\bigr)$ on $\R^2$. Its Jacobian,

$$
J_f = \begin{pmatrix}
\eu^x\cos y & -\eu^x\sin y\\
\eu^x\sin y & \eu^x\cos y
\end{pmatrix},
\qquad
\det J_f = \eu^{2x} > 0 ,
$$

never vanishes: by [Theorem 15.14](#thm-b2-diffcalc-inverse), $f$ is a local $C^1$ diffeomorphism at *every* point of the plane. Yet $f$ is far from injective: $f(x, y + 2\pi) = f(x, y)$, so every value is taken infinitely often; and it is not surjective either, since $\norm{f(x, y)} = \eu^x > 0$ misses the origin. Closing insight: the inverse function theorem is irreducibly *local* — invertibility of every $\dd f_a$ yields a patchwork of local inverses that need not assemble into one. (Readers who know complex numbers will recognize $z \mapsto
\eu^z$; the patchwork is the family of logarithm branches.) Compare [Exercise 15.12](#exo-b2-diffcalc-12), where a quantitative global hypothesis does force one global inverse.

**Remark 15.18 (Common pitfalls).**

*(i) Directional derivatives are cheap, [differentials](#def-b2-diffcalc-differential) are not:* all directional derivatives may exist — and even fail to depend linearly on the direction — without [differentiability](#def-b2-diffcalc-differential) ([Exercise 15.2](#exo-b2-diffcalc-2)); only the $C^1$ criterion ([Theorem 15.2](#thm-b2-diffcalc-c1)) upgrades partials to a [differential](#def-b2-diffcalc-differential). *(ii) Critical does not mean extremal:* saddles ([Example 15.12](#ex-b2-diffcalc-quarticrun)) and the silent singular case ([Theorem 15.11](#thm-b2-diffcalc-extrema) (3)) both hide behind $\nabla f = 0$. *(iii) No vector-valued mean value equality:* only the *inequality* of [Theorem 15.5](#thm-b2-diffcalc-mvi) survives ([Exercise 15.9](#exo-b2-diffcalc-9)); never write $f(b) - f(a) = \dd
f_c(b-a)$ for $f$ with values in $\R^m$, $m \geq 2$. *(iv) Local invertibility is not injectivity:* [Example 15.17](#ex-b2-diffcalc-localnotglobal). *(v) The gradient belongs to the inner product:* $\nabla f$ is the vector representing $\dd f_a$ in a chosen inner product; change the product (as in the weighted example of the quadratic-forms chapter) and the gradient rotates, while the [differential](#def-b2-diffcalc-differential) — the intrinsic object — does not move.

**Remark 15.19 (Where this is used).**

Everything downstream of this chapter is differential calculus applied: the differential-equations chapter linearizes flows and uses Liouville’s [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) formula (proved in this chapter’s weekend problem); the chapters on curves and surfaces study level sets and parametrizations through the implicit function theorem; multiple integrals change variables through Jacobians. The weekend problem develops the calculus *on the space of matrices itself* — [differential](#def-b2-diffcalc-differential) of the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det), of the inverse, the [matrix exponential](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp), and the orthogonal group as a smooth level set — the Year 2 shadow of what the Year 3 volume formalizes as manifolds and Lie groups.

## 15.4 Exercises

**Exercise 15.1 ★.**

Compute the Jacobian matrices of $f(x,y) = (x^2 - y^2,\, 2xy)$ and of $\Phi(r,\theta,z) = (r\cos\theta, r\sin\theta, z)$; where are the [differentials](#def-b2-diffcalc-differential) invertible?

**Solution of Exercise 15.1.**

$J_f = \begin{pmatrix} 2x & -2y\\ 2y & 2x\end{pmatrix}$, $\det J_f
= 4(x^2 + y^2)$: invertible away from the origin. (This $f$ is $z
\mapsto z^2$ in complex disguise.)

$J_\Phi = \begin{pmatrix} \cos\theta & -r\sin\theta & 0\\
\sin\theta & r\cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}$, $\det = r$: invertible for $r \neq 0$ (cylindrical coordinates).

**Exercise 15.2 ★.**

Let $f(x,y) = \frac{x^3}{x^2 + y^2}$ ($f(0,0) = 0$). Prove that all directional derivatives of $f$ at $0$ exist, but that $f$ is not [differentiable](#def-b2-diffcalc-differential) at $0$ *(the map $v \mapsto$ directional derivative is not linear)*.

**Solution of Exercise 15.2.**

For $v = (a, b) \neq 0$: $\frac{f(tv) - 0}{t} = \frac{t^3a^3}{t\cdot
t^2(a^2+b^2)} = \frac{a^3}{a^2 + b^2}$: every directional derivative exists, with value $D_v = \frac{a^3}{a^2+b^2}$. But $v
\mapsto D_v$ is not linear ($D_{(1,0)} = 1$, $D_{(0,1)} = 0$, $D_{(1,1)} = \frac12 \neq 1$): no linear map can produce these values, so $f$ is not [differentiable](#def-b2-diffcalc-differential) at $0$ (the [differential](#def-b2-diffcalc-differential) would have to be $v \mapsto D_v$).

**Exercise 15.3 ★.**

Find and classify the critical points of $f(x, y) = x^3 + y^3 -
3xy$ using [Theorem 15.11](#thm-b2-diffcalc-extrema), and of $g(x,y) = x^4 +
y^4 - 2(x - y)^2$.

**Solution of Exercise 15.3.**

$f = x^3 + y^3 - 3xy$: critical points $(0,0)$ and $(1,1)$ (Year 1 computation). Hessians: $H = \begin{pmatrix} 6x & -3\\ -3 &
6y\end{pmatrix}$. At $(0,0)$: eigen-signs mixed ($\det = -9 < 0$): saddle. At $(1,1)$: $\det = 27 > 0$, trace $> 0$: positive definite, strict local minimum — now justified by [Theorem 15.11](#thm-b2-diffcalc-extrema) rather than decreed.

$g = x^4 + y^4 - 2(x-y)^2$: $\nabla g = (4x^3 - 4(x - y),\; 4y^3 +
4(x-y))$; critical points $(0,0)$, $(\sqrt2, -\sqrt2)$, $(-\sqrt2, \sqrt2)$ (Year 1). At $(\pm\sqrt2, \mp\sqrt2)$: $H =
\begin{pmatrix} 12\cdot2 - 4 & 4\\ 4 & 20\end{pmatrix} =
\begin{pmatrix} 20 & 4\\ 4 & 20\end{pmatrix}$: positive definite (diagonally dominant; [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $24, 16$): strict local minima. At $(0,0)$: $H = \begin{pmatrix} -4 & 4\\ 4 & -4\end{pmatrix}$, singular negative semidefinite: the test is silent; the directional study ($g(x,x) = 2x^4 > 0$, $g(x,-x) = 2x^4 - 8x^2 < 0$ small) shows a saddle-like point — no extremum.

**Exercise 15.4 ★★.**

Let $f \colon \R^n \to \R$ be $C^1$ and *homogeneous of degree $p$*: $f(tx) = t^pf(x)$ for $t > 0$. Prove Euler’s identity

$$
\langle \nabla f(x), x\rangle = p\,f(x) ,
$$

and its converse for $C^1$ functions on $\R^n\setminus\{0\}$.

**Solution of Exercise 15.4.**

Differentiate $t \mapsto f(tx)$ at $t = 1$: by the chain rule, $\langle \nabla f(x), x\rangle$; by homogeneity the same function is $t^pf(x)$, of derivative $pf(x)$ at $t = 1$: Euler’s identity.

Converse: fix $x \neq 0$ and let $\varphi(t) = f(tx) - t^p f(x)$ on $t > 0$. Then $\varphi'(t) = \langle\nabla f(tx), x\rangle -
pt^{p-1}f(x) = \frac1t\bigl(\langle \nabla f(tx), tx\rangle -
p\,t^pf(x)\bigr)$. The hypothesis — Euler’s identity at the point $tx$ — evaluates the bracket as $p\,f(tx) - p\,t^pf(x) =
p\,\varphi(t)$. So $\varphi' = \frac{p}{t}\varphi$ with $\varphi(1)
= 0$: the linear ODE’s unique solution is $\varphi \equiv 0$ (Year 1 uniqueness), i.e. $f(tx) = t^pf(x)$.

**Exercise 15.5 ★★.**

Let $A$ be [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) and $f(x) = \frac12\langle Ax, x\rangle -
\langle b, x\rangle$. Compute $\nabla f$ and $H_f$; when is $f$ convex? Assuming $A$ positive definite, show that $f$ has a unique global minimum at the solution of $Ax = b$ — gradient descent’s raison d’être.

**Solution of Exercise 15.5.**

Expanding $f(x + h) - f(x) = \langle Ax - b, h\rangle +
\frac12\langle Ah, h\rangle$ (symmetry of $A$): $\nabla f(x) = Ax -
b$ and $H_f = A$ everywhere. $f$ is convex iff $A$ is positive semidefinite (the Hessian test, global here since $H$ is constant: the second-order Taylor formula is exact). If $A$ is positive definite: the unique critical point is $x^* = A^{-1}b$, and $f(x^*
+ h) - f(x^*) = \frac12\langle Ah, h\rangle \geq
\frac{\lambda_{\min}}{2}\norm h^2 > 0$ for $h \neq 0$: strict global minimum.

**Exercise 15.6 ★★.**

(Lagrange multiplier, one constraint, proved by hand) Let $f, g$ be $C^1$ on $\R^2$, and suppose $f$ attains, at $a$, a local extremum on the level set $\{g = 0\}$, with $\nabla g(a) \neq 0$. Prove that $\nabla f(a) = \lambda\nabla g(a)$ for some $\lambda$. *(Parametrize the level set near $a$ by the implicit function theorem and differentiate $t \mapsto f(\gamma(t))$.)* Application: extrema of $f(x,y) = xy$ on the circle $x^2 + y^2 = 1$.

**Solution of Exercise 15.6.**

Since $\nabla g(a) \neq 0$, one partial, say $\frac{\partial
g}{\partial y}(a) \neq 0$: the implicit function theorem ([Theorem 15.14](#thm-b2-diffcalc-inverse)’s companion) parametrizes $\{g =
0\}$ near $a = (a_1, a_2)$ as $\gamma(t) = (t, y(t))$ with $y$ $C^1$, $y'(t) = -\frac{\partial_x g}{\partial_y g}(\gamma(t))$ (differentiate $g(t, y(t)) = 0$). The one-variable function $t
\mapsto f(\gamma(t))$ has a local extremum at $t = a_1$:

$$
0 = \frac{\dd}{\dd t}f(\gamma(t))\Big|_{a_1}
= \partial_x f(a) + \partial_y f(a)\,y'(a_1)
= \partial_x f(a) - \partial_yf(a)\frac{\partial_x
g(a)}{\partial_y g(a)} :
$$

the vectors $\nabla f(a)$ and $\nabla g(a)$ have proportional coordinates: $\nabla f(a) = \lambda \nabla g(a)$ with $\lambda =
\frac{\partial_y f(a)}{\partial_y g(a)}$.

Application: on the circle, $\nabla(xy) = (y, x)$ parallel to $(2x, 2y)$ forces $y^2 = x^2$; with the constraint, the candidates are $\pm\bigl(\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\bigr)$ (value $\frac12$) and $\pm\bigl(\tfrac{1}{\sqrt2},
-\tfrac{1}{\sqrt2}\bigr)$ (value $-\frac12$): max $\frac12$, min $-\frac12$ (attained: the circle is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact)).

**Exercise 15.7 ★★.**

Determine the extrema of $f(x, y) = x^2 + y^2 - xy + x - y$ on $\R^2$, then its maximum and minimum on the closed triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$ *(interior critical points, then the three edges, then the vertices)*.

**Solution of Exercise 15.7.**

$\nabla f = (2x - y + 1,\; 2y - x - 1) = 0$: solving, $x =
-\frac13$, $y = \frac13$. Hessian $\begin{pmatrix} 2 & -1\\ -1 &
2\end{pmatrix}$, positive definite: global minimum of the quadratic $f$, value $f\bigl(-\frac13, \frac13\bigr) = -\frac13$.

On the triangle $T$: the interior critical point $(-\frac13,
\frac13) \notin T$ (negative $x$). Edges: on $y = 0$, $x \in
\intcc{0}{1}$: $f = x^2 + x$, increasing: extremes $0$ and $2$. On $x = 0$: $f = y^2 - y$, minimum $-\frac14$ at $y = \frac12$, values $0$ and $0$ at the ends. On $x + y = 1$: substitute $y = 1 - x$, $f = x^2 + (1-x)^2 - x(1-x) + x - (1-x) = 3x^2 - x$; on $\intcc{0}{1}$: minimum $-\frac{1}{12}$ at $x = \frac16$, values $0$ (at $x=0$) and $2$ (at $x=1$). Vertices: $f(0,0) = 0$, $f(1,0)
= 2$, $f(0,1) = 0$. Global on $T$: minimum $-\frac14$ at $(0,
\frac12)$, maximum $2$ at $(1, 0)$.

**Exercise 15.8 ★★★.**

Let $f \colon \R^2 \to \R^2$, $f(x, y) = (x + y^2,\; y + x^2)$. Show that $f$ is a local diffeomorphism near $0$, compute $\dd(f^{-1})_{(0,0)}$, and find the largest $r$ such that $\dd f$ is invertible on the ball $\norm{(x,y)}_2 < r$ *(compute $\det J_f$)*.

**Solution of Exercise 15.8.**

$J_f = \begin{pmatrix} 1 & 2y\\ 2x & 1\end{pmatrix}$, $\det J_f =
1 - 4xy$. At $0$: $\det = 1 \neq 0$: local diffeomorphism ([Theorem 15.14](#thm-b2-diffcalc-inverse)), with

$$
\dd(f^{-1})_{(0,0)} = (J_f(0))^{-1} = I_2 .
$$

Invertibility on a ball: need $4\abs{xy} < 1$ throughout; on $\norm{(x,y)}_2 < r$, $\abs{xy} \leq \frac{x^2 + y^2}{2} <
\frac{r^2}{2}$, so $r = \frac{1}{\sqrt2}$ works; and it is largest: at $(x, y) = \bigl(\tfrac12, \tfrac12\bigr)$, of [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $\frac{1}{\sqrt2}$, $\det J_f = 0$.

**Exercise 15.9 ★★★.**

(Rolle fails, mean value survives) Give $f \colon \R \to \R^2$, $C^1$, with $f(0) = f(2\pi)$ but $f'(t) \neq 0$ for all $t$ (no vector-valued Rolle). Then verify on your example the mean value *inequality* of [Theorem 15.5](#thm-b2-diffcalc-mvi).

**Solution of Exercise 15.9.**

$f(t) = (\cos t, \sin t)$: $f(0) = f(2\pi) = (1, 0)$, yet $f'(t) =
(-\sin t, \cos t)$ has [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $1$, never zero: no point where the derivative vanishes — Rolle has no vector analogue. The mean value inequality holds comfortably: $\norm{f(2\pi) - f(0)} = 0 \leq
2\pi \cdot \sup\norm{f'} = 2\pi$.

**Exercise 15.10 ★.**

Compute the [differential](#def-b2-diffcalc-differential) and the gradient of $f(x) = \norm
x_2^2$ and of $g(x) = \langle Ax, x\rangle$ on $\R^n$ ($A$ a square matrix, not assumed [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint)), and the Hessian of each. For which $A$ is $g$ convex?

**Solution of Exercise 15.10.**

$f(x+h) - f(x) = 2\langle x, h\rangle + \norm h^2$: $\dd f_x =
2\langle x, \cdot\rangle$, $\nabla f(x) = 2x$, Hessian $2I$ (constant). For $g$:

$$
g(x + h) - g(x) = \langle Ax, h\rangle + \langle Ah, x\rangle
+ \langle Ah, h\rangle
= \bigl\langle (A + A^{\mathsf T})x,\ h\bigr\rangle +
O(\norm h^2),
$$

so $\nabla g(x) = (A + A^{\mathsf T})x$ and $H_g = A +
A^{\mathsf T}$, constant. By [Exercise 15.11](#exo-b2-diffcalc-11), $g$ is convex iff $A + A^{\mathsf T}$ is positive semidefinite — only the [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) part of $A$ matters, as indeed $g(x) =
\langle \frac{A + A^{\mathsf T}}2 x, x\rangle$.

**Exercise 15.11 ★★.**

Let $f \colon \R^n \to \R$ be $C^2$. Prove that $f$ is convex if and only if its Hessian $H_x$ is positive semidefinite at every $x$ *(reduce to one variable: $t \mapsto f(a + t(b-a))$; use Taylor–Lagrange in one direction, and for the converse evaluate $\varphi''$)*.

**Solution of Exercise 15.11.**

$f$ is convex iff its restriction to every segment is convex, i.e. iff every $\varphi(t) = f(a + tv)$ is convex. By the chain rule $\varphi''(t) = \langle H_{a+tv}\,v,\ v\rangle$.

If all Hessians are positive semidefinite: $\varphi'' \geq 0$, so each $\varphi$ is convex (Year 1 volume) and $f$ is convex. Conversely if $f$ is convex, each $\varphi$ is convex, so $\varphi''(0) \geq 0$: $\langle H_a v, v\rangle \geq 0$ for every $a$ and every direction $v$: all Hessians are positive semidefinite.

**Exercise 15.12 ★★★.**

(A global inverse theorem) Let $g \colon \R^n \to \R^n$ be $C^1$ with $\vertiii{\dd g_x} \leq k < 1$ for all $x$, and $f =
\mathrm{id} + g$.

1. Show $\norm{f(x) - f(y)} \geq (1 - k)\norm{x - y}$ : $f$ is injective, with [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) inverse on its image.
2. Show that for each $y \in \R^n$ the map $x \mapsto y -  g(x)$ is a contraction of the [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) space $\R^n$ , and conclude by the Banach fixed point theorem ( [Theorem 4.12](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-banach) ) that $f$ is surjective.
3. Conclude that $f$ is a bijection of $\R^n$ with $(1-k)^{-1}$ -Lipschitz inverse — a *global* counterpart of [Theorem 15.14](#thm-b2-diffcalc-inverse) (which, by contrast, is purely local).

**Solution of Exercise 15.12.**

1. By the mean value inequality ([Theorem 15.5](#thm-b2-diffcalc-mvi)) applied to $g$: $\norm{g(x) - g(y)} \leq k\norm{x-y}$, so $$\norm{f(x) - f(y)} \geq \norm{x - y} - \norm{g(x) -  g(y)} \geq (1 - k)\norm{x - y} :$$ $f$ is injective and $f^{-1}$ (defined on the image) is $\frac{1}{1-k}$-Lipschitz.
2. Fix $y$ ; $T(x) = y - g(x)$ satisfies $\norm{T(x) -  T(x')} = \norm{g(x') - g(x)} \leq k\norm{x - x'}$ : a contraction of the [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) space $\R^n$ . Banach ( [Theorem 4.12](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-banach) ) gives a fixed point $x^*  = y - g(x^*)$ , i.e. $f(x^*) = y$ : $f$ is surjective.
3. $f$ is thus a bijection of $\R^n$ with $\frac{1}{1-k}$ -Lipschitz inverse: a global inverse theorem, where the smallness of $\dd g$ everywhere replaces the local invertibility hypothesis of [Theorem 15.14](#thm-b2-diffcalc-inverse) .

## 15.5 Problem: The calculus of matrices — Jacobi, exponential, and the orthogonal group

**Problem 15.1.**

The cleanest playground for differential calculus is the space $\mathcal M_n(\R) \simeq \R^{n^2}$ itself: its most natural maps — product, inverse, [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det), exponential — have [differentials](#def-b2-diffcalc-differential) of striking elegance. This problem computes them all: the *Neumann series*, the [differential](#def-b2-diffcalc-differential) of the inverse, *Jacobi’s formula* for the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) with *Liouville’s formula* as dividend, the *matrix exponential* with $\det\eu^A = \eu^{\operatorname{tr}A}$, and finally the orthogonal group $O_n$ as a smooth level set with the antisymmetric matrices as tangent space — differential geometry in embryo. Throughout, $\vertiii\cdot$ is the [operator norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-continuouslinear) subordinate to $\norm\cdot_2$, and $\langle X, Y\rangle =
\operatorname{tr}(X^{\mathsf T}Y)$ the Frobenius inner product.

**Part I — The Neumann series.**

1. Prove submultiplicativity, $\vertiii{AB} \leq  \vertiii A\,\vertiii B$ , and deduce that polynomial maps of $A$ (matrix products, [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) , trace) are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\mathcal M_n(\R)$ .
2. For $\vertiii X < 1$, show that $\sum_{k\geq0}X^k$ converges [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) in $\mathcal M_n(\R)$ ([Theorem 5.21](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-absoluteconvergence)), that its sum is $(I - X)^{-1}$, and that $$\vertiii{(I - X)^{-1}} \leq \frac{1}{1 - \vertiii X},  \qquad  (I - X)^{-1} = I + X + O\bigl(\vertiii X^2\bigr) .$$
3. Deduce that $GL_n(\R)$ is *[open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology)* : if $A$ is invertible and $\vertiii H <  \frac{1}{\vertiii{A^{-1}}}$ , then $A + H$ is invertible. Deduce also that $GL_n(\R)$ is *dense* in $\mathcal M_n(\R)$ *(perturb $A$ by $\varepsilon I$: $\det(A + \varepsilon I)$ is a nonzero polynomial in $\varepsilon$)* .
4. Show that the inversion map $\Phi(A) = A^{-1}$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $GL_n(\R)$ .

**Part II — First [differentials](#def-b2-diffcalc-differential).**

5. Show that the squaring map $A \mapsto A^2$ is [differentiable](#def-b2-diffcalc-differential) with [differential](#def-b2-diffcalc-differential) $H \mapsto AH + HA$ , and more generally that $A \mapsto A^k$ has [differential](#def-b2-diffcalc-differential) $H \mapsto \sum_{i=0}^{k-1}  A^iHA^{k-1-i}$ . Why can one *not* write $kA^{k-1}H$ in general?
6. Prove that $\Phi(A) = A^{-1}$ is [differentiable](#def-b2-diffcalc-differential) on $GL_n(\R)$ with $$\dd\Phi_A(H) = -A^{-1}HA^{-1}$$ *(write $(A + H)^{-1} = (I + A^{-1}H)^{-1}A^{-1}$ and expand by question 2)*. Check the formula against the scalar case $n = 1$.
7. For a $C^1$ curve $t \mapsto A(t) \in GL_n(\R)$ , deduce $\bigl(A(t)^{-1}\bigr)' = -A^{-1}A'A^{-1}$ , and expand $t \mapsto (I + tB)^{-1}$ to first order at $t = 0$ .
8. Compute the [differential](#def-b2-diffcalc-differential) of $f(A) =  \operatorname{tr}(A^k)$ and identify its gradient for the Frobenius inner product: $$\dd f_A(H) = k\operatorname{tr}\bigl(A^{k-1}H\bigr),  \qquad  \nabla f(A) = k\,\bigl(A^{k-1}\bigr)^{\mathsf T} .$$
9. Same questions for $f(A) = \operatorname{tr}  (A^{\mathsf T}A) = \norm A_F^2$ : [differential](#def-b2-diffcalc-differential) , gradient, and the (constant) Hessian; conclude that $\norm\cdot_F^2$ is strictly convex.

**Part III — Jacobi’s formula.**

10. Prove $$\det(I + H) = 1 + \operatorname{tr}H +  O\bigl(\vertiii H^2\bigr)$$ *(expand $\det(e_1 + h_1, \dots, e_n + h_n)$ by multilinearity in the columns: terms with at least two $h$-columns are $O(\vertiii H^2)$)*: $\dd(\det)_I =  \operatorname{tr}$.
11. For $A$ invertible, deduce $$\dd(\det)_A(H) = \det(A)\,  \operatorname{tr}\bigl(A^{-1}H\bigr) .$$
12. Show that for *every* $A$ (invertible or not), $\frac{\partial\det}{\partial a_{ij}}(A) = C_{ij}$, the $(i,j)$ cofactor *(Laplace expansion along row $i$)*, so that with the adjugate $\operatorname{adj}A =  \operatorname{com}(A)^{\mathsf T}$: $$\dd(\det)_A(H) =  \operatorname{tr}\bigl(\operatorname{adj}(A)\,H\bigr),  \qquad  \nabla(\det)(A) = \operatorname{com}(A) ,$$ recovering question 11 when $A$ is invertible ($\operatorname{adj}A = \det(A)A^{-1}$). This is *Jacobi’s formula*: $\bigl(\det  A(t)\bigr)' = \operatorname{tr}\bigl(  \operatorname{adj}(A(t))\,A'(t)\bigr)$.
13. (Liouville’s formula) Let $A(t)$ be a $C^1$ curve of matrices satisfying the linear differential equation $A'(t) = M(t)A(t)$. Prove $$\bigl(\det A(t)\bigr)' =  \operatorname{tr}\bigl(M(t)\bigr)\,\det A(t),  \qquad\text{hence}\qquad  \det A(t) = \det A(0)\,  \exp\Bigl(\int_0^t\operatorname{tr}M\Bigr)$$ *(use $\operatorname{adj}(A)\,A = \det(A)I$ and the cyclic invariance of the trace)* — the Wronskian identity that the differential-equations chapter will use constantly.
14. Show that $SL_n(\R) = \{\det = 1\}$ is a smooth level set: at every $A \in SL_n(\R)$ the [differential](#def-b2-diffcalc-differential) $\dd(\det)_A$ is a *surjective* linear map onto $\R$ *(evaluate it at $H = \frac1nA$)* .

**Part IV — The [matrix exponential](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp).**

15. Show that $\eu^A = \sum_{k\geq0}\frac{A^k}{k!}$ converges [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) for every $A$ , [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on every ball, with $\vertiii{\eu^A} \leq  \eu^{\vertiii A}$ ; and that $\eu^A$ depends [continuously](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $A$ .
16. Prove that $AB = BA$ implies $\eu^{A+B} =  \eu^A\eu^B$ *([Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini), legitimate by absolute convergence)* ; deduce that $\eu^A$ is always invertible, with inverse $\eu^{-A}$ : $\exp$ maps $\mathcal M_n(\R)$ into $GL_n(\R)$ .
17. Show that $t \mapsto \eu^{tA}$ is $C^1$ (indeed $C^\infty$) with $$\frac{\dd}{\dd t}\,\eu^{tA} = A\,\eu^{tA} =  \eu^{tA}A$$ *(differentiate the series term by term on segments: the derived series converges [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series))*.
18. Prove the identity $$\det\bigl(\eu^{A}\bigr) = \eu^{\operatorname{tr}A}$$ *(apply Liouville’s formula, question 13, to $A(t) = \eu^{tA}$)*. Sanity checks: $n = 1$; nilpotent $A$; and the trace-zero matrices land in $SL_n(\R)$.
19. Show $\eu^H = I + H + O(\vertiii H^2)$ , so $\exp$ is [differentiable](#def-b2-diffcalc-differential) at $0$ with $\dd(\exp)_0 =  \mathrm{id}$ ; conclude with the inverse function theorem ( [Theorem 15.14](#thm-b2-diffcalc-inverse) ) that $\exp$ is a $C^1$ diffeomorphism from a neighborhood of $0$ onto a neighborhood of $I$ : every matrix close to the identity has a logarithm.
20. Show that $\exp$ maps [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) matrices to [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) *positive definite* matrices, bijectively *(diagonalize; the inverse is the spectral logarithm)* .

**Part V — The orthogonal group as a level set.**

21. Let $F(A) = A^{\mathsf T}A$ , from $\mathcal M_n(\R)$ to the [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) matrices $S_n$ . Compute $\dd F_A(H) =  A^{\mathsf T}H + H^{\mathsf T}A$ and show that at every $A \in O_n = F^{-1}(I)$ this [differential](#def-b2-diffcalc-differential) is *surjective* onto $S_n$ *(given $S \in S_n$, try $H = \frac12 AS$)* : $O_n$ is a smooth level set, of dimension $n^2 - \frac{n(n+1)}2 = \frac{n(n-1)}2$ .
22. Show that every $C^1$ curve $A(t) \in O_n$ with $A(0) =  I$ has an *antisymmetric* velocity $A'(0)$ , and conversely that for antisymmetric $K$ the curve $\eu^{tK}$ stays in $O_n$ : the tangent space of $O_n$ at $I$ is exactly the antisymmetric matrices.
23. Show $\det\eu^{K} = 1$ for antisymmetric $K$ (question 18): the exponential curve lives in the rotation group $SO_n$. Compute it fully for $n = 2$: with $J =  \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$, prove $$\eu^{\theta J} = \begin{pmatrix} \cos\theta &  -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix} :$$ the [matrix exponential](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp) *is* the rotation by $\theta$, and the series definitions of cosine and sine reappear inside a matrix.
24. (The density trick) Using the density of $GL_n(\R)$ (question 3) and [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), extend from invertible to all matrices the identity $$\operatorname{adj}(AB) =  \operatorname{adj}(B)\operatorname{adj}(A)$$ *(for invertible $A, B$ both sides equal $\det(AB)(AB)^{-1}$; both sides are polynomial in the entries)*.
25. Synthesis. In one sentence each: (i) which earlier chapters provided the engine of each Part ( [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) and normed algebras; the spectral theorem; the inverse function theorem); (ii) which formula of this problem the differential-equations chapter will lean on, and where; (iii) what $\dd(\det)_I = \operatorname{tr}$ and $\det\eu^A =  \eu^{\operatorname{tr}A}$ say about trace and [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) as “infinitesimal and global volume”; (iv) what the Year 3 volume makes of questions 21–23 (Lie groups and their Lie algebras).

**Solution of Problem 15.1.**

**1.** $\norm{ABx} \leq \vertiii A\norm{Bx} \leq
\vertiii A\vertiii B\norm x$: take the sup over $\norm x = 1$. Matrix products, $\det$ and $\operatorname{tr}$ are polynomial functions of the entries, hence [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ($\mathcal M_n(\R)
\simeq \R^{n^2}$, all [norms](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) equivalent: [Theorem 5.13](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-finitedim)).

**2.** $\sum\vertiii{X^k} \leq \sum\vertiii X^k <
\infty$: the series converges [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def), hence converges ([Theorem 5.21](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#thm-b2-nvs-absoluteconvergence)). From $(I -
X)\sum_{k\leq N}X^k = I - X^{N+1} \to I$: the sum is $(I -
X)^{-1}$. [Norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm): $\leq \sum\vertiii X^k = \frac{1}{1 -
\vertiii X}$; and

$$
(I - X)^{-1} - I - X = \sum_{k\geq2}X^k = X^2(I - X)^{-1},
\qquad
\vertiii{X^2(I-X)^{-1}} \leq
\frac{\vertiii X^2}{1 - \vertiii X} = O(\vertiii X^2).
$$

**3.** $A + H = A(I + A^{-1}H)$ with $\vertiii{A^{-1}H}
\leq \vertiii{A^{-1}}\vertiii H < 1$: invertible by question 2: the [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) ball of radius $\vertiii{A^{-1}}^{-1}$ around $A$ lies in $GL_n(\R)$. Density: $\det(A + \varepsilon I)$ is a degree-$n$ polynomial in $\varepsilon$ with leading coefficient $1$: it has finitely many roots, so there are $\varepsilon_k \to 0$ with $A + \varepsilon_kI$ invertible, converging to $A$.

**4.** For $\vertiii H < \frac{1}{2\vertiii{A^{-1}}}$:

$$
(A + H)^{-1} - A^{-1}
= \bigl[(I + A^{-1}H)^{-1} - I\bigr]A^{-1},
$$

of [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) at most $\frac{\vertiii{A^{-1}H}}{1 -
\vertiii{A^{-1}H}}\,\vertiii{A^{-1}} \leq
2\vertiii{A^{-1}}^2\vertiii H \to 0$: $\Phi$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at every $A \in GL_n(\R)$.

**5.** $(A+H)^2 = A^2 + AH + HA + H^2$: the map $H
\mapsto AH + HA$ is linear and the error $H^2$ is $O(\vertiii H^2)$. Expanding $(A + H)^k$ and sorting by the number of $H$-factors: the linear part is $\sum_{i=0}^{k-1}A^iHA^{k-1-i}$, and the terms with $\geq 2$ factors $H$ are bounded by $\binom k2$-many products of [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $\leq \vertiii A^{k-2}\vertiii H^2$-scale: $O(\vertiii H^2)$. One cannot collapse the sum to $kA^{k-1}H$ because $H$ and $A$ need not commute — the sum is the correct noncommutative derivative.

**6.** For small $H$:

$$
(A+H)^{-1} = (I + A^{-1}H)^{-1}A^{-1}
= \bigl(I - A^{-1}H + O(\vertiii H^2)\bigr)A^{-1}
= A^{-1} - A^{-1}HA^{-1} + O(\vertiii H^2) :
$$

$\dd\Phi_A(H) = -A^{-1}HA^{-1}$, linear in $H$. For $n = 1$: $\dd(1/a)(h) = -h/a^2$, the familiar derivative.

**7.** Chain rule along the curve: $\bigl(A(t)^{-1}\bigr)' = \dd\Phi_{A(t)}(A'(t)) =
-A(t)^{-1}A'(t)A(t)^{-1}$. At $A(t) = I + tB$, $t = 0$: $(I +
tB)^{-1} = I - tB + O(t^2)$.

**8.** By question 5 and cyclic invariance of the trace:

$$
\dd f_A(H) = \operatorname{tr}\Bigl(\sum_{i=0}^{k-1}
A^iHA^{k-1-i}\Bigr) = k\operatorname{tr}\bigl(A^{k-1}H\bigr) .
$$

Against the Frobenius product, $\dd f_A(H) =
\operatorname{tr}\bigl((\nabla f)^{\mathsf T}H\bigr)$ requires $(\nabla f)^{\mathsf T} = kA^{k-1}$: $\nabla f(A) =
k\,(A^{k-1})^{\mathsf T}$.

**9.** $f(A + H) - f(A) = 2\operatorname{tr}
(A^{\mathsf T}H) + \operatorname{tr}(H^{\mathsf T}H)$: the [differential](#def-b2-diffcalc-differential) is $H \mapsto 2\operatorname{tr}(A^{\mathsf T}H) =
2\langle A, H\rangle$, so $\nabla f(A) = 2A$; the second-order term is exactly $\norm H_F^2$: the Hessian is twice the identity [quadratic form](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def), positive definite and constant, so $\norm\cdot_F^2$ is strictly convex (the Taylor formula is exact here).

**10.** By multilinearity in the columns, $\det(I + H) =
\sum_{S\subseteq\{1,\dots,n\}}\det(M_S)$ where $M_S$ has column $h_j$ for $j \in S$ and $e_j$ otherwise. $S = \varnothing$ gives $1$; $S = \{j\}$ gives the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) of $I$ with column $j$ replaced by $h_j$, namely its $j$-th entry $h_{jj}$, summing to $\operatorname{tr}H$; each term with $\abs S \geq 2$ is a [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) with at least two columns of size $O(\vertiii H)$, hence $O(\vertiii H^2)$ (multilinear maps on a finite-dimensional space are bounded), and there are finitely many. So $\det(I + H) = 1 + \operatorname{tr}H +
O(\vertiii H^2)$: $\dd(\det)_I = \operatorname{tr}$.

**11.** $\det(A + H) = \det A\,\det(I + A^{-1}H) =
\det A\,\bigl(1 + \operatorname{tr}(A^{-1}H) +
O(\vertiii H^2)\bigr)$: the [differential](#def-b2-diffcalc-differential) is $H \mapsto
\det(A)\operatorname{tr}(A^{-1}H)$.

**12.** Laplace expansion along row $i$: $\det A =
\sum_j a_{ij}C_{ij}$, and the cofactors $C_{ij}$ do not involve row $i$: $\frac{\partial\det}{\partial a_{ij}} = C_{ij}$. Hence

$$
\dd(\det)_A(H) = \sum_{i,j}C_{ij}h_{ij}
= \operatorname{tr}\bigl(\operatorname{com}(A)^{\mathsf T}
H\bigr)
= \operatorname{tr}\bigl(\operatorname{adj}(A)H\bigr),
\qquad
\nabla(\det)(A) = \operatorname{com}(A) .
$$

For invertible $A$, $\operatorname{adj}A = \det(A)A^{-1}$ recovers question 11. Along a $C^1$ curve, the chain rule reads $(\det A(t))' = \operatorname{tr}(\operatorname{adj}
(A(t))\,A'(t))$: Jacobi’s formula.

**13.** With $A' = MA$ and $\operatorname{adj}(A)A = \det(A)I$:

$$
(\det A)' = \operatorname{tr}\bigl(\operatorname{adj}(A)MA
\bigr)
= \operatorname{tr}\bigl(A\operatorname{adj}(A)M\bigr)
= \det A\;\operatorname{tr}M
$$

(cyclicity; $A\operatorname{adj}A = \det(A) I$ as well). The scalar linear ODE $y' = \operatorname{tr}(M(t))\,y$ has the unique solution $y(t) = y(0)\exp\bigl(\int_0^t
\operatorname{tr}M\bigr)$ (Year 1): Liouville’s formula.

**14.** At $A \in SL_n(\R)$, take $H = \frac1nA$: $\dd(\det)_A\bigl(\tfrac1nA\bigr) = \frac1n\det(A)
\operatorname{tr}(A^{-1}A) = \frac1n\cdot1\cdot n = 1 \neq 0$: the [differential](#def-b2-diffcalc-differential) is a nonzero linear form, hence surjective onto $\R$ at every point of the level set: $SL_n(\R)$ is a smooth level set (of dimension $n^2 - 1$).

**15.** $\sum_k\vertiii{A^k/k!} \leq
\sum\vertiii A^k/k! = \eu^{\vertiii A}$: absolute convergence (question 2’s [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) argument), with [normal convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on every ball $\vertiii A \leq R$ (bound $R^k/k!$ independent of $A$). Each partial sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) (polynomial); the uniform limit on balls is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity): $A \mapsto \eu^A$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), with $\vertiii{\eu^A} \leq \eu^{\vertiii A}$.

**16.** Both series converge [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def), so the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) is legitimate ([Theorem 7.14](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini)):

$$
\eu^A\eu^B = \sum_{n\geq0}\frac{1}{n!}\sum_{k=0}^n\binom
nkA^kB^{n-k}
= \sum_{n\geq0}\frac{(A+B)^n}{n!} = \eu^{A+B},
$$

the binomial identity requiring $AB = BA$. With $B = -A$: $\eu^A\eu^{-A} = \eu^0 = I$: every $\eu^A \in GL_n(\R)$.

**17.** The series $\sum t^kA^k/k!$ and its derived series $\sum t^{k-1}A^k/(k-1)! = A\sum t^{k-1}A^{k-1}/(k-1)!$ converge [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on every segment $\abs t \leq T$ (bounds $T^k\vertiii A^k/k!$): the differentiation theorem for series ([Theorem 10.11](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#thm-b2-funcseq-seriestransfer), applied entry by entry) gives $\frac{\dd}{\dd t}\eu^{tA} = A\eu^{tA}$; factoring $A$ on the right instead gives $\eu^{tA}A$. Iterating: $C^\infty$.

**18.** $A(t) = \eu^{tA}$ satisfies $A'(t) = A\,A(t)$: Liouville’s formula (question 13) with constant $M = A$ gives $\det\eu^{tA} = \eu^{t\operatorname{tr}A}$ (value $1$ at $t =
0$); at $t = 1$, $\det\eu^A = \eu^{\operatorname{tr}A}$. Checks: $n = 1$ is the exponential itself; a nilpotent $A$ has $\operatorname{tr}A = 0$ and $\eu^A$ unipotent of [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) $1$; $\operatorname{tr}A = 0$ gives $\det\eu^A = 1$: the trace-zero matrices are sent into $SL_n(\R)$.

**19.** $\eu^H - I - H = \sum_{k\geq2}H^k/k!$, of [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $\leq \vertiii H^2\eu^{\vertiii H} = O(\vertiii H^2)$: $\dd(\exp)_0 = \mathrm{id}$, invertible. Moreover $\exp$ is $C^1$: by question 5, the [differential](#def-b2-diffcalc-differential) candidate $H \mapsto
\sum_k\frac1{k!}\sum_iA^iHA^{k-1-i}$ is a [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) convergent series of linear maps depending [continuously](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $A$ (bounds $\vertiii A^{k-1}/(k-1)!$ on balls), so the partials exist and are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Theorem 15.2](#thm-b2-diffcalc-c1) and the series transfer theorem). The inverse function theorem ([Theorem 15.14](#thm-b2-diffcalc-inverse)) applies at $0$: $\exp$ is a $C^1$ diffeomorphism of a neighborhood of $0$ onto a neighborhood of $I$ — matrices near $I$ have logarithms.

**20.** For [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) $S = PDP^{\mathsf T}$ (spectral theorem): $\eu^S = P\eu^DP^{\mathsf T}$ is [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) with [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\eu^{\lambda_i} > 0$: positive definite. Surjectivity: a positive definite $Q =
P\operatorname{diag}(\mu_i)P^{\mathsf T}$ ($\mu_i > 0$) is $\eu^S$ for $S = P\operatorname{diag}(\ln\mu_i)P^{\mathsf T}$. Injectivity: $\eu^S$ determines its [eigenspaces](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen), which are exactly those of $S$ (on each [eigenspace](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $S$ for $\lambda$, $\eu^S$ acts as $\eu^\lambda$; distinct $\lambda$ give distinct $\eu^\lambda$), and taking $\ln$ of the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) recovers $S$. So $\exp$ is a bijection from [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) matrices onto positive definite ones.

**21.** $F(A + H) = A^{\mathsf T}A + A^{\mathsf T}H +
H^{\mathsf T}A + H^{\mathsf T}H$: $\dd F_A(H) =
A^{\mathsf T}H + H^{\mathsf T}A$ (values in $S_n$; error $O(\vertiii H^2)$). At $A \in O_n$ and for $S \in S_n$, the choice $H = \frac12AS$ gives

$$
A^{\mathsf T}\cdot\tfrac12AS + \tfrac12(AS)^{\mathsf T}A
= \tfrac12 S + \tfrac12 S^{\mathsf T} = S :
$$

surjective. $O_n = F^{-1}(I)$ is a smooth level set of dimension $n^2 - \dim S_n = \frac{n(n-1)}2$.

**22.** Differentiating $A(t)^{\mathsf T}A(t) = I$ at $t
= 0$ (with $A(0) = I$): $A'(0)^{\mathsf T} + A'(0) = 0$: antisymmetric. Conversely, for $K^{\mathsf T} = -K$: $(\eu^{tK})^{\mathsf T}\eu^{tK} = \eu^{tK^{\mathsf T}}
\eu^{tK} = \eu^{-tK}\eu^{tK} = I$ ([transpose](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-transpose) the series termwise; the exponents commute): the curve stays in $O_n$, with velocity $K$ at $t = 0$. Tangent space at $I$ $=$ the antisymmetric matrices, of the expected dimension $\frac{n(n-1)}2$.

**23.** $\operatorname{tr}K = 0$ for antisymmetric $K$, so $\det\eu^K = \eu^0 = 1$ (question 18): the exponential lands in $SO_n$. For $n = 2$: $J^2 = -I$, so $J^{2m} =
(-1)^mI$, $J^{2m+1} = (-1)^mJ$, and

$$
\eu^{\theta J}
= \Bigl(\sum_m\frac{(-1)^m\theta^{2m}}{(2m)!}\Bigr)I
+ \Bigl(\sum_m\frac{(-1)^m\theta^{2m+1}}{(2m+1)!}\Bigr)J
= \cos\theta\,I + \sin\theta\,J ,
$$

the rotation by $\theta$: the sine and cosine series live inside the [matrix exponential](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp).

**24.** For invertible $A, B$: $\operatorname{adj}(AB) =
\det(AB)(AB)^{-1} = \det(B)\det(A)B^{-1}A^{-1} =
\operatorname{adj}(B)\operatorname{adj}(A)$. Both sides of the identity are polynomial (hence [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)) maps of the entries of $(A, B)$; they agree on the dense subset $GL_n\times GL_n$ of $\mathcal M_n\times\mathcal M_n$ (question 3: approximate each factor), so they agree everywhere.

**25.** (i) Part I ran on [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) of finite-dimensional normed spaces ([absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent series converge), Part IV on the same plus the spectral theorem for question 20, Part V’s local logarithm on the inverse function theorem. (ii) The differential-equations chapter leans on Liouville’s formula (question 13) for the Wronskian of linear systems, and on $\frac{\dd}{\dd t}\eu^{tA} = A\eu^{tA}$ (question 17), which is the statement that $\eu^{tA}$ solves $X' = AX$. (iii) $\dd(\det)_I = \operatorname{tr}$ says the trace is the infinitesimal rate of volume change, and $\det\eu^A =
\eu^{\operatorname{tr}A}$ integrates that statement globally. (iv) The Year 3 volume names the structures: $O_n$ and $SL_n(\R)$ are Lie groups, their tangent spaces at $I$ (antisymmetric and trace-zero matrices) are Lie algebras, and $\exp$ is the bridge between them.