What is Hermitian endomorphism?

Example 13.7 (A skew-Hermitian matrix, diagonalized)

$A = \begin{pmatrix} 0 & -2\\ 2 & 0\end{pmatrix}$ satisfies $A^\dagger = A^{\mathsf T} = -A$: skew-Hermitian (also real antisymmetric — over $\R$ it has no eigenvalues at all). Characteristic polynomial $X^2 + 4$: eigenvalues $\pm2\iu$, purely imaginary, as Exercise 13.9 predicts in general. Eigenvectors: $(A - 2\iu I)v = 0$ gives $v_1 = \frac{1}{\sqrt2}(1, \iu)$, and $v_2 = \frac{1}{\sqrt2}(1, -\iu)$ for $-2\iu$; they are orthogonal:

So $A = U\operatorname{diag}(2\iu, -2\iu)\,U^\dagger$ with $U = (v_1\ v_2)$ unitary. Closing insight: $H = -\iu A = \begin{pmatrix} 0 & 2\iu\\ -2\iu & 0\end{pmatrix}$ is Hermitian with the real spectrum $\{\pm2\}$ and the same eigenvectors — the bijection $u \mapsto \iu u$ between Hermitian and skew-Hermitian endomorphisms (Exercise 13.9), seen matrix by matrix; over $\R$ the same $A$ is a rotation-scaling with no eigenvectors at all, and only the passage to $\C$ reveals its normal form.

Example 13.9

$A = \begin{pmatrix} 0 & -\iu\\ \iu & 0\end{pmatrix}$ is Hermitian ($A^\dagger = A$): eigenvalues from $\chi_A = X^2 - 1$: $\pm 1$ (real, as promised), with orthonormal eigenvectors $\frac{1}{\sqrt2}(1, \iu)^{\mathsf T}$ and $\frac{1}{\sqrt2}(1, -\iu)^{\mathsf T}$. (Physicists know $A$ as a Pauli matrix; the reality of Hermitian spectra is why quantum observables are modeled by Hermitian operators.)

Example 13.10 (A positive definite Hermitian matrix, worked)

$A = \begin{pmatrix} 2 & 1-\iu\\ 1+\iu & 3\end{pmatrix}$: Hermitian, since the diagonal is real and the off-diagonal entries are conjugates. Characteristic polynomial:

spectrum $\{1, 4\}$, real and positive — $A$ is positive definite. Eigenvectors: for $\lambda = 4$, the system $(A - 4I)v = 0$ gives $v_4 = (1 - \iu,\ 2)$ (check the second row: $(1+\iu)(1-\iu) - 2 = 0$); for $\lambda = 1$, $v_1 = (1 - \iu,\ -1)$. Orthogonality, with the conjugate in the first slot:

Normalizing ($\norm{v_4}^2 = 2 + 4 = 6$, $\norm{v_1}^2 = 2 + 1 = 3$) gives the unitary $U = \bigl(\frac{v_4}{\sqrt6}\ \frac{v_1}{\sqrt3}\bigr)$ with $A = U\operatorname{diag}(4,1)U^\dagger$. Closing insight: the Rayleigh reading is immediate — on the unit sphere of $\C^2$, $\langle x, Ax\rangle$ ranges over $\intcc{1}{4}$, attained at the two eigenvectors; this is the $n = 2$ germ of the Courant–Fischer theory built in the weekend problem. Spot-check that the form is real off the eigenvectors too: at $x = (1, \iu)$,

as Proposition 13.6’s proof mechanism (conjugate-symmetry against $A^\dagger = A$) guarantees for every $x$.