What is change of basis?

Example 21.9 (Reading the change-of-basis matrix)

In $\R^2$, from the canonical $\mathcal B$ to $\mathcal B' = \bigl((1,1), (1,-1)\bigr)$:

(new vectors written in old coordinates, column by column). The vector of old coordinates $X = (3, 1)^{\mathsf T}$ has new coordinates $X' = P^{-1}X = \frac12(3 + 1,\ 3 - 1)^{\mathsf T} = (2, 1)^{\mathsf T}$: indeed $2(1,1) + 1(1,-1) = (3,1)$. Mind the direction — the matrix $P$ is built from the new basis but converts new to old coordinates ($X = PX'$); passing from old to new costs the inverse. Writing the sanity check $2(1,1) + (1,-1) = (3,1)$ after every conversion catches the inverted-$P$ error, which is the most common mistake of the chapter.

Example 21.11 (A good basis makes a map transparent)

Let $u(x, y) = (y, x)$ (swap), with matrix $A = \begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$ in the canonical basis. In the basis $\mathcal B' = \bigl((1,1), (1,-1)\bigr)$:

No matrix product was really needed: $u$ fixes $(1,1)$ and reverses $(1,-1)$, so in $\mathcal B'$ its matrix must be $\operatorname{diag}(1, -1)$ — the swap is the reflection across the line $y = x$. Finding, for a given endomorphism, a basis in which its matrix becomes diagonal is the central problem of the Year 2 volume (reduction theory); the weekend problem below shows how far polynomial identities alone already go.

Example 21.12 (Change of basis, run in reverse)

The projection onto $F = \operatorname{Vect}(1,1)$ along $G = \operatorname{Vect}(1,-1)$ has, in the adapted basis $\mathcal B' = \bigl((1,1),(1,-1)\bigr)$, the transparent matrix $A' = \operatorname{diag}(1, 0)$. To get its canonical-basis matrix, run Theorem 21.10 backwards, $A = P A' P^{-1}$:

Check: $A^2 = A$ (idempotent), $\operatorname{tr} A = 1 = \operatorname{rk} A$, and $A\binom{1}{1} = \binom11$, $A\binom{1}{-1} = 0$, as prescribed. This reverse direction — design the matrix in the good basis, then conjugate back — is how rotation, reflection and projection matrices are actually produced in practice.