I'm reading a textbook where the Spectral Theorem for symmetric matrices is proven. I understand almost everything about the proof except for one thing. The theorem is stated as follows:
Theorem: Let $A \in \mathbb{R}^{n \times n}$. Then $A$ is orthogonally diagonalizable if and only if $A$ is symmetric.
The first implication is easy. The converse is proven by induction by the author. Here is part of the proof:
We want to prove that for any symmetric matrix $A$, there is an orthogonal matrix $P$ and a diagonal matrix $D$ such that $P^T AP = D$. We prove this by induction. Any $1 \times 1$ symmetric matrix is already diagonal, so we can take $P = I$ and the basic step is proven.
Now assume the theorem holds for $(n -1) \times (n-1)$ symmetric matrices, with $n \geq 2$. Then we now prove it also holds for $n$. So let $A$ be an $ n \times n$ symmetric matrix. We know that $A$ has only real eigenvalues (he concludes this on the basis of a preceding theorem). Let $\lambda_1$ be any eigenvalue of $A$, and let $v_1$ be the corresponding eigenvector which satisfies $||v_1 || = 1 $. Then we can extend the set $\left\{v_1 \right\}$ to a basis $\left\{ v_1, x_1, x_2, \ldots, x_n \right\}$ of $\mathbb{R}^n$. We can then use the Gram-Schmidt process to transform into an orthonormal basis $B = \left\{v_1, v_2, \ldots, v_n \right\}$ of $\mathbb{R}^n$.
Let $P$ be the matrix whose columns are the vectors in $B$, with the first column being $v_1$. Then $P$ is orthogonal because its column vectors are all orthonormal. Now $P^T A P = P^{-1} AP$ represents the linear transformation $T: x \mapsto Ax $ in the basis $B$. But we know that the first column of $P^T AP$ will be the coordinate vector of $T(v_1)$ with respect to the basis $B$. Now, $T(v_1) = Av_1 = \lambda_1 v_1$, so this coordinate vector is \begin{align*} \begin{pmatrix} \lambda_1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}. \end{align*} It follows that...
He then shows $P^T A P$ is diagonal by making use of induction hypothesis on a smaller block matrix.
But here is what I don't understand. He says $P^T A P$ represents the linear transformation $T: x \mapsto Ax$. What does he mean here? Does he mean the linear transformation $L_A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ ? This doesn't seem right to me, since the matrixrepresentation of $L_A$ is just $A$. Also, what he says after that doesn't really make sense to me, i.e. that the first column $P^T A P$ is the coordinate vector $T(v_1)$ with respect to $B$. Maybe someone can clarify this, or provide an example?