4

We say that a matrix $A$ in $\mathbb{R}^{n\times n}$ is symmetric if $A^T=A$, and that $U\in \mathbb{R}^{n\times n}$ is orthogonal if $U^TU=UU^T=I$. Show the following.

(a) Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and let $g$ be an inner product on $V$. Let $E$ and $B$ be two orthonormal bases on $V$. Show that the change of basis matrix $[I]^E_B$ is orthogonal.

(b) Let $A\in\mathbb{R}^{n\times n}$ be symmetric. Show that there is an orthogonal matrix $U\in\mathbb{R}^{n\times n}$ such taht $U^TAU$ is diagonal.

I've done part (a), but I cannot do part (b).

I had a plan to somehow use the spectral theorem, but I cannot do it.

Any help!

Community
  • 1
james1395
  • 643

1 Answers1

4

If you are allowed to use the spectral theorem, the proof of (v) is easy:

The spectral theorem says that for a real symmetric matrix $A$ there is an orthonormal set of eigenvectors $\{ \bf{e}_i\}$ such that $$ \begin{array}{c} A{\bf e}_i = \lambda_i {\bf e}_i \\ {\bf e}_i^T A = \lambda_i {\bf e}_i^T \\ \left< {\bf e}_i^T , {\bf e}_j \right> = \delta_{ij} \end{array} $$ Now consider $U$ such that $$ U_{ij} = [{\bf e}_i]_j $$ Then $AU$ is a square matrix with column $i$ equal to ${\bf e}_i$ for each $i$. From which: $$ [U^T(AU)]_{ki} = \lambda_i \left< {\bf e}_k^T , {\bf e}_i \right> = \lambda_i \delta_{ki} $$ so $U^TAU = U^T(AU)$ is a diagonal matrix with elements $\lambda_i$.

Mark Fischler
  • 41,743