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So the spectral theorem says that : A symmetric matrix $A \in \mathcal{R}^{n\times n} :\mathcal{V} \to \mathcal{V}$ can be written as $M= VDV^T$.

Here is my attempt to prove this:

Assume that there is a eigenvalue $\lambda$ for A with unit eigenvector $v \in \mathcal{V}$. the subspace $U\subset \mathcal{V}$ orthogonal to $v$ is invariante under $A$. So let $\{u_1,..,u_{n-1}\}$ be an orthonormal basis for $U$, and consider matrix $M=[v,u_1,...,u_{n-1}]$ where $v$ and $u_i's$ are column vectors in $M$. Then AM= MB.

Where $B=\left[ \begin{array}{c|c} \lambda & 0...0 \\ \hline 0's & S \end{array} \right]$

$S$ is symmetric because $A=MBM^T$ and $A$ is symmetric. We apply induction on $B$ and diagonalize $B$. Then I'm not sure how to connect the diagonalized form of B to form a diagonalized form for $A$.

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    $\begingroup$ The statement confuses a matrix with a lnear transformation. Is A a matrix or a linear transformation? How are A and M related? What is V? What is D? Please state EXACTLY what your version of the spectral theorem is. $\endgroup$ – P. Lawrence Mar 14 '20 at 2:41
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You have $S=WEW^T$, with $W$ orthogonal and $E$ diagonal. Then $$ A=MBM^T=M\begin{bmatrix} \lambda &0\\ 0& WEW^T\end{bmatrix} M^T=M\begin{bmatrix} 1&0\\0&W\end{bmatrix} \begin{bmatrix} \lambda&0\\0&E\end{bmatrix} \begin{bmatrix} 1&0\\0&W\end{bmatrix} ^TM^T $$

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