# clarification in the proof of spectral theorem

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 $$A$$.

• 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. – P. Lawrence Mar 14 '20 at 2:41

You have $$S=WEW^T$$, with $$W$$ orthogonal and $$E$$ diagonal. Then $$A=MBM^T=M\begin{bmatrix} \lambda &0\\ 0& WEW^T\end{bmatrix} M^T=M\begin{bmatrix} 1&0\\0&W\end{bmatrix} \begin{bmatrix} \lambda&0\\0&E\end{bmatrix} \begin{bmatrix} 1&0\\0&W\end{bmatrix} ^TM^T$$