How can we get an invertible matrix $W \in$ $\mathbb{R}$$^{3\times3}$ to make $W\hspace{-1mm}AW^{T}$ a diagonal matrix? Here,
$$ A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 3 & 1 & 3 \end{bmatrix} $$
I try to compute the eigenvalues of $A$, but it is difficult to factor the polynomial $t^3 - 5t^2 - 5t - 1 = 0$. I've also attempted to compute its SVD, but the polynomial becomes more complicated. Do we have other ways to solve it?
You only need to find the eigenvectors if you insist that $W$ must be orthogonal. Basically here we are completing the square. First we find a matrix $$M=\pmatrix{1&0&0\\*&1&0\\*&0&1}$$ such that $$MAM^T=B=\pmatrix{2&0&0\\0&*&*\\0&*&*}.$$ We can take $$M=\pmatrix{1&0&0\\-1/2&1&0\\-3/2&0&1}$$ and then $$B=\pmatrix{2&0&0\\0&-1/2&-1/2\\0&-1/2&-3/2}.$$ Now, we seek a matrix $$N=\pmatrix{1&0&0\\0&1&0\\0&*&1}$$ to make $NBN^T$ diagonal. Then $W=NM$ suffices.
