I need to find the singular value decomposition of $$\pmatrix{a&-b&0\\b&a&-b\\0&b&a}$$
I already determined sigma to be: $$\Sigma = \pmatrix{\sqrt{a^2+b^2}&0&0\\0&a&0\\0&0&\sqrt{a^2-b^2}}$$
Now I was wondering whether there some something else that I was missing to this question other than just continuing as usual, to find $V$ and $U$?
EDIT
So trying the old method:
The the matrix be $A$. Then $A^TA$ is $$\pmatrix{a^2+b^2&0&-b^2\\0&a^2&0\\-b^2&0&a^2+b^2}$$
The first two eigenvectors are
$$v_1=\pmatrix{\frac{4}{\sqrt{17}}\\0\\\frac{1}{\sqrt{17}}}\\ v_2=\pmatrix{\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}} $$ However the third eigenvector seems to be $(0,0,0)^T$ which cannot be right? Right?
We write $$A^TA = \begin{pmatrix}m&0&-n\\0&m+n&0\\-n&0&m\end{pmatrix}$$with $m=a^2+b^2$ and $n=b^2$.
The obvious eigenvectors are $(1,0,1)^T$ for the eigenvalue $m-n=a^2$; $(1,0,-1)^T$ and $(0,1,0)^T$ for the eigenvalue $m+n=2b^2+a^2$.