Singular Value Decomposition and Square matrices

Question

Hi,

I understand how to get part a. As:

$$v^{t}v = I_{\alpha}$$

$$ A= Uv^{t}v \Sigma v^{t} $$

Then it is easy to show that $$ v \Sigma v^{t} $$ is symmetric and $$ Uv^{t} $$ is orthogonal.

However for the second part, I am unsure of the answer. What I thought to so is think about the polar decomposition of A transpose, which is :

$$ A^{t} = Q*S $$ $$ (A^{t} )^{t} = (Q*S)^{t} $$ $$ (A^{t} )^{t} = S^{t}Q^{t} $$ $$ A = S_{1}Q^{t} $$ $$ Q{t} = Q_{1} $$ As Q transpose remains orthogonal $$ A = S_{1}Q_{1} $$

But I am not sure if this is correct? I have saw somethings online that make me think it is not. Can someone clarify if I am incorrect, and if I am why?

Thanks!

Answer 1

You can use the same trick as in the hint, just on the other side of the $\Sigma$.

$A = U\Sigma V^T = U\Sigma U^TUV^T = (U\Sigma U^T)\cdot (UV^T)$