I have the following matrix:
$ C'^* = \left( \begin{array}{ccc} -0.0045 & -0.0059 & 0 \\ -0.0059 & -1.0000 & 0 \\ 0 & 0 & 0 \end{array} \right)$
I put it through the SVD algorithm and got the following matrices:
$U = \left( \begin{array}{ccc} 0.0059 & 1.0000 & 0 \\ 1.0000 & -0.0059 & 0 \\ 0 & 0 & 1.0000 \end{array} \right), V = \left( \begin{array}{ccc} -0.0059 & -1.0000 & 0 \\ -1.0000 & 0.0059 & 0 \\ 0 & 0 & 1.0000 \end{array} \right)$ and finally:
$\Sigma = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0.0045 & 0 \\ 0 & 0 & 0 \end{array} \right)$
The problem is that I would like $\Sigma$ to be in a canonical form, i.e:
$\Sigma = \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \end{array} \right)$
I suppose I would need to change the matrices U and V in order to change $\Sigma$ to the canonical form since $C'^* = USV^{T}$. Does anyone have any clue how I could do it?
Thank you
I'm not sure what you mean here by canonical form? But it's trivial to do what you're asking...
Define matrix $M$: $$ M = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{.0045}} & 0 \\ 0 & 0 & 1 \end{array} \right] $$ That is, the diagonal entries are $\Sigma_{i,i}^{-\frac{1}{2}}$ for all non-zero singular values $\Sigma_{i,i}$ and $1$ otherwise. Trivially, we have:
$$M\Sigma M = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] = \hat{\Sigma} $$
Also observe that $M$ is invertible hence we can write: $$ U \Sigma V = \underbrace{\left(UM^{-1}\right)}_{\hat{U}} \underbrace{M\Sigma M}_{\hat{\Sigma}} \underbrace{\left( M^{-1}V\right)}_{\hat{V}} $$
Observe that $\hat{U}$ and $\hat{V}$ still form an orthogonal basis, but the basis vectors are no longer of norm 1.