SVD: why are the results different: hand-calculation and Matlab computation?

Question

I am solving a SVD computatiob question.

The following matrix is given: A=: $$\begin{bmatrix} 1.5 & 0.5 & 0&0 \\ 0.5 & 1.5 & 0 & 0\\ 0 & 0 & 1.5 & 0.5\\ 0&0 & 0.5 & 1.5 \end{bmatrix}$$ I have taken the SVD(A) = [ U S V]. My computation of the matrix S is the same with Matlab, V and U are different. I have checked a solution in the book, my result is correct.

$$V=\begin{bmatrix} 0.5 & 0.5 & 0.5&0.5 \\ 0.5 & 0.5 & -0.5 & -0.5\\ 0.5 & -0.5 & -0.5& 0.5\\ 0.5&-0.5 & 0.5 & -0.5 \end{bmatrix}$$

Is there another way, how to find V matrix from SVD in Matlab?

I don't have access to MATLAB right now, but the U and V in the SVD decomposition are not unique, so it is possible that you and Matlab are both correct. — slimmerikko, Oct 21 '19 at 07:39
Yes you can for example use the SVD of $I_2$ and $\begin{bmatrix}1.5&0.5\0.5&1.5\end{bmatrix}$ combined with the properties of Kronecker products. — mathreadler, Oct 21 '19 at 08:02
@NoelMiller https://en.wikipedia.org/wiki/Kronecker_product#Abstract%20properties — mathreadler, Oct 21 '19 at 08:43
@mathreadler Do you mean: $ kron(V_{svd(I_2)},V_{A_{part}}]$? — Noel Miller, Oct 21 '19 at 14:09
Ehm, yeah, but also singular values themselves are products like this. — mathreadler, Oct 21 '19 at 14:13

score 1 · Answer 1 · answered Oct 21 '19 at 07:51

When looking at the singular values of $A$, we see $\sigma(A) = (2,2,1,1)$ so there are two sets of double singular values.

Given two singular vectors $u_1$ and $u_2$ with $A u_1 = \sigma_1 v_1$ and $A u_2= \sigma_1 v_2$, then $\tilde{u} = c_1u_1 + c_2 u_2$ is a singular vector to that singular vlaue $\sigma_1$ as well. This can be seen by $$A\tilde{u} = c_1Au_1 + c_2 Au_2 = \sigma_1 \cdot (c_1 v_1 + c_2 v_2) = \sigma_1 \tilde{v}_1$$

SVD: why are the results different: hand-calculation and Matlab computation?

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