Doubt on singular value decomposition

Question

I am studying control systems and I am studying the topic of Singular Value Decomposition(SVD).

I have found a transfer matrix of a systemm, and computed its DC gain. Then I have performed the syngular value decomposition, and found that it is:

\$V=\begin{bmatrix} 0.1316 & 0.9913 & 0\\ 0& 0& 1\\ 0.9913& -0.1316 & 0 \end{bmatrix}\$

\$S=\begin{bmatrix} 8.7936 & 0& 0 & 0\\ 0& 0.8200& 0& 0\\ 0& 0 & 0 & 0 \end{bmatrix}\$

\$U=\begin{bmatrix} 0.5786 & 0.4065& 0& 0.707\\ -0.1127& 0.1605 & 0.9806& 0\\ -0.5636 & 0.8024& -0.1961 & -0.0000\\ -0.5786& -0.4065& -0.0000& 0.7071 \end{bmatrix}\$

Now, I am not sure I have understand what does this implies for a control system.

So far, I have understood that I the matrix \$V\$ is the matrix of the input directions, so if I have an input, it can have in this case three directions, which are the three columns of the matrix $V$. Then, the input in the chosen direction gets amplified byt the singular value which corresponds to that direction.

In my case, I don't understand few things. First of all, I see that the last two columns of the singular value matrix are zeros, what does this mean? And also I can see that inputs directions are three, while the output directions are \$4\$, and I have only two singular values, since one is zero, so I have at most three singular values.

What does it means? Can somebody help me clarify this situation?

Answer 1

I think this is way more fit to MathSE, but...

First of all, I see that the last two columns of the singular value matrix are zeros, what does this mean?

The singular values matrix can always be seen as the actual singular values \$\Sigma\$ and some "padding" zeros \$\mathbf{0}\$, so that you could match the size of \$U\$ and \$V\$. You could have either:

\$ S = \begin{bmatrix} \Sigma & \mathbf{0}\end{bmatrix}\$, \$ S = \begin{bmatrix} \Sigma \\ \mathbf{0}\end{bmatrix}\$, \$ S = \begin{bmatrix} \Sigma\end{bmatrix}\$.

in your case, the last columns of \$S\$ are the "padding" zeros. while the first 3 columns make up the diagonal matrix with the singular values \$\Sigma\$, it just happens that you have a zero singular value.

And also I can see that inputs directions are three, while the output directions are 4, and I have only two singular values, since one is zero, so I have at most three singular values.

Not sure what you mean by this, but that's what happens when you do the SVD of a matrix \$A \in \mathbb{R}^{4\times 3}\$, you get a \$U \in \mathbb{R}^{4\times 4}\$ and \$V \in \mathbb{R}^{3\times 3}\$. And since those matrices should be orthogonal (\$V^\top V = I\$) some of the rows/columns are associated with the left null space or the right null space.

For \$A = USV^\top\$, some of the columns of \$U\$ would be associated with the left null space, and for \$V\$ some of its columns (rows of \$V^\top\$) would be associated with the (right) null space.