Say I have a frequency response of a $2 \times 2$ system at a fixed frequency: $G(i \omega$). It can be anything, for example $\begin{bmatrix} 1 & 3 \\ 4 & 2 \end{bmatrix}$. If I compute the SVD, I get the following: $U = \begin{bmatrix} -0.53 & -0.85 \\ -0.85 & 0.53 \end{bmatrix}$, $V = \begin{bmatrix} -0.77 & 0.64 \\ 0.64 & -0.77 \end{bmatrix}$ and $\Sigma = \begin{bmatrix} 5.12 & 0 \\ 0 & 1.95 \end{bmatrix}$. Usually $\bar{u}$ and $\bar{v}$ are defined as the first columns of $U$ and $V$, respectively. These two are related to "strongest" singular value, which is obvious from this example. However, I don't know how to interpret this computation: \begin{equation} \bar{v} \cdot \bar{u}^T\end{equation} \begin{equation} \begin{bmatrix} -0.77 \\ -0.64 \end{bmatrix} \cdot \begin{bmatrix} -0.53 \\ -0.85 \end{bmatrix}^T = \begin{bmatrix} 0.40 & 0.65 \\ 0.34 & 0.55 \end{bmatrix}\end{equation} If I do this, what am I exactly computing and what is the interpretation? The norm of the outcome is always equal to 1, I guess because U and V are unitary.


1 Answer 1


If $A=U\Sigma V^*$ is the SVD, $\Sigma$ is the diagonal matrix of singular values and $u_k, v_k$ are the columns of $U,V$ respectively then we have $A = \sum_k \sigma_k u_k v^*_k$.

Then if $\sigma_1 $ is much bigger than the other $\sigma_k$s then $\sigma_1 u_1 v^*_1$ is the 'dominant' part of $A$.

Note that you probably need the Hermitian transpose not just the transpose if you are dealing with frequency response. (And what is this $i$ in $G(i \omega)$?)

  • $\begingroup$ Regarding the parenthetical, $G(s)$ is usually used to refer to the transfer function (Laplace transoform) so that plugging in $s = i \omega$ (where $i^2 = -1$) gives us the Fourier transform, i.e. frequency response. $\endgroup$ Commented Jun 21, 2020 at 9:32
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    $\begingroup$ @Omnomnomnom: Sorry, that was tongue-in-cheek; I am an electrical engineer, usually it is written $G(j \omega)$, and is typically complex, hence my eyebrow raise :-). $\endgroup$
    – copper.hat
    Commented Jun 21, 2020 at 15:01
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    $\begingroup$ Ah, I guess I was a bit slow on the uptake there :) $\endgroup$ Commented Jun 21, 2020 at 15:24

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