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It is possible to show that SVD of block diagonal matrix is equivalent to independent SVDs of each block. I am wondering if there is something interesting to say on the case where the matrix is composed of 2 blocks of size $N$, with $k$ rows and $k$ columns of size $2N+k$ (overlapping in a $k\times k$ corner) added on. I am particularly interested in the influence of the “shared” rows on the reconstruction of the entire matrix when using truncated SVD.

An example:

\begin{pmatrix} \begin{matrix} X_1 \end{matrix} & 0 & \vdots \\ \ 0& X_2 & \vec{u} \\ \cdots & \vec{v} &\vdots \end{pmatrix}

Here, $X_1$ and $X_2$ are the two diagonal blocks; each block is $N \times N$. I concatenate to them a single row $\vec{v}$ with $2N+1$ entries and a single column $\vec{u}$ with $2N+1$ entries, and aim to understand how the SVD of the complete matrix relates to the SVD of the block diagonal sub-matrix.

Thanks

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    $\begingroup$ It's a bit difficult for me to understand what you mean by "$N+k$ dense rows of size $2N$". Could you explain a bit more or give an example? $\endgroup$
    – Max
    Dec 18, 2020 at 2:01
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    $\begingroup$ I added an example, I hope it's easier to understand now. $\endgroup$
    – user1767774
    Dec 18, 2020 at 22:42
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    $\begingroup$ Yes, I think so. I edited a bit more, check if you agree. Now I think I understand what you are asking. (Still don't know if there is any useful relation of the kind you are asking for, though. Seems challenging). $\endgroup$
    – Max
    Dec 18, 2020 at 22:56
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    $\begingroup$ I believe that the matrices you're describing are called block arrowhead matrices. $\endgroup$
    – NoName
    Dec 19, 2020 at 17:27
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    $\begingroup$ Cauchy's interlacing theorem for singular values gives inequalities. The discussion here explains why the interlacing inequality holds for singular values. $\endgroup$
    – eepperly16
    Dec 21, 2020 at 7:35

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