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Given a fat matrix $A \in \mathbb{R}^{m \times n}$ (where $m<n$) with singular values $\sigma_1, \dots, \sigma_m$, can I express the singular values $\tilde{\sigma}_1, \dots, \tilde{\sigma}_{m+n}$ of the following block lower triangular matrix

$$T = \begin{pmatrix} I & 0 \\ A & I \end{pmatrix}$$

in terms of the singular values of $A$? Or can I give some upper bounds on the singular values of $T$?

Kind regards and thanks in advance!

EDIT:

I think I got a solution myself: on a standard textbook on matrix analysis I found the property $ \sigma_i(C) - \sigma_i(B) \leq \Vert C - B \Vert $ for all $i$ with $A,B \in \mathbb{R}^{m \times n}$.

Now, defining $C = T$ and $$ B = \begin{pmatrix} 0 & 0 \\ A & 0 \end{pmatrix} $$ we have $C - B = I$ and $\sigma_{max}(B) = \sigma_{max}(A)$. Since $I$ has norm 1, then, we have

$$ \sigma_{max}(T) = \sigma_{max}(C) \leq 1 + \sigma_{max}(A).$$

Am I forgetting anything? It looks too beautiful to be true.

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  • $\begingroup$ I think I got a solution myself: on a standard textbook on matrix analysis I found the property $ \sigma_i(C) - \sigma_i(B) \leq \Vert C - B \Vert $ for all $i$ with $A,B \in \mathbb{R}^{m \times n}$. Now, defining $C = T$ and $$ B = \begin{pmatrix} 0 & 0 \\ A & 0 \end{pmatrix} $$ we have $C - B = I$ and $ \sigma_{max}(B) = \sigma_{max}(A)$. Since $I$ has norm 1, then, we have $$ \sigma_{max}(T) = \sigma_{max}(C) \leq 1 + \sigma_{max}(A).$$ Am I forgetting anything? It looks too beautiful to be true. $\endgroup$
    – Trb2
    Commented Nov 28, 2020 at 0:11
  • $\begingroup$ Feel free to edit that in your question. $\endgroup$
    – Argyll
    Commented Nov 28, 2020 at 0:12
  • $\begingroup$ @Argyll what should I edit? $\endgroup$
    – Trb2
    Commented Nov 28, 2020 at 0:13
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    $\begingroup$ Eigenvalues of $T T^\top$. $\endgroup$
    – Rodrigo de Azevedo
    Commented Nov 28, 2020 at 22:32
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    $\begingroup$ @Trb2 I obtained the same (or close). I am not very happy with it either. Please consider editing your question again and including the work in your comment. $\endgroup$
    – Rodrigo de Azevedo
    Commented Dec 5, 2020 at 10:52

1 Answer 1

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The upper bound is trivial since

$$ \sigma_{\max}(B) = \|B\|_2 = \|I + E \|_2 \le 1 + \|E\|_2 = 1 + \|A\|_2 = 1 + \sigma_{\max}(A). $$

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