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For an $n\times n$ diagonizable matrix $A$, is there a relation between the eigenvalues of $A$ and the eigenvalues of $A^TA$?

I ask this because I am looking into the relation between $A$ and $A+cI$, specifically their condition numbers:

$$\kappa(A)=\sqrt{\frac{\lambda_{\mathrm{max}}\left(A^TA\right)}{\lambda_{\mathrm{min}}\left(A^TA\right)}}\quad \mbox{ and }\quad \kappa(A+cI)=\sqrt{\frac{\lambda_{\mathrm{max}}\Big((A+cI)^T(A+cI)\Big)}{\lambda_{\mathrm{min}}\Big((A+cI)^T(A+cI)\Big)}}$$

I know for SPD matrices this reduces to

$$\kappa(A)=\frac{\lambda_{\mathrm{max}}(A)}{\lambda_{\mathrm{min}}(A)}\quad \mbox{ and }\quad \kappa(A+cI)=\frac{\lambda_{\mathrm{max}}(A)+c}{\lambda_{\mathrm{min}}(A)+c},$$ such that I can derive stuff like $$\kappa(A)<\kappa(A+cI)\quad \mathrm{ or } \quad \kappa(A)>\kappa(A+cI) $$ depending on $c$.

However, for general $A$ when I try rewriting $\kappa(A)$ I don't know what to do with the $\lambda(A^TA)$ part.

Can anyone help me? Or provide another trick to relate $\kappa(A)$ and $\kappa(A+cI)$ for general $A$, like I did for SPD matrices.

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    $\begingroup$ Are you aware that for $A = [0\ 1;1\ 0]$ it holds $\kappa(A) = 1$ but $\kappa(A+I) = \infty$? $\endgroup$
    – Dirk
    Commented Feb 3, 2016 at 14:40
  • $\begingroup$ Yes. That is because in your example $1$ is an eigenvalue of $A$. Note that this falls under my remark "such that I can derive stuff like [...] depending on $c$". $\endgroup$
    – Eric S.
    Commented Feb 3, 2016 at 14:49
  • $\begingroup$ @Dirk, but why do you point this out? I don't see how it contributes to answering my question... $\endgroup$
    – Eric S.
    Commented Feb 5, 2016 at 10:43
  • $\begingroup$ I was under the impression that you had in mind to make the matrix $A$ "numerically better" by adding $cI$. My remark should only point out that it's not that simple. $\endgroup$
    – Dirk
    Commented Feb 5, 2016 at 10:57
  • $\begingroup$ @Dirk, Ah okay. That is indeed what I intend to do. I am looking for which values of $c$ the addition $cI$ is favorable. For the SPD case I have done this into great detail, as the condition number of $A$ and $A+cI$ have attractive forms. For the general case, as stated, I don't know how to relate the condition numbers, due to the $\lambda(A^TA)$ term. BTW, my previous comment might have come across a bit blunt, that was not my intention. Thank you for thinking along with me :) $\endgroup$
    – Eric S.
    Commented Feb 5, 2016 at 11:26

2 Answers 2

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Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ and $s_1,\ldots,s_n$ be those of $\sqrt{A^*A}$, ordered by $$|\lambda_1|\ge\cdots\ge|\lambda_n|,\qquad s_1\ge\cdots\ge s_n.$$ Then it holds $$\prod_{j=1}^k|\lambda_j|\le\prod_{j=1}^ks_j,\qquad \forall\,k=1,\ldots,n.$$ For instance, if $k=1$, this is $\rho(A)\le\|A\|$ where we use the operator norm. For $k\ge1$, this can be viewed as the same inequality applied to he $k$-th exterior power of $A$.

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A simple relation between singular values and eigenvalues does not exist in general, as far as I know.

This is an old question, e.g.

A. Horn, On the eigenvalues of a matrix with prescribed singular values Proc. Am. Math. Soc 5 4-7 (1954)

H. Weyl H, Inequalities between the two kinds of eigenvalues of a linear transformation Proc. Natl. Acad. Sci. USA 35 408-11 (1949)

For a more recent paper, that treats this problem from a statistical point of view, you can try this

On the mean density of complex eigenvalues for an ensemble of random matrices with prescribed singular values, Yi Wei and Yan V Fyodorov, J. Phys. A: Math. Theor. 41 502001 (2008)

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