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.
Best Answer
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 $A$.