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A is an $m\times m$ square matrix with (possibly complex) eigenvalues $\lambda_i$. Suppose $\max |\lambda_i(A)|<r$ (for some positive no. $r$ say). Then can we conclude any upper-bound of $\ \lambda_{\max} (A'A)$ (in terms of $r$) ? Thanks in advance.

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  • $\begingroup$ Is $\;A'=A^*=\;$ the adjoint of $\;A\;$ (the transpose conjugate of $\;A\;$?) $\endgroup$ – DonAntonio Sep 17 '16 at 20:13
  • $\begingroup$ For the time being let's assume A is real matrix. $\endgroup$ – Robert Sep 18 '16 at 7:49
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The question you are asking is deeply related to the relation between spectral norm $\|A\|_2$ and spectral radius $\rho(A)$. If $\lambda_i$'s are eigenvalues of $A$, then spectral radius is defined as: $$ \rho(A)=\max|\lambda_i|. $$ Spectral norm is defined as: $$ \|A\|_2=\max_{\|x\|_2=1}\|Ax\|_2. $$ See that $\|Ax\|_2^2=x^*(A^*A)x$ and since $A^*A$ is positive semidefinite, we can see: $$ \|A\|_2^2= \lambda_{\max}(A^*A). $$ In general $\|A\|_2\geq \rho(A)$, which means that the upper bound that you have does not tell that much about spectral norm. In general the inequality can be strict.

For instance consider $A=\left[\begin{array}{cc} 0&1\\ 0&0 \end{array} \right]$. The eigenvalue is $0$ which give $\rho(A)=0$. However: $$ A^* A=\left[\begin{array}{cc} 0&0\\ 1&0 \end{array} \right] \left[\begin{array}{cc} 0&1\\ 0&0 \end{array} \right]= \left[\begin{array}{cc} 0&0\\ 0&1 \end{array} \right] $$ which has eigenvalues $0,1$ and hence $\|{{A}}\|_{2}=1$. Hence $\rho(A)<\|A\|_2$.

However for some cases these two are equal. A very general class of these matrices are normal matrices. In this case any upper bound on spectral radius is an upper bound on spectral norm.


To see why $\|A\|_2\geq \rho(A)$, pick a unit norm eigenvector $y$ of $A$ corresponding to the eigenvalue with maximum absolute value. It is easy to see that $\|Ay\|_2=\rho(A)$. From the definition of $\|A\|_2$, we have $\rho(A)\leq \|A\|_2$.

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  • $\begingroup$ No, something's wrong here. $\endgroup$ – Robert Sep 18 '16 at 7:49
  • $\begingroup$ Please care to explain. $\endgroup$ – Arash Sep 18 '16 at 7:49
  • $\begingroup$ in this step $\|Ax\|_2^2= \lambda_{\max}(A^*A).$, LHS depends on x $\endgroup$ – Robert Sep 18 '16 at 7:52
  • $\begingroup$ That's right. Just a typo. corrected. $\endgroup$ – Arash Sep 18 '16 at 7:55
  • $\begingroup$ Yes, and it also I found another post from here math.stackexchange.com/questions/965950/… $\endgroup$ – Robert Sep 18 '16 at 7:56
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Since $\;\lambda\;$ is an eigenvalue of $\;A\implies \overline\lambda\;$ is an eigenvalue of $\;A^*\;$ , and since $\;|\lambda|=|\overline\lambda|\;$we get that

$$\left|\lambda_\max(A^*A)\right|=|\lambda_\max(A)|^2<r^2$$

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  • $\begingroup$ I don't get how you write the step $|\lambda_{\max} (A^*A)|=(\max |\lambda_i(A)|)^2$. I know non-zero eigenvalues of AB and BA are same, but I suppose this does not apply here. I am missing something else. Would you please elaborate? $\endgroup$ – Robert Sep 18 '16 at 7:48

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