Max Eigenvalue of A'A A is an $m\times m$ square matrix with (possibly complex) eigenvalues $\lambda_i$. Suppose $\max |\lambda_i(A)|&lt;r$ (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.
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 $A=\left[\begin{array}{cc}
 0&amp;1\\
0&amp;0 \end{array}
\right]$. The eigenvalue is $0$ which give $\rho(A)=0$. However:
 $$
 A^* A=\left[\begin{array}{cc}
 0&amp;0\\
1&amp;0  \end{array}
\right]
\left[\begin{array}{cc}
 0&amp;1\\
0&amp;0  \end{array}
\right]=
\left[\begin{array}{cc}
 0&amp;0\\
0&amp;1  \end{array}
\right]
$$
which has eigenvalues $0,1$ and hence $\|{{A}}\|_{2}=1$. Hence $\rho(A)&lt;\|A\|_2$.

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 $\rho(A)\leq \|A\|_2$.
Since $\;\lambda\;$ is an eigenvalue of $\;A\implies \overline\lambda\;$ is an eigenvalue of $\;A^*\;$ , and since $\;|\lambda|=|\overline\lambda|\;$we get that

$$\left|\lambda_\max(A^*A)\right|=|\lambda_\max(A)|^2&lt;r^2$$