Absolute of all eigenvalues are always bounded by maximal singular value There are many discussions about the singular values and eigenvalues, such as What is the difference between Singular Value and Eigenvalue?. I want to ask the particular one in title. 
Usually, for a general square matrix, singular valure are not equal to eigenvalues. But singular values are alwyas nonnegative. My claim is 

$A\in \mathbb{R}^{n\times n}$. $|\lambda_i(A)|\leq \sigma_\max(A)$, for all $i$. 

The reason is the definition of the maximal singular value of $A$, which is $$\sigma_\max(A) = \|A\|_2 = \max_{\|x\|=1} \|Ax\|.$$ 
It reflects the maximal gain of $A$. And this $x$ does not have to be the eigenvector of $A$. However, the eigenvalue of $A$ is $$Av = \lambda v, \ \ \ \ \|v\|=1.$$ To get $\lambda_\max$, we have to have the extra constraint $\|Av\| 
= \|\lambda_\max v\|$. So $|\lambda(A)|\leq \sigma_\max (A)$.
Here $A$ is any real matrix not necessarily symmetric.  I am not sure if I am correct. 

I think is odd that the maximal singular value is definite in this way, but it coincides with the maximal eigenvector of $A^{*}A$.
Let $\sigma_1^2 > \dots > \sigma_n^2 $ be the eigenvalues of $A^{*}A$ and let $v_1, \dots v_n$ be the corresponding eigenvectors, which can also be a ortonormal basis. Take $u$ a vector and write as $u = \alpha_1 v_1 + \dots + \alpha_n v_n$ so
$|Au|^2 = <Au,Au> = <A^{*}Au,u> = <\sum_{i} \sigma_i^{2}\alpha_iv_i, \sum_{i}\alpha_iv_i>= \sum_{i} \sigma_i^{2}|\alpha_iv_i|^2 \leq \sum_{i} \sigma_1^{2}|\alpha_iv_i|^2 = \sigma_1^2 |u|$
So $|Au| \leq \sigma_1$ for every $u$ unitary. So your claim is true.
