Matrix inequality: Eigenvalues A is a $n \times n$ non symmetric matrix. Can $\sigma_1(A)$ - the greatest singular value of A -  be upper bounded by some function of the eigenvalues of A?
 A: You can not have such a boundary for a generic matrix basing only on the its eigenvalues. For example, $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Both eigenvalues are zeros, yet $\sigma_1(A)=1$. Take now $B=\begin{pmatrix}0&0\\0&0\end{pmatrix} $. Both eigenvalues of $B$ are also zeros, yet $\sigma_1(B)=0$.
You can, however, write $$\sigma_1(A)=\sup_{x\ne 0}\frac{\|Ax\|^2}{ \|x\|^2} = \|A\|^2\ge \max_j(|\lambda_j|^2),$$where $\lambda_j$ are the eigenvalues of $A$.
Edit to answer the comment
You are essentially asking to estimate the norm of the matrix by a function of its spectral radius, which is impossible without any additional info on the matrix (its eigenvectors and generalised eigenvectors, for example).
You can rewrite the matrix $A=PJP^{-1}$ where $J$ is its Jordan norm. You can write the norm of $J$ in terms of eigenvalues of $A$ and the structure of eigenvectors and generalised eigenvectors, then say that $\sqrt{\sigma_1(A)}=\|A\|\le cond(P)\|J\|$. However, the conditionment number $cond(P)=\|P\|\|P^{-1}\|$ is, in general case, very large; this, coupled with quite unstable behaviour of structure of Jordan form and of the matrix $P$ itself (with respect to coefficients of $A$), the above estimation on $\|A\|$ has little to no viable applications. On top of that, in real life we don't really have a reliable method of finding eigenvalues for non-symmetric matrices.
