Element with maximum magnitude in $A \leq \max(\sigma_{i})$, where $\sigma_{i}$ is singular values of A

Question

Let $A$ be a matrix with real values. Is it true that element with maximum magnitude in $A$ is less than $\max(\sigma_{i})$, where $\sigma_{i}$ is singular values of A? That is, is $$ \max_{ij} |A_{ij}| \leq \sigma_{\max}(A)?$$

Answer 1

Yes. Suppose $A $ is $m\times n $. Let $e_1,\ldots,e_n $ and $f_1,\ldots,f_m $ be the canonical bases of $\mathbb R^n $ and $\mathbb R^m $ respectively. Write $A=USV $ the singular value decomposition. Let $v=Ve_j $; as $V $ is a unitary, $\|Ve_j\|=1$. Then, since $S_{pq}=0 $ if $p\ne q $, \begin{align} |A_{ij}| & =|\langle Ae_j,f_i\rangle| =|\langle USVe_j,f_i\rangle| =|\langle Sv,U^*f_i\rangle|\\ \ \\ &\leq\|Sv\|\,\|U^*f_i\|=\|Sv\| =\left (\sum_{k=1}^m(Sv)_k^2\right)^{1/2}\\ \ \\ &=\left (\sum_{k=1}^m\left(\sum_{h=1}^nS_{kh}v_h\right)^2\right)^{1/2} =\left (\sum_{k=1}^mS_{kk}^2v_k^2\right)^{1/2}\\ \ \\ &\leq \left (\sum_{k=1}^m\sigma_\max(A)^2v_k^2\right)^{1/2} = \sigma_\max (A)\,\|v\|=\sigma_\max (A). \end{align} (note that, if $k>n $, $S_{kk} $ does not exist; but it can be taken as zero and the inequalities still hold)