Double inequality involving matrix

Prove that

$\max_{1\leq j\leq p}\sqrt{\sum_{i=1}^{q}a^2_{ij}}\leq \left \| \begin{pmatrix} a_{11} &\cdots &a_{1p} \\ \vdots& \ddots &\vdots \\ a_{q1}&\cdots &a_{qp} \end{pmatrix} \right \|_2 \leq \sqrt{\sum_{i=1}^{q}\sum_{j=1}^{p}a^2_{ij}}$

Solution

Fix $j$ . Apply the matrix $A$ on $e_j$ thus:

$\Vert Ae_j\Vert _2 \le \Vert A\Vert _2 \Vert e_j\Vert _2= \Vert A\Vert _2$

Since $Ae_j$ is exactly the $j$ – th column of $A$ the previous equality can be rerwritten as

$\sqrt{\sum_{i=1}^{q}a^2_{ij}}\leq \Vert A \Vert _2$

Since this holds for all $j=1, 2, \dots, n$ we get $\max$ and the left inequality follows.

For a random unit vector $x=(x_1, x_2, \dots, x_n)$ the $i$ coordinate of the vector $Ax$ is $\sum \limits_j a_{ij}x_j$ . It follows from Cauchy – Schwartz that

$\left |\sum _ja_{ij}x_j\right |^2 \le \sum _j|a_{ij}|^2\sum _j|x_j|^2 = \sum _j|a_{ij}|^2$

Summing over all $i$ ‘s till we find $\Vert Ax \Vert _2$ we conclude that, for every unit vector $x$ , it holds that $\Vert Ax \Vert _2$ is less than the right hand side. Taking supremum with respect to all $x$ the right hand side inequality follows.

Double inequality involving matrix

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