Home » Uncategorized » Double inequality involving matrix

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.

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network