What is the two norm of $\text{diag}\{a_1,a_2,\dots,a_n\}\left[\begin{smallmatrix}\exp(x_1)\\\exp(x_2)\\\vdots\\\exp(x_n)\end{smallmatrix}\right]$

Question

I know that the two norm of a matrix is the largest singular value.

I am presented with an operator $U$ where $U: \mathbb{R}^n \to \mathbb{R}^n$

$U = A\exp(x) = \text{diag}\{a_1, a_2, \ldots, a_n\}\begin{bmatrix} \exp(x_1) \\ \exp(x_2) \\ \vdots \\ \exp(x_n) \end{bmatrix}$ where $a_i > 0$

How do I evaluate $\|U\|_2$ in this case?

Answer 1

I'm afraid your operator does not have a finite norm. Consider, for example, the case $x_1=x_2=\ldots=x_n=c_1$. Then $\left\|x\right\|_2=c_1\sqrt{n}$. However, by your definition of $U$, we have $$\left\|Ux\right\|_2 = \sqrt{\sum_k a_k^2 {\rm e}^{2c_1}} = C_2 \cdot {\rm e}^{c_1},$$ where $C_2 = \sqrt{\sum_k a_k^2}>0$. Therefore, $$\frac{\left\|Ux\right\|_2}{\left\|x\right\|_2} = \frac{C_2}{c_1} {\rm e}^{c_1},$$ which grows unbounded for $c_1\rightarrow \infty$. Therefore, the supremum over $x$ does not exist.

Answer 2

You have $U= diag (a_1 \exp(x_1),...,a_n \exp(x_n))$, thus $U^T=U$ and therefore

$U^TU=U^2=diag (a_1^2 \exp(2x_1),...,a_n^2 \exp(2x_n))$

The largest singular value 0f $U$ is given by

$$\max\{a_1^2 \exp(2x_1),...,a_n^2 \exp(2x_n)\}$$