Let $v \in \mathbb{R}^n$ and $A=v^T$. Prove that $\| A \|_2=\| v\|_2$
I know this question has already been posted, but its answer has not convinced me. The norm from $A$ is the matrix 2-norm, and the second is the standard euclidean vector 2-norm. Some auxiliary results that I thought could help are:
- $\| M \|_2 = \sqrt{\|M^TM\|_2}, \ A \in \mathbb{R}^{m \times n}$
- $\| M \|_2 = \lambda_{max}, \ M$ is a symmetric matrix
I will let $||\cdot||_2$ denote the operator $2$-norm and $|\cdot|_2$ denote the vector $2$-norm.
Let $A = v^T$. Then we have $$||A||_2 = \sqrt{||(v^T)^Tv^T||_2} = \sqrt{||vv^T||_2}.$$ $vv^T$ is a matrix with rank 1. Indeed, $(vv^T)x \in \mathrm{span}\{v\} \ \forall x \in\mathbb{R}^n$. This means that $vv^T$ only has a single nonzero eigenvalue and, since $vv^T$ is real and symmetric, it will be positive. We find this eigenvalue and eigenvector by inspection: $$ (vv^T)v = |v|_2^2 v. $$ Therefore $|v|_2^2$ is an eigenvalue of $vv^T$. By our previous argument, it is the only nonzero eigenvalue, and it is positive, so it is the largest eigenvalue of $vv^T$. Therefore, we have $$ ||v^T||_2 = \sqrt{||vv^T||_2} = \sqrt{|v|_2^2} = |v|_2. $$