I'm trying to prove that $\|v^T\|_2=\|v\|_2$ for $v\in \mathbb{R}^n$. The first norm is the matrix norm, and the second norm is the usual euclidian norm.
I already prove that:
$\|v\|_2\leq \|v\|_1\leq \sqrt{n} \|v\|_2;$
$\|v\|_\infty\leq \|v\|_2\leq \sqrt{n} \|v\|_\infty; $
$\frac{1}{\sqrt{n}}\|v\|_\infty\leq \|v\|_1\leq \sqrt{n} \|v\|_\infty;$
and, for matrix norms, if $A\in\mathbb{R}^{m\times n}$:
$\frac{1}{\sqrt{n}}\|A\|_\infty\leq \|A\|_2\leq \sqrt{m} \|A\|_\infty;$
$\frac{1}{\sqrt{m}}\|A\|_1\leq \|A\|_2\leq \sqrt{n} \|A\|_1.$
Finally, I also proved that
$\|v^T\|_1=\|v\|_\infty$;
$\|v^T\|_\infty=\|v^T\|_1$.
With all these relations I managed to prove that
$$ \frac{1}{\sqrt{n}}\|v\|_2\leq \|v^T\|_2\leq \sqrt{n}\|v\|_2, $$ and I don't know how to "remove" these $\sqrt{n}$ and prove the equality.
I assume that by Matrix norm you mean $$\tag{1} \|v^\top\|_2=\sup_{w\not=0}\frac{v\cdot w}{\|w\|_2}\,. $$ The Cauchy-Schwarz inequality says $$ |v\cdot w|\le \|v\|_2\,\|w\|_2 $$ with equality holding if and only if $v$ and $w$ are linearly dependent. This means that the sup on the RHS of (1) is smaller than $\|v\|_2$ and is attained for $w=v$. It follows that $$\tag{2} \|v^\top\|_2=\frac{v\cdot v}{\|v\|_2}=\|v\|_2\,. $$