Two norms ${∥∙∥}_{1}$ and ${∥∙∥}_{2}$ are equivalents $\iff$ $\exists$ $c_{1}$, $c_{2} > 0$ such that $c_{1}{∥x∥}_{1}\le {∥x∥}_{2} \le c_{2}{∥x∥}_{1}$
Let $X$ - Hilbert space over $\mathbb{R}$, $\dim X=n < \infty$
${∥x∥}_{1}=\sqrt{\left<x\mid x\right>}$
${∥x∥}_{2}=\sqrt{\left<Ax\mid Ax\right>}$
where matrix $A:X \to X$ is nondegenerate on $X$.
I showed that ${∥x∥}_{1}$, ${∥x∥}_{2}$ are norms, but I can't find the $c_{1}$, $c_{2}$ to show that they are equivalent. They must be, since two norms on the same finite dimension space are equivalent. Any ideas how to solve the problem?
In general, as long as $A:X\to X$ is a bounded linear operator with bounded inverse, the two norms are equivalent. Boundedness means there is $c>0$ with $\|Ax\|\le c\|x\|$. Here, $\|x\|_2=\|Ax\|_1\le c\|x\|$. Also as $A^{-1}$ is bounded, there is a $c'>0$ with $\|A^{-1}x\|\le c'\|x\|$, equivalently $\|x\|\le c'\|Ax\|$. Then $\|x\|_1\le c'\|x\|_2$.
For $X$ finite-dimensional, all operators are bounded. If we use the usual inner product on vectors, and $A$ is a matrix, then the $j$-th entry of $Av$ is $\left<a_j,v\right>$ where $a_j$ is the $j$-th row of $A$ and so is $\le\|a_j\|\|v\|$ by Cauchy-Schwarz. A crude estimate now gives $\|Av\|\le\|v\|\sum_{j=1}^n\|a_j\|$.