# Thread: Equivalence of vector norms

1. ## Equivalence of vector norms

Hi.

problem:
For each of the following, verify the inequality and give an example of a nonzero vector or matrix for which equality is achieved. In this problem x is an m-vector and A is an $m \times n$ matrix.

(a) $||x||_{\infty} \leq ||x||_2$,

(b) $||x||_2 \leq \sqrt{m} ||x||_{\infty}$,

(c) $||A||_{\infty} \leq \sqrt{n} ||A||_2$,

(d) $||A||_2 \leq \sqrt{m} ||A||_{\infty}$.

attempt:

(a) I rewrite the inequality so it is easier for me to read:
$(max_i |x_i|)^2 \leq (x^2_1+\cdots +x^2_m)$
Let $x_k$ be the largest component of $x$. Then $(max_i|x_i|)^2=x^2_k$.
If the remaining components are zero, the inequality becomes an equality. If one or more of the remaining components are nonzero, it becomes an inequality.
Example of vector: $x=[x_k,0,\cdots,0]$.

I'm sure there are better ways to verify this.

(b) Again, I rewrite:

$x^2_1+\cdots+x^2_m \leq m(max_i|x_i|)^2$.
Let $x_k$ be the largest component of $x$.
Since $mx^2_k=x^2_k+\cdots +x^2_k$ (m-times) and since $x_i$ for $i\neq k$ are smaller than $x_k$, the inequality holds.
It's an equality for all vectors $x$ with $m=1$

I am not at all sure what to do with (c) and (d).
Any hints are appreciated.

2. Is $||A||_p$ the regular p norm when A is regarded as a vector in $\mathbb{R}^{nm}$?

Is $||A||_p$ the regular p norm when A is regarded as a vector in $\mathbb{R}^{nm}$?
$||A||_p=\left( \sum^m_{i=1} \sum^n_{j=1} |a_{ij}|^p \right)^{1/p}$
4. Yes. A matrix is a block of nm numbers, so it can be identified with a vector in R^{nm}, and its norm is the same as the norm of this vector, so this makes (a) stronger than (c) I believe. For (d), consider an n by 1 matrix all of whose entries are C>0. Then (d) asserts $nmC^2 \leq mC^2$ which is false if n>1, so perhaps there was a typo? If m is replaced by nm then (d) follows from (b) in the same way that (c) follows from (a).
5. If I look at the matrix A as a $mn$-dimensional vector and then calculate the, say 2-norm, I am calculating the Forbenius norm. $||A||_2$ is the matrix norm induced by vector norms, and so is not the same as the Forbenius norm.