Supnorms are equivalent with respect to two different bases

Question

If $F$ is a finite dimensional vector spaces over $\mathbb R$, I have to show that suprimum norm with respect to two different bases are equivalent.

I know how to prove that in a finite dimentional vector space any two norms are equivalent. In that proof we use induction on dimention.

Edit: without using that (in finite dimentional spaces two norms are equivalent) How can one prove this.

I tried to do this but no result. Any help

If you know that any two norms are equivalent then there is nothing left to prove. — Kavi Rama Murthy, Dec 22 '19 at 12:31
See this answer for proving that two norms on a finite-dimensional vector space are equivalent: https://math.stackexchange.com/a/599866/38584 — Math1000, Dec 22 '19 at 13:00

mechanodroid · Accepted Answer · 2019-12-26T17:01:22.220

Let $B = \{b_1, \ldots, b_n\}$ and $C = \{c_1, \ldots, c_n\}$ be two bases for $F$ which define norms $$\left\|\sum_{i=1}^n \beta_ib_i\right\|_B := \max_{1\le i \le n}|\beta_i|, \qquad \left\|\sum_{i=1}^n \gamma_ic_i\right\|_C := \max_{1\le i \le n}|\gamma_i|$$ For $x := \sum_{i=1}^n \beta_ib_i \in F$ we have

$$\left\|\sum_{i=1}^n \beta_ib_i\right\|_C \le \sum_{i=1}^n|\beta_i|\|b_i\|_C \le \left(\sum_{i=1}^n\|b_i\|_C\right)\left(\max_{1\le i \le n} |\beta_i|\right) = \left(\sum_{i=1}^n\|b_i\|_C\right)\left\|\sum_{i=1}^n \beta_ib_i\right\|_B$$

or $\|x\|_C \le \left(\sum_{i=1}^n\|b_i\|_C\right)\|x\|_B$.

Analogously we get $\|x\|_B \le \left(\sum_{i=1}^n\|c_i\|_B\right)\|x\|_C$ so we conclude that the norms $\|\cdot\|_B$ and $\|\cdot\|_C$ are equivalent.

Supnorms are equivalent with respect to two different bases

1 Answers1