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

  • 2
    $\begingroup$ If you know that any two norms are equivalent then there is nothing left to prove. $\endgroup$ – Kavi Rama Murthy Dec 22 '19 at 12:31
  • $\begingroup$ Yes but iwnt to prove this without using that . $\endgroup$ – thomson Dec 22 '19 at 12:34
  • $\begingroup$ See this answer for proving that two norms on a finite-dimensional vector space are equivalent: math.stackexchange.com/a/599866/38584 $\endgroup$ – Math1000 Dec 22 '19 at 13:00

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.


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