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I'm need a little help here :

using the parallelogram law: $$ 2||u||^2 + 2||v||^2 = ||u + v||^2 + ||u - v||^2 $$ I'm need to show that the next norm can not be satisfy by inner product:

The norm $||A|| = sup||Av||_\mathbb{R^2}$ that defined all over the real matrixes 2 x 2 when the norm $|| . ||_\mathbb{R^2}$ is the standard norm over $\mathbb{R^2} ( ||v||_\mathbb{R^2}=\sqrt{v_1^2+v_2^2}) $ and the supermum is taken over all the vectors $ v \in \mathbb{R^2} $ that satisfy $||v||_\mathbb{R^2}=1$

I think that the good way to prove it ,is by finding example that break this law. but my problem is that I'm even can't understand the norm and how it function.

Thanks in advance

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2 Answers 2

This norm is called the operator norm and is extremely useful. Since $A$ is a linear map, the following definition is equivalent: $\| A \| = \sup \left\{ \frac{\| Av \|}{\| v \|} : v \not= 0 \right\}$, so you can think of it as "the largest amount $A$ lengthens a vector by".

It has the property that $\| Av \|_{\mathbb R^2} \le \| A \| \| v \|_{\mathbb R^2}$, and it's the maximal value such that that inequality always holds: from this you can prove that for any matrices $A$ and $B$, $\| ABv \|_{\mathbb R^2} \le \| A \| \| Bv \|_{\mathbb R^2} \le \| A \| \| B \| \| v \|_{\mathbb R^2}$, so $\| AB \| \le \| A \| \| B \|$, so the norm plays nicely with matrix multiplication.

To get a feel for how this works, perhaps try going through the proof that it is a norm:

  • $\| A \| \ge 0$: this is obvious, since it is a supremum over a set of positive numbers.
  • $\| A \| = 0 \implies A = 0$: well, if $A$ ever has $Av \not= 0$, then $\| Av \| > 0$, so the supremum will be strictly positive.
  • $\| \lambda A \| = |\lambda| \| A \|$: this follows from the same property for the vector norm, and from the fact that $\sup$ lets positive constants through unchanged.
  • $\| A + B \| \le \| A \| + \| B \|$: intuitively this ought to be true because on the LHS we have to give the same $v$ to $A$ and $B$ whereas on the RHS we maximise each separately. See if you can work out the details just by expanding the definitions.
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The question asks whether this norm is an inner product norm, which is equivalent to the parallelogram law being satisfied. –  1015 Feb 15 '13 at 0:28
    
@julien: correctly observed! But it also says "my problem is that I'm even can't understand the norm and how it function." – my hope is that once that issue is addressed, the questioner will be able to solve the problem. –  Ben Millwood Feb 15 '13 at 0:34
1  
Sorry, you're right. I had not noticed the last sentence of the post. –  1015 Feb 15 '13 at 0:37

First show that the norm of $\rm A$ is equal to the greatest module of the eigenvalues of $\rm A$.

Second $u = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} ; v = \begin{bmatrix} 1 & 0 \\0 & 0 \end{bmatrix}$ is a counter-example.

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Note that the norm is equal to the spectral radius when $A$ is normal, not in general. This is fine here since your examples are normal. –  1015 Feb 15 '13 at 0:14

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