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Consider the induced matrix p-norm. My question is:

Can these norms be induced by inner product?

It sufficed to check that if it satisfies the parallelogram law. But I don't find an immediate way to either find a counterexample or prove that it is true. I generated some examples using MATLAB. It seems that the answer should be no. Any idea?

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Norbert's answer in a nutshell: Consider $A_1 = \begin{bmatrix} 1 & 0 \cr 0 & 0\end{bmatrix}$ and $A_2 = \begin{bmatrix} 0 & 0 \cr 0 & 1\end{bmatrix}$. –  Martin Mar 1 '13 at 10:03
    
@Martin: the induced $1,2,\infty$ norms are easy to check with your example. How about the general indued $p$ norms? –  Jack Mar 1 '13 at 14:40
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This works for all the operator norms induced by the $p$-norms: just check that $\lVert A_1 \pm A_2\rVert_p = 1$ and $\lVert A_i \rVert_p = 1$. The main point in Norbert's answer is writing these matrices in a different way. In theorem 3 it is assumed that $\mathcal{B}(X)$ is equipped with the operator norm, by the way. You can always equip $\mathcal{B}(\mathbb{R}^n)$ with the Hilbert-Schmidt norm which is an inner-product norm, but not an induced operator norm. –  Martin Mar 1 '13 at 14:47
    
@Martin: Ah, I see. Thank you! –  Jack Mar 1 '13 at 15:08

1 Answer 1

up vote 2 down vote accepted

Theorem 1. Let $X$ be a normed space. For a given $z\in X$ and $f\in X^*$ define operator $$ z\bigcirc f:X\to X:x\mapsto f(x) z $$ Then we have the following equalities $$(\lambda_1z_1+\lambda_2z_2)\bigcirc f=\lambda_1(z_1\bigcirc f)+\lambda_2(z_2\bigcirc f)$$ $$z\bigcirc (\mu_1 f_1+\mu_2 f_2)=\mu_1(z\bigcirc f_1)+\mu_2(z\bigcirc f_2)\tag{1}$$ $$\Vert z\bigcirc f\Vert=\Vert z\Vert\Vert f\Vert$$

Proof. Straightforward computation.

Theorem 2. Let $X$ be a normed space. Fix $f\in X^*$ with $\Vert f\Vert=1$, the map $$ i:X\to\mathcal{B}(X):x\mapsto x\bigcirc f $$ is an isometry.

Proof. Immediate consequence of theorem 1.

Theorem 3. If $X$ is a normed space with $\mathrm{dim}(X)>1$ then $\mathcal{B}(X)$ is not an inner product space.

Proof. If $\mathcal{B}(X)$ is an inner product space then does its every subspace. In particular $i(X)$ is a inner product space. Since $i$ is an isometry, then $X$ is an inner product space too. Now recall that we have isometric embedding $$ j:X\to X^*:x\mapsto(z\mapsto\langle z, x\rangle) $$ Since $\mathrm{dim}(X)>1$ we have two orthogonal vectors $e_1, e_2\in X$. It is not hard to check that for operators $A_1=e_1\bigcirc j(e_1)$, $A_2=e_2\bigcirc j(e_2)$ we have $$ \Vert A_1+A_2\Vert^2+\Vert A_1-A_2\Vert^2\neq 2\Vert A_1\Vert^2+2\Vert A_2\Vert^2 $$ so $\mathcal{B}(X)$ is not an inner product space. Contradiction.

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What's ${\mathcal B}(X)$? –  Jack Mar 1 '13 at 13:10
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$\mathcal{B}(X)$ is the space of all bounded linear opeartors on $X$. If $X$ is finite dimensional then $\mathcal{B}(X)$ is equal to the space of all linear operators on $X$. –  no identity Mar 1 '13 at 13:55
    
I'm a little bit confused with theorem 3. Let $X={\Bbb R^n}$. Then ${\mathcal B}(X)$ is all the n by n matrices. Can one identify ${\mathcal B}(X)$ as ${\Bbb R}^{n^2}$ which can be an inner product space? –  Jack Mar 1 '13 at 14:26
    
We can identify them as linear spaces, but they ae not identical as normed spaces - their norms are different! –  no identity Mar 1 '13 at 14:50
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I asssume that $\mathcal{B}(X)$ is equipped with operator norm, defined as $$\Vert T\Vert=\sup\{\Vert Tx\Vert_X:x\in X,\;\Vert x\Vert_X\leq 1\}$$ –  no identity Mar 1 '13 at 15:11

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