Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $v \in \mathbb{R}^k$, and let $A \in \mathbb{R}^{m \times k}$ and let $B \in \mathbb{R}^{m \times n}$ such that each column of $B$, $B_i$, has $$||B_i||_2 \le 1.$$

Is it true that:

  1. $||v A^{\top} B||_{\infty} \le ||v A^{\top}||_2$ ?

  2. If the spectral norm of $A$ is such that $||A||_{\mathrm{spectral}} \le 1$, is it true that $||v A^{\top}||_2 \le ||v||_2$ ?

Thanks.

share|improve this question
    
In finite dimensional spaces spaces, norms are equivalent. –  Mhenni Benghorbal Sep 4 '12 at 16:35
3  
@Mhenni I am not sure how this answers my question? –  kloop Sep 4 '12 at 16:47
add comment

2 Answers

up vote 2 down vote accepted

1) Cauchy-Schwarz says $$|(v A^T B)_i| = |v A^T B_i| \le \|v A^T\|_2 \|B_i\|_2 \le \|v A^T\|_2$$

2) Yes because the spectral norm is the operator norm corresponding to the $2$-norm on vectors, and $\|A\|_{\text{spectral}} = \|A^T\|_{\text{spectral}}$.

share|improve this answer
    
could you explain the first equality equation you wrote? (and the reference please) –  sara Sep 4 '12 at 17:19
    
For any row vector $w$ and matrix $M$ with compatible dimensions, the $i$'th entry of $wM$ is the product of $w$ and the $i$'th column of $M$. –  Robert Israel Sep 4 '12 at 17:21
    
thanks Robert -- your answers are always very concise and accurate. –  kloop Sep 7 '12 at 1:50
add comment

I give it a try with your first question: If $A\in\mathbb{R}^{m\times n}$ then $$\frac{1}{\sqrt{n}}\|A\|_\infty\leq \|A\|_2\leq \sqrt{m}\|A\|_\infty$$ $$\frac{1}{\sqrt{m}}\|A\|_1\leq \|A\|_2\leq \sqrt{n}\|A\|_1$$

Also $$\|AB\|_a\leq \|A\|_a\|B\|_a$$

Note that $\|B_i\|_2<1$ implies that absolute value of the every element of matrix B is less than 1, $|B_{ij}|<1$, then $\|B\|_\infty<n$ (infinity norm of a matrix is the maximum absolute row sum of the matrix). And with the same argument $\|B\|_1<m$.

Then using these properties we have $$\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq \sqrt{nm} \|vA^T\|_2\|B\|_\infty\leq n\sqrt{nm} \|vA^T\|_2$$ or $$\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq n \|vA^T\|_2\|B\|_1\leq nm \|vA^T\|_2$$ (these are conservative bounds you can probably obtain a tighter bound)

share|improve this answer
    
I think the answer to the second question is yes. Spectral norm is the 2-induced norm for a square matrix then $\|A\|_{spectral}=\|A^T\|_{spectral}$. $\|vA^T\|_{spectral}\leq \|v\|_2\|A^T\|_2=\|v\|_2\|A\|_2\leq \|v\|_2$ –  sara Sep 4 '12 at 17:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.