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Im prooving the inequality: $\|AB\|_F \leq \|A\|_2 \|B\|_F$. To prove this I need to know, if the following is true:

  1. Lets $B_{n \times r}~=~(\mathbf{b_1}, \ldots, \mathbf{b_r})$ is a matrix, $\mathbf{b_1}$, $\ldots$, $\mathbf{b_r}$ are vectors $n \times 1$, then \begin{equation*} \|B\|_2^2~=~\|\mathbf{b_1}\|_2^2~+\ldots~+~\|\mathbf{b_r}\|_2^2. \end{equation*}

  2. Lets $A_{m \times n}, B_{n \times r}~=~(\mathbf{b_1}, \ldots, \mathbf{b_r})$ are matrices, $\mathbf{b_1}$, $\ldots$, $\mathbf{b_r}$ are vectors $n \times 1$, then \begin{equation*} \|AB\|_F^2~=~\|A\mathbf{b_1}\|_F^2~+\ldots~+~\|A\mathbf{b_r}\|_F^2. \end{equation*}

If these two equations are true, then I can finish the proof. In other case, its bad. Can anybody say me, whether they are true or not and in the case they are true, why?

Thank you very much. Eva

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1 Answer 1

  1. It should be an inequality since for example if $n=r$ and $B=Id$ we don't have the result. It can be shown by Cauchy-Schwarz inequality: let $x$ of norm $1$ such that $\lVert Bx\rVert=\lVert B\rVert_2$. Then $$\lVert B\rVert^2_2=\lVert Bx\rVert_2^2=\sum_{i=1}^n\left(\sum_{j=1}^rb_{ij}x_j\right)^2\leq \sum_{i=1}^n\sum_{j=1}^rb_{ij}^2$$ and we are done.
  2. It's true, in fact $Ab_j$ are vectors, just write the corresponding sums to check it.
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What should be $N = Id$? –  Eva Mar 15 '12 at 21:10
    
Sorry, I meant $B$. –  Davide Giraudo Mar 15 '12 at 21:11
    
Thx a lot, I understand the second part now. But I still dont get why $Bx = \|x\| = 1$, when $B$ is a matrix and $x$ is a vector. How in this case can $Bx$ be a number? –  Eva Mar 15 '12 at 22:17
    
And Im not sure, if still holds $n = r$ and $B = Id$. –  Eva Mar 15 '12 at 22:26
    
$Bx$ is a vector, not a number. If you take $n=r$ and $B$ the identity matrix we get that LHS is $1$ but the RHS is $n$. –  Davide Giraudo Mar 15 '12 at 22:28

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