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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

1 Answers 1

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  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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    But if the 1. equation isnt true, then its a problem. :(2012-03-20