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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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    What should be $N = Id$?2012-03-15
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    Sorry, I meant $B$.2012-03-15
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    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?2012-03-15
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    And Im not sure, if still holds $n = r$ and $B = Id$.2012-03-15
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    $Bx$ 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 $n$.2012-03-15
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    You wrote: let $x$ of norm $1$ such that $Bx=\|x\|$. I think it means $Bx=\|x\| = 1$, doesnt it? If the answer is yes, how can be $Bx=1$? If the anwer is no, I dont understand the sentence. :)2012-03-15
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    Sorry I will edit it.2012-03-16
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    Doesnt the definiton of the 2-norms imply $\|B\|_2^2 \geq \|Bx\|_2^2$?2012-03-20
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    Yes, it's implied.2012-03-20
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    So why are you writing that $\|B\|_2^2 = \|Bx\|_2^2$?2012-03-20
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    Yes, it's true for a particular $x$ (not all) since the unit ball of $\mathbb R^n$ is compact.2012-03-20
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    Ok, thx a lot. So then I have a problem prove the inequality: $\|AB\|_F \leq \|A\|_2 \|B\|_F$.2012-03-20
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    I thought this way:2012-03-20
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    Let $\mathbf{b}$ is a vector $n \times 1$, $A$ is a matrix $m \times n$. Because $\|\mathbf{b}\|_F^2~=~\|\mathbf{b}\|_2^2$ and $\|A\mathbf{b}\|_2^2~\leq~\|A\|_2^2 \cdot \|\mathbf{b}\|_2^2$ then $$\|A\mathbf{b}\|_F^2~=~\|A\mathbf{b}\|_2^2~\leq~\|A\|_2^2 \cdot \|\mathbf{b}\|_2^2~=~\|A\|_2^2 \cdot \|\mathbf{b}\|_F^2$$.2012-03-20
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    Then I need to prove that $\|A\|_2^2 \cdot \|B\|_2^2 \geq \|AB\|_F^2$ and if the 1. equation is true, then I can write \begin{equation*} \|A\|_2^2 \cdot \|B\|_2^2 = \|A\|_2^2 \cdot \[ \ {\|\mathbf{b_1}\|_2^2~+\ldots~+~\|\mathbf{b_r}\|_2^2} \ \] \end{equation*}2012-03-20
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    then its easy to prove that \begin{equation*} \|A\|_2^2 \cdot \[ \ {\|\mathbf{b_1}\|_2^2~+\ldots~+~\|\mathbf{b_r}\|_2^2} \ \] \geq \|AB\|_F^2 \end{equation*}2012-03-20
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    But if the 1. equation isnt true, then its a problem. :(2012-03-20