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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

Eva
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1 Answers1

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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.
Davide Giraudo
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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
  • 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. :) – Eva Mar 15 '12 at 22:46
  • Sorry I will edit it. – Davide Giraudo Mar 16 '12 at 18:09
  • Doesnt the definiton of the 2-norms imply $|B|_2^2 \geq |Bx|_2^2$? – Eva Mar 20 '12 at 20:51
  • Yes, it's implied. – Davide Giraudo Mar 20 '12 at 21:13
  • So why are you writing that $|B|_2^2 = |Bx|_2^2$? – Eva Mar 20 '12 at 21:16
  • Yes, it's true for a particular $x$ (not all) since the unit ball of $\mathbb R^n$ is compact. – Davide Giraudo Mar 20 '12 at 21:17
  • Ok, thx a lot. So then I have a problem prove the inequality: $|AB|_F \leq |A|_2 |B|_F$. – Eva Mar 20 '12 at 21:25
  • I thought this way: – Eva Mar 20 '12 at 21:26
  • 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 $$|A\mathbf{b}|_F^2~=~|A\mathbf{b}|_2^2~\leq~|A|_2^2 \cdot |\mathbf{b}|_2^2~=~|A|_2^2 \cdot |\mathbf{b}|_F^2$$. – Eva Mar 20 '12 at 21:30
  • 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} – Eva Mar 20 '12 at 21:39
  • 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} – Eva Mar 20 '12 at 21:46
  • But if the 1. equation isnt true, then its a problem. :( – Eva Mar 20 '12 at 21:47