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Let $M_{d \times d} (\Bbb R)$ be the space of real $d \times d$ matrices. For $A = (a_{ij} ) \in M^{d×d}$, define $$ \| A \|_2 := \left( {\sum_{i,j=1}^{d}|a_{ij}|^2} \right)^\frac{1}{2} $$ Show that $$\|AB\|_2 \leq \|A\|_2 \|B\|_2$$

I don't know where to start this step. How do I show that $I+A+A^2+\cdots$ converges to a limit if $\|A\|_2 \leq 1$? If anyone could please help me, I would appreciate it.

Rodrigo de Azevedo
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Dumb Mathematician
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    This is Cauchy Schwarz inequality on $\mathbb{R}^{d^2}$ – Matthew H. Jan 08 '22 at 20:56
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    How is the question in the block quote related to the one underneath? – user1551 Jan 08 '22 at 20:57
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    One way to prove it is $|AB|2^2 \leq |A|^2|B|_2^2 \leq |A|_2^2|B|_2^2$, where $|A| = \sup{|x| = 1}|Ax|$ is the operator norm of $A$. All of these follow easily once you realize that for any orthonormal basis $u_1, \dots, u_d$ of $\mathbb{R}^d$, $$|A|_2^2 = |Au_1|^2 + \dots + |Au_d|^2.$$ – Mason Jan 08 '22 at 21:37

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In my humble opinion brute-force calculation should suffice.

Denote $a_{ij}$ to be the element on the $i$-th row and $j$-th column of matrix $\mathbf{A}$; similar for $b_{ij}$. Then $$\mathbf{A B} = \begin{bmatrix} \sum a_{1j} b_{j1} & \sum a_{1j} b_{j2} & \cdots & \sum a_{1j} b_{jd} \\ \sum a_{2j} b_{j1} & \sum a_{2j} b_{j2} & \cdots & \sum a_{2j} b_{jd} \\ \vdots & \vdots & \ddots & \vdots \\ \sum a_{dj} b_{j1} & \sum a_{dj} b_{j2} & \cdots & \sum a_{dj} b_{jd} \end{bmatrix} \, . $$ Then $$ \begin{split} \left\Vert \mathbf{A B} \right\Vert_2^2 & = \sum_i \sum_k \left\vert \sum_j a_{ij} b_{jk} \right\vert ^ 2 \\ & \leq \sum_i \sum_k \left[ \left( \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_j \left\vert b_{jk} \right\vert ^ 2 \right) \right] \ (*)\\ & = \sum_i \sum_k \left[ \left( \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_{\color{red}{l}} \left\vert b_{\color{red}{l} k} \right\vert ^ 2 \right) \right] \\ & = \sum_i \sum_k \sum_j \sum_l \left\vert a_{ij} \right\vert ^ 2 \left\vert b_{lk} \right\vert ^ 2 \\ & = \left( \sum_i \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_k \sum_l \left\vert b_{lk} \right\vert ^ 2 \right) \\ & = \left\Vert \mathbf{A} \right\Vert_2^2 \left\Vert \mathbf{B} \right\Vert_2^2 \, . \end{split} $$ $\left(*\right)$ follows from Cauchy-Schawarz inequality. So there is $$ \left\Vert \mathbf{A B} \right\Vert_2 \leq \left\Vert \mathbf{A} \right\Vert_2 \left\Vert \mathbf{B} \right\Vert_2 \, . $$

To show that the equality sign can hold, just take this example: \begin{align} \mathbf{A} & = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \, , \quad \mathbf{B} = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix} \\ \mathbf{A B} & = \begin{bmatrix} 6 & 6 & 6 \\ 6 & 6 & 6 \\ 6 & 6 & 6 \end{bmatrix} \, . \end{align} Apparently $\left\Vert \mathbf{A} \right\Vert_2 = 3$, $\left\Vert \mathbf{B} \right\Vert_2 = 6$, $\left\Vert \mathbf{A B} \right\Vert_2 = 18 = \left\Vert \mathbf{A} \right\Vert_2 \left\Vert \mathbf{B} \right\Vert_2$.

user952196
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