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Why does for every matrix norm $\lVert \mathbb{\cdot }\lVert $

$$\lVert \mathbb{I } \rVert \geq 1$$ hold (where $\mathbb{I }$ is the identity matrix)? I tried to prove it just by the definitions of a matrix norm but I didn't succeed. Can anybody help me?

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    $\begingroup$ $\|A B\|\le \|A\|\ \|B\|$? $\endgroup$
    – user99914
    Oct 11, 2015 at 23:23
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    $\begingroup$ I was going to say, they probably assume only sub-multiplicative norms. $\endgroup$
    – Brian Tung
    Oct 11, 2015 at 23:27
  • $\begingroup$ $A = \mathbb{I}$ so, $\lVert \mathbb{I} \rVert \lVert B \rVert \geq \lVert \mathbb{I } B \rVert$. What is B? $\endgroup$
    – Breaking M
    Oct 11, 2015 at 23:34
  • $\begingroup$ $B = A$... ${}{}{}$ $\endgroup$
    – user99914
    Oct 12, 2015 at 2:17

1 Answer 1

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Matrix norms are usually defined to be submultiplicative (see for example here), that is for any two matrices $A, B$ such that $AB$ exists, we have $$ \def\norm#1{\left\|#1\right\|}\norm{AB} \le \norm A \norm B\tag{$*$} $$ Now, let's look at $\mathrm{id}$, as $\mathrm{id} \ne 0$, we have $\norm{\mathrm{id}} >0$ and, as $\mathrm{id}^2 = \mathrm{id}$ by ($*$), $$ \norm{\mathrm{id}} = \norm{\mathrm{id}^2} \le \norm{\mathrm{id}}^2 $$ Dividing by $\norm{\mathrm{id}}\ne 0$, gives $1 \le \norm{\mathrm{id}}$.

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