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I am not sure how to even start this proof. I know that for any induced matrix norm, $||Ax||$ is less than or equal to $||Ax||$ $||x||$.

IamKnull
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Vishnu P
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  • Do you know the definition of Matrix norm? Based on the description you give, I'll help you to prove. – IamKnull Sep 08 '19 at 17:02
  • The definition is a set of 4 properties. ||A|| >= 0, ||aA|| = |a|||A||, ||A+B|| <= ||A|| + ||B||, ||AB|| <= ||A||||B|| – Vishnu P Sep 08 '19 at 17:08
  • There are equivalent definitions for this, that's why I asked what definition do you know? If you want I can give you a hint based on this definition. – IamKnull Sep 08 '19 at 17:13

2 Answers2

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If $\Vert A\Vert$ is the norm of the matrix induced by the norm $\Vert \cdot \Vert$, it means that

$$\Vert A \Vert =\sup \{\Vert Ax \Vert \mid \Vert x \Vert =1\}$$

If $A$ is the identity matrix, then $Ix=x $ for all $x \in V$ the associated vector space. Hence $\Vert Ix \Vert =1$ for all $x $ such that $\Vert x \Vert =1$.

Therefore $\Vert I \Vert =1$.

mathcounterexamples.net
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If we take the usual definition of norm, $\|A\| = sup \{\|Ax\|:\|x\| = 1\}$

By definition of the induced norm of $A$, we have $\|A\| \leq \|Ax\|$. It follows that $\|I\| \leq \|Ix\|= \|x\| = 1$ for all $x$ such that $\|x\| = 1$

And as you say, $\|Ix\| \leq \|I\| \|x\|$, then if we take some $x\not=0$, we have $\frac{\|Ix\|}{\|x\|} \leq \|I\|$, thus $1 = \frac{\|x\|}{\|x\|} = \frac{\|Ix\|}{\|x\|}\leq \|I\|$

Then $1 \leq \|I\| \leq 1 \implies \|I\| = 1$

Anton Vrdoljak
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ZAF
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