I am trying to understand why the identity matrix $I$ must have a norm $1$, for any choice of matrix-norm $|\cdot|$? How would i show this?
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What is your definition of matrix-norm? It isn't true for any arbitrary norm on some space space of matrices. – tharris Apr 17 '13 at 22:44
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Not only it must have norm $1$. It has norm $1$. – Mariano Suárez-Álvarez Apr 17 '13 at 22:46
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yes it has norm 1. – user67411 Apr 17 '13 at 22:47
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If your norm satifies $N(a \cdot b) = N(a) N(b)$, then $N(I) = N(I \cdot I) = N(I)^2$, so $N(I) = 1$ (as it can't be 0). – vonbrand Apr 17 '13 at 23:20
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@vonbrand It doesn't have to satisfy that property, only $N(ab) \leq N(a)N(b)$. – tharris Apr 17 '13 at 23:25
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1@user67411 Can you give us a link to the text you are using / state in full the definition of "matrix norm" that the text gives? – tharris Apr 17 '13 at 23:59
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its on a question sheet from uni for a badly organised new course! maybe just a bad question, no extra conditions apart from it is a matrix norm on K^(n*n) – user67411 Apr 18 '13 at 00:08
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Unless I am missing something, this is not necessarily the case. Take your favorite matrix norm $|\cdot|$ (suppose that it gives $|I| = 1$), and define a new norm $\|\cdot\| = \pi|\cdot|$. Then $\|\cdot\|$ is a matrix norm with $\|I\| = \pi$.
$\textbf{Edit:}$ If you are including the submultiplicativeness in the definition of your matrix norm (The definition of a matrix norm does not need to include the submultiplicative property), again pick your now favorite submultiplicative matrix norm $|\cdot|$ with $|I| = 1$. Then, $\|\cdot\| = \pi|\cdot|$ is a submultiplicative matrix norm ($\|AB\| = \pi|AB| \leq \pi |A||B| \leq \pi |A| \pi|B| = \|A\|\|B\|$) with $\|I\| = \pi$.
Lord Soth
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2I think matrix norms have to at least be sub-multiplicative $|AB| \leq |A||B|$ – tharris Apr 17 '13 at 22:58
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i was wondering why i cannot find an further information on this... so confused been bugging me for a few days now – user67411 Apr 17 '13 at 23:00
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@TomHarris But the norm $\pi|\cdot|$ is submultiplicative whenever $|\cdot|$ is submultiplicative? – Lord Soth Apr 17 '13 at 23:11
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@LordSoth Yeah, sorry I thought you were using $\pi$ for an arbitrary constant (no idea why, it's the least arbitrary greek letter!). I was trying to make the point that we at least require $|I| \geq 1$. I think the OP must have some extra conditions in their definition of matrix norm that they're not telling us. – tharris Apr 17 '13 at 23:45
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As many others have pointed out, the matrix norm of $I$ is not necessarily equal to $1$. In fact, if $\|\cdot\|$ is a matrix norm $c\|\cdot\|$ is also a matrix norm for any $c\ge1$.
Yet, if your matrix norm is submultiplicative, we have $0<\|I\|=\|I^2\|\le\|I\|^2$ and hence $\|I\|\ge1$, but strict inequality may still occur. E.g. the Frobenius norm $\|\cdot\|_F$ is submultiplicative, but $\|I_n\|_F=\sqrt{n}>1$ when $n>1$.
If your textbook is not erred, it is perhaps talking about an induced matrix norm, i.e. one defined by $\|A\|=\sup_{x\neq0}\frac{\color{red}{\|}Ax\color{red}{\|}}{\color{red}{\|}x\color{red}{\|}}$ for some vector norm $\color{red}{\|\cdot\|}$ defined on $K^n$. In this case, it is straight from the definition that $\|I\|=1$.
user1551
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