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?

  • $\begingroup$ What is your definition of matrix-norm? It isn't true for any arbitrary norm on some space space of matrices. $\endgroup$ – tharris Apr 17 '13 at 22:44
  • $\begingroup$ Not only it must have norm $1$. It has norm $1$. $\endgroup$ – Mariano Suárez-Álvarez Apr 17 '13 at 22:46
  • $\begingroup$ yes it has norm 1. $\endgroup$ – user67411 Apr 17 '13 at 22:47
  • $\begingroup$ 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). $\endgroup$ – vonbrand Apr 17 '13 at 23:20
  • 1
    $\begingroup$ @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? $\endgroup$ – tharris Apr 17 '13 at 23:59

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 $\|I\| = \pi$.

  • 2
    $\begingroup$ I think matrix norms have to at least be sub-multiplicative $|AB| \leq |A||B|$ $\endgroup$ – tharris Apr 17 '13 at 22:58
  • $\begingroup$ the text i am reading says it is 1! $\endgroup$ – user67411 Apr 17 '13 at 22:59
  • $\begingroup$ i was wondering why i cannot find an further information on this... so confused been bugging me for a few days now $\endgroup$ – user67411 Apr 17 '13 at 23:00
  • $\begingroup$ @TomHarris But the norm $\pi|\cdot|$ is submultiplicative whenever $|\cdot|$ is submultiplicative? $\endgroup$ – Lord Soth Apr 17 '13 at 23:11
  • $\begingroup$ @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. $\endgroup$ – tharris Apr 17 '13 at 23:45

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, by submultiplicativity of matrix norm, we always have $\|I\|=\|I^2\|\le\|I\|^2$. It follows that $\|I\|$ is always bounded below by $1$.

If your textbook is not erred, it is perhaps talking about an induced matrix norm, i.e. one defined by $\|A\|=\sup_{x\neq0}\frac{\|Ax\|}{\|x\|}$ for some vector norm $\|x\|$ defined on $K^n$. In this case, it is straight from the definition that $\|I\|=1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.