2

If I have a matrix A = $\alpha$I where I is the identity matrix and a is some non-zero real number.

How would I go about taking the induced matrix norm of it?

I'm leaning towards using the submultiplicativity property, but I'm just not really sure how to approach this one, new to norms ;p. thanks

nuni
  • 23

1 Answers1

0

I don't believe that the submultiplicativity is needed here. Indeed, an induced matrix norm is by definition, shaped as $$\||M\|| = \sup_{x\neq 0}\frac{\|Ax\|}{\|x\|}$$ where $M$ is squared matrix, i.e. mapping vectors from a vector space $V$ to vectors of $V$, and $\|\cdot\|$ is a norm on $V$. Now, by homogeneity of the norm $\|\cdot \|$ we have that $\|\alpha I x\|=|\alpha | \|Ix\|=|\alpha | \|x\|$, and therefore, if $A=\alpha I$, for every $x\neq 0, x \in V$, we have $$\frac{\|Ax\|}{\|x\|}=\frac{|\alpha| \|x\|}{\|x\|}=|\alpha|,$$ implying $\||A\||=|\alpha|$.

Surb
  • 55,662
  • But isn't the property

    ||Ax|| $\le$ ||A|| * ||x||, how can we assume that it will be equal?

     – nuni Oct 23 '23 at 20:56
  • $\sup_{x\neq 0}\frac{|Mx|}{|x|} = \inf{\lambda >0 \mid |Ax|\leq \lambda |x|, \forall x}$, anyway, as stated in my post, $|\alpha I x|= |\alpha| |IIx|=|\alpha| |x|$ for every $x$, so $|\alpha| |x|\leq |\alpha I x |\leq |\alpha| |x|$ for all $x$. – Surb Oct 23 '23 at 20:57
  • Starting from $|Ax|\leq |A| |x|$, if $x\neq 0$ then $|x|> 0$ so, we have $\frac{|Ax|}{|x|}\leq |A|$ for all $x\neq 0$. Taking the supremum on the latter inequality shows $\sup_{x\neq 0}\frac{|Ax|}{|x|}\leq |A|$. For the reverse inequality: since $x\mapsto |Ax|$ is a continuous function on the compact set ${x:|x|=1}$, it attains its maximum. This maximum is precisely $|A|$. – Surb Oct 23 '23 at 21:02