Show that the identity matrix $I$ must have norm $1$. 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?
 A: 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 $\|I\|=1$.
A: 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 $\|I\| = \pi$.
