My try :
we know that $||I||\geq1$ and by submultiplicativity of matrix norm $||I||=||I^2||\leq\|I\|^2 \implies ||I|| \leq 1$ so $||I|| = 1$
my problem is that I don't know if I used any propreties of an operator norm also I don't know what 'operator norm' means I tried understanding by reading the wikipedia link but it is too abstract for me.
I'm really confused please help me.
Given a norm on the vector space $V,$ $\|\cdot\|_{V},$ the operator norm is defined on the vector space of linear operators from $V\rightarrow V,$ and is defined by $$\|A\|=\sup_{x\neq0}\frac{\|Ax\|_{V}}{\|x\|_{V}}.$$ Then using the definition, we have that $$\|I\|=\sup_{x\neq0}\frac{\|Ix\|_{V}}{\|x\|_{V}}=\sup_{x\neq0} 1=1.$$
In fact, the proof you gave is incorrect. We may consider the norm on matrices (which is not an operator norm) given by $\|A\|=(\mathrm{tr}(A^{T}A))^{1/2}.$ Then the identity matrix of order $n$ will have norm $\mathrm{tr}(I^{2})=\mathrm{tr}(I)=n,$ which is not $1$ as soon as $n>1.$
