Let $A=\begin{pmatrix}4 & 0 \\1 & \frac{1}{4}\end{pmatrix}$. What is the supremum norm $\|A^2\|$? Recall that $\|A\|:=\sup\{\frac{|Av|}{|v|}: v\neq 0\}$.
Attempt: $A^2=\begin{pmatrix}4^2 & 0 \\4+\frac{1}{4} & \frac{1}{4^2}\end{pmatrix}$. I thought I should calculate the first singular value, which is the equal of the norm, to find the value of the norm. I was wondering if there was the simple way as I want to calculate the higher power, e.g., $A^{15}$.
In general, I want to know whether this is true for $SL(2)$ matrices, one should look at the diagonal entries or not to calculate the norm.