Compute the matrix norm of $A$= $\begin{pmatrix}2 &2\\ -1& 2\\\end{pmatrix}$
My attempt
I found $A^T$ = $\begin{pmatrix}2 & -1\\ 2& 2\\\end{pmatrix}$ and $A^TA$= $\begin{pmatrix}5 & 2\\ 2& 8\\\end{pmatrix}$.
Then det($A^TA$ - $\lambda$$I_2$) of $A^TA$.
From this I got $\lambda^2$- 13$\lambda$ +36.
I put this into the quadratic formula and got the roots $\frac{13+\sqrt25}{2}$. Is this my answer? Or is my answer $\sqrt9$?
Any help is appreciated!
You take the square root, not of the trace of the product, but of its spectral radius which is the largest absolute value of any eigenvalue. Here, we see that the eigenvalues solve the quadratic equation
$\lambda^2-13\lambda+36=0.$
If we try different rational roots that divide $36$ and must be positive by Descartes'Rule of Signs, we find roots of $4$ and $9$ (thus $(\lambda-4)(\lambda-9)=0$). So the largest absolute value of any eigenvalue of the product is $9$, and the square root is $3$ which is the (Euclidean) norm of the original matrix.