Let $A(x,y) = \begin{bmatrix} 1 & -y\\ -x & 1 \end{bmatrix}$, where $x,y\gt0$. Find the spectral radius $\rho (x,y)$ of $A$ as a function of $x,y$.
What I did is I went ahead and found the characteristic polynomial of $A$, being: $$(1-\lambda)^2-xy$$ So in order for $\lambda$ to be an eigenvalue, it has to be a solution of $$(1-\lambda)^2-xy=0 \iff 1-\lambda = \sqrt{xy} \iff \lambda = 1-\sqrt{xy}$$
At that point, I'm unsure how to proceed? In order to find the spectral radius, I have to find the highest $\lambda$ (in absolute value) that satisfies the above equation. Any input is appreciated.
You missed out the negative root.
$(1-\lambda)^2=xy\implies1-\lambda=\pm\sqrt{xy}\implies\lambda=1\pm\sqrt{xy}$ and the spectral radius is $1+\sqrt{xy}$ since $|1-\sqrt{xy}|\le1+\sqrt{xy}$.