Let $A$ be a $2\times 2$ squared matrix. Prove that max $tr\ (A)$ s.t. $AA^{T}=I$ has solution. 
Let a squared matrix $A=\bigl(\begin{smallmatrix}
x_{11} & x_{12} \\ 
x_{21} & x_{22}
\end{smallmatrix}\bigr)$. Prove that the following optimization problem:
max $tr\ (A)$  s.t. $AA^{T}=I$
Where $tr$ is the trace of matrix A, $A^{T}$ is the transpose of matrix A and $I$ is the identity matrix.

My attempt
It follows the Berge's Maximum Theorem. But I have to prove two things:
a.) $tr(A)$ is a continuous function. That is $tr(A)=x_{11}+x_{22}$ is continuous.
b.) $AA^{T}=I$ is a compact set. That is, the following set is compact:
$\{\begin{matrix}
x_{11}^2+x_{12}^2=1
\\ x_{21}^2+x_{22}^2=1
\\ x_{11}.x_{21}+x_{12}.x_{22}=0
\end{matrix} $
But I don't know what else to do. Any suggestion?

I might be missing something but the condition that $AA^T = I$ implies that $A$ is an orthogonal matrix. Thus, the columns (or rows) of $A$ form an orthonormal basis for $\mathbb{R}^2$ (I'm assuming this is over the reals). Thus $x_{ij} \leq 1$ for all $i,j$. 
Therefore, $tr(A) \leq 2$. Since $tr(I) = 2$ we have that the maximum is attained. 
