Let $A$ be a $2\times 2$ squared matrix. Prove that max $tr\ (A)$ s.t. $AA^{T}=I$ has solution.

Question

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?

Answer 1

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.

Answer 2

One way is to note that $\|Ax\|_2^2 = \langle Ax, Ax \rangle = \langle x, A^TAx \rangle = \|x\|^2$, and so $\|A\|_2 =1$.

Alternatively, one could use the Frobenius norm to get $\|A\|_F^2 = \operatorname{tr} (A^T A) = \operatorname{tr} I = n$. Hence $\|A\|_F = \sqrt{n}$ and so the set $F=\{ A | g(A) = I \}$ is bounded.

Since the function $g(A) = A^T A -I$ is continuous, and $F$ is bounded, we see that $F$ is closed and bounded, hence compact.

The function $f(A) = \operatorname{tr}A$ is linear, hence continuous.

Hence $\max \{ f(A) | g(A) = 0 \}$ has a solution.

Since all eigenvalues of $A$ are bounded by $1$ (since $\|A\|_2 = 1$) we have $f(A) \le n$, and since $I \in F$ and $f(I) = n$ we see that the $\max$ is $n$.