I found an elementary proof of a problem I saw on mathoverflow. Mateusz Wasilewski gave a nice proof, but I was hoping to find one that was solely linear algebraic. The problem states that if $\pi, \sigma: G \rightarrow U_n(\mathbb{C})$ are two equivalent unitary representations of a finite group, then $\pi$ and $\sigma$ are unitarily equivalent.

The proof goes as follows. If $\pi$ and $\sigma$ are equivalent, then there exists $T \in GL_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for every $g \in G$. Let $T = W \Sigma V^*$ be the singular value decomposition of $T$, so that $T = (W V^*) (V \Sigma V^*)$ is the polar decomposition of $T$. We will show that $A = V \Sigma V^* \pi(g)$ is normal, that is $A^* A = A A^*$. $$\begin{align*}

A A^* &= (V \Sigma V^* \pi(g)) (\pi(g)^* V \Sigma^* V^*) \\

&= \sum |\lambda_i|^2 \\

A^* A &= (\pi(g)^* V \Sigma^* V^*)(V \Sigma V^* \pi(g)) \\

&= \sum |\lambda_i|^2,

\end{align*} $$ where $\lambda_i$ are the diagonal entries of $\Sigma$.

The components of the polar decomposition of a normal matrix commute, so $(V \Sigma V^*) \pi(g) = \pi(g) (V \Sigma V^*)$. Then $$\begin{align*}

T \pi(g) &= \sigma(g) T \\

(W V^*) (V \Sigma V^*) \pi(g) &= \sigma(g) (W V^*) (V \Sigma V^*) \\

(W V^*) \pi(g) (V \Sigma V^*) &= \sigma(g) (W V^*) (V \Sigma V^*) \\

(W V^*) \pi(g) &= \sigma(g) (W V^*).

\end{align*}$$ Since $WV^*$ is unitary by construction, we have completed the proof.

The proof goes as follows. If $\pi$ and $\sigma$ are equivalent, then there exists $T \in GL_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for every $g \in G$. Let $T = W \Sigma V^*$ be the singular value decomposition of $T$, so that $T = (W V^*) (V \Sigma V^*)$ is the polar decomposition of $T$. We will show that $A = V \Sigma V^* \pi(g)$ is normal, that is $A^* A = A A^*$. $$\begin{align*}

A A^* &= (V \Sigma V^* \pi(g)) (\pi(g)^* V \Sigma^* V^*) \\

&= \sum |\lambda_i|^2 \\

A^* A &= (\pi(g)^* V \Sigma^* V^*)(V \Sigma V^* \pi(g)) \\

&= \sum |\lambda_i|^2,

\end{align*} $$ where $\lambda_i$ are the diagonal entries of $\Sigma$.

The components of the polar decomposition of a normal matrix commute, so $(V \Sigma V^*) \pi(g) = \pi(g) (V \Sigma V^*)$. Then $$\begin{align*}

T \pi(g) &= \sigma(g) T \\

(W V^*) (V \Sigma V^*) \pi(g) &= \sigma(g) (W V^*) (V \Sigma V^*) \\

(W V^*) \pi(g) (V \Sigma V^*) &= \sigma(g) (W V^*) (V \Sigma V^*) \\

(W V^*) \pi(g) &= \sigma(g) (W V^*).

\end{align*}$$ Since $WV^*$ is unitary by construction, we have completed the proof.