Specifically, the question is as follows:
Let $B$ be the standard basis of $\mathbb{R}^2$. Let $T\in\mathcal{L}(\mathbb{R}^2)$ be such that
$$\mathcal{M}(T,B)=\begin{bmatrix}2&3\\0&2\end{bmatrix}$$
Find the polar decomposition $T=S\sqrt{T^*T}$ of $T$. (Hint: The eigenvectors of $T^*T$ are $(1,2)$ and $(-2,1)$.)
It seems like this should be very straightforward. $T$ and $T^*$ are both invertible, so it should follow that $S=\sqrt{T(T^*)^{-1}}$, but this is false, as is easily checked (https://www.wolframalpha.com/input/?i=sqrt(%7B%7B2,3%7D,%7B0,2%7D%7D%7B%7B2,0%7D,%7B3,2%7D%7D%5E-1)sqrt(%7B%7B2,0%7D,%7B3,2%7D%7D%7B%7B2,3%7D,%7B0,2%7D%7D)).
I'm not quite sure why the eigenvectors have been given considering the eigenvalues haven't been given, but the eigenvalues of $T^*T$ are $1$ and $16$. My other thought was to use the singular value decomposition to find the polar decomposition, but I run into similar computational issues. These attempts make me think I'm missing something theoretically. This is a homework question, so I request assistance in finding my error/hints to get to the true solution, but not an actual solution.
Best Answer
I believe there's a more elegant solution, but this works:
Observe that $T^*T=\begin{bmatrix}4&6\\6&16\end{bmatrix}$ and $\sqrt{T^*T}=\begin{bmatrix}2&\sqrt{6}\\\sqrt{6}&\sqrt{13}\end{bmatrix}$. Also, as is easily verified with a computational algebra engine, $$\sqrt{T^*T}^{-1}=\begin{bmatrix}\frac{\sqrt{13}}{-6+2\sqrt{13}}&-\frac{\sqrt{6}}{-6+2\sqrt{13}}\\-\frac{\sqrt{6}}{-6+2\sqrt{13}}&\frac{2}{-6+2\sqrt{13}}\end{bmatrix}.$$ Similarly, $$S=T\sqrt{T^*T}^{-1}=\begin{bmatrix}-\frac{3\sqrt{6}-2\sqrt{13}}{2(-3+\sqrt{13})}&-\frac{-3+\sqrt{6}}{-3+\sqrt{13}}\\ -\frac{\sqrt{6}}{-3+\sqrt{13}}&\frac{2}{-3+\sqrt{13}}\end{bmatrix}.$$ It is easily verified that this is the desired $S$.