Finding If $u\in \operatorname{Im}(T)$

Question

Let $V\in\mathbb{C}^{2\times 2}$ inner product space defined by $\langle A,B \rangle= \operatorname{Tr}(AB^*)$.

Let $T:V\to V$ linear operator be defined $T(X)=AX-XA$ where $A=\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$

Does $Y=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ belong to $ \operatorname{Im}(T)$ if so, find pre-image for $Y$ If not find the matrix in $ \operatorname{Im}(T)$ that is the closest to $Y$.

It is a my first time encoring this question, any idea where to start?

Answer 1

The identity matrix $Y$ cannot belong to $\operatorname{Im} T$ because $\operatorname{Tr} (AB - BA) = 0$ for all $X \in \mathbb{C}^{2\times 2}$.

To find the closest approximation to $Y$ within $\operatorname{Im} T$, notice that

\begin{align} T\pmatrix{x_{11} & x_{12} \\ x_{21} & x_{22}} &= \pmatrix{-ix_{12} +ix_{21} & -ix_{11} +ix_{22} \\ ix_{11} -ix_{22} & ix_{12} -ix_{21}} \\ &= (x_{12} - x_{21}) \pmatrix{-i & 0 \\ 0 & i} + (x_{11} - x_{22}) \pmatrix{0 & -i \\ i & 0} \end{align}

so $\left\{\frac1{\sqrt2}\pmatrix{-i & 0 \\ 0 & i}, \frac1{\sqrt2}\pmatrix{0 & -i \\ i & 0}\right\}$ is an orthonormal basis for $\operatorname{Im} T$.

Now construct the orthogonal projection onto $\operatorname{Im} T$.