Prove that $L_A(X)=AX-XA$ is symmetric if and only if $A$ is symmetric.

Question

Define for a fixed $A \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ the mapping:

$$L_A : \mathbb{M}^{2 \times 2}(\mathbb{R}) \to \mathbb{M}^{2 \times 2}(\mathbb{R}) : X \mapsto AX-XA. $$

Define on $\mathbb{M}^{2 \times 2}(\mathbb{R})$ the dotproduct $\langle \cdot , \cdot \rangle$ as follows: $ \langle X, Y \rangle = [X]_{\xi}^t [Y]_{\xi}$.

Here is $[\cdot ]_\xi$ the coordinate map that belongs to the standard basis

$$\xi = \{ E_1 = {\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}}, E_2={\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}}, E_3={\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}}, E_4={\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}} \}$$ of $\mathbb{M}^{2 \times 2}(\mathbb{R})$.

Write $M_A$ for the matrix such that for all $X \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ it satisfies $[L_A (X)]_\xi=M_A [X]_\xi$.

Prove that $L_A$ is symmetric if and only if $A$ is symmetric.

In a previous exercise I had to determine all the matrices $M_{E_1}, M_{E_2}, M_{E_3}, M_{E_4}$. And this is the result:

$M_{E_1} = {\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}},~M_{E_2}={\begin{pmatrix} 0 & 0 & -1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}}, ~M_{E_3}={\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}}, \\M_{E_4}={\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}}$

How am I to use this to prove the given problem? I already showed that $L_A$ is symmetric if and only if $M_A$ is symmetric, meaning $\langle L_A(X), Y \rangle = \langle X, L_A(Y) \rangle$ if and only if $M_A = M_A^t$.

I am at a loss. Thanks in advance.

Answer 1

Let $A=\pmatrix{a&b\\ c&d}$. Then $M_A=aM_{E_1}+cM_{E_2}+bM_{E_3}+dM_{E_4}$. Here are some entries of $M_A$: $$ M_A=\pmatrix{ 0 &b &-c &0\\ c &d-a &0 &-c\\ -b &0 &\ast &\ast\\ 0 &-b &\ast &\ast}.\tag{1} $$ And I will leave the rest to you. Now, the given inner product $\langle X,Y\rangle$ is the usual dot product of $[X]_{\xi}$ and $[Y]_{\xi}$ in $\mathbb{R}^4$. Therefore, $L_A$ is symmetric w.r.t. $\langle\cdot,\cdot\rangle$ if and only if $M_A$ is a symmetric matrix. More specifically, \begin{align*} L_A \text{ is symmetric } &\Leftrightarrow \langle L_A(X),Y\rangle=\langle X,L_A(Y)\rangle \text{ for all } X,Y\in M^{2\times2}(\mathbb{R})\\ &\Leftrightarrow [X]_{\xi}^tM_A^t [Y]_{\xi}=[X]_{\xi}^t M_A[Y]_{\xi} \text{ for all } X,Y\in M^{2\times2}(\mathbb{R})\\ &\Leftrightarrow u^tM_A^tv=u^t M_Av \text{ for all } u,v\in \mathbb{R}^4\\ &\Leftrightarrow M_A^t=M_A. \end{align*}

From $(1)$, we see that for $M_A$ to be symmetric, a necessary condition is $b=c$. If you could determine the unspecified entries of $M_A$ correctly (note that your $M_{E_3}$ is wrong; there's perhaps a typo), you will see that $b=c$ is also a sufficient condition. Hence the result.

Answer 2

Note: I am sorry this is a little far from your notations. But lots of people would define this inner product as $\mbox{Tr}(X^*Y)$. This idea of using the trace to create bilinear objects is frequent and natural (see the Killing form, to mention one famous example). Last but not least, observe that the argument below applies to $M_n(\mathbb{R})$ and even $M_n(\mathbb{C})$ for any $n$. This is my main motivation for doing it the following way. If you are only interested in the $2\times 2$ case, then you should keep your notations and do like @user1551 did.

Observe that the inner product you are given on the matrices has a nice and tractable matrix trace expression:

$$ (X,Y)=\mbox{Tr}(X^*Y) $$

where $X^*$ denotes the adjoint of $X$, i.e. the transpose in the real case.

Now, using the commutativity of the trace $\mbox{Tr}(MN)=\mbox{Tr}(NM)$, straightforward computations show that $$ (L_A(X),Y)=\mbox{Tr}(A^*(YX^*-X^*Y)) $$ while $$ (X,L_A(Y))=\mbox{Tr}(A(YX^*-X^*Y)). $$

Easy direction: it follows at once that if $A=A^*$, then $(L_A(X),Y)=(X,L_A(Y))$ for every $X,Y$. Hence $L_A=L_A^*$, i.e. $L_A$ is symmetric.

Conversely: assume that $L_A=L_A^*$. Then the matrix $B=A-A^*$ satisfies $$\mbox{Tr}(B(YX^*-X^*Y))=\mbox{Tr}((BY-YB)X^*) =0$$ for every $X,Y$. First, this implies, with $X=BY-YB$, $$ \|BY-YB\|^2=\mbox{Tr}((BY-YB)^*(BY-YB))=\mbox{Tr}((BY-YB)(BY-YB)^*) =0. $$ Where the norm is the one induced by your inner product. It follows that $BY-YB=0$ for every $Y$, i.e. $B$ commutes with every other matrix. It is well-know that the only such matrices are the scalar matrices (nice exercise, just compute the products with matrices of the canonical basis $E_{ij}$). So $B=A-A^*=\lambda I_2$. But taking the trace of the latter, we find $2\lambda=\mbox{Tr}(A)-\mbox{Tr}(A^*)=0$. Finally, $B=A-A^*=0$, so $A=A^*$ is symmetric.

Answer 3

"Another" way is the following:

Show that $\langle L_{E_2}(X),Y\rangle=\langle X,L_{E_3}(Y)\rangle$ and that $\langle L_{E_3}(X),Y\rangle=\langle X,L_{E_2}(Y)\rangle$ (actually these two are equivalent).
Therefore for $A=\begin{pmatrix} a_1&a_3\\a_2&a_4\end{pmatrix}$ we have (note that $M_{E_1},M_{E_4}$ are symmetric) $$ \langle L_A(X),Y\rangle=\sum_{i=1}^4a_i\langle L_{E_i}(X),Y\rangle=\\ a_1\langle X,L_{E_1}(Y)\rangle+a_2\langle X,L_{E_3}(Y)\rangle+a_3\langle X,L_{E_2}(Y)\rangle+a_4\langle X,L_{E_4}(Y)\rangle=\\ \langle X,L_{A^t}(Y)\rangle. $$ It follows that $L_A$ is symmetric $\iff \langle X,L_{A}(Y)\rangle=\langle X,L_{A^t}(Y)\rangle, \ \forall X,Y\in M^{2\times2}(\mathbb R)\ldots$