Show that the given two linear mappings commute

Question

Given $A\in M_n(\mathbb C)$, show that the mappings

$$\alpha_A(B) = \frac{1}{2}(AB + BA^{*})$$ $$\beta_A(B) = \frac{1}{2i}(AB - BA^{*})$$

define $\mathbb R$-linear maps $HERM_n(\mathbb C) → HERM_n(\mathbb C)$.
Also show that $\alpha_A, \beta_A$ commute with each other.

What I did -

1. First I proved that $\alpha_A$ and $\alpha_B$ are Hermitian.
2. Then I tried to prove that $\alpha_A*\alpha_B=\alpha_B*\alpha_A$.
3. The final expression was $AB^2A^*=BA^*AB$ but I can't see how they are equal.

.

Answer 1

You have misunderstood the meanings of $\alpha_A\beta_A$ and $\beta_A\alpha_A$.

$\alpha_A$ and $\beta_A$ are linear maps. The product $\alpha_A\beta_A$ means the function composition $\alpha_A\circ\beta_A$, i.e. $f=\alpha_A\circ\beta_A$ is the linear map defined by $f(B)=\alpha_A\left(\beta_A(B)\right)$.

When you are asked to prove that $\alpha_A$ commutes with $\beta_A$, what you need to prove is not that $$ \alpha_A(B)\beta_A(B)=\beta_A(B)\alpha_A(B)\quad\forall B, $$ but that $\alpha_A\circ\beta_A=\beta_A\circ\alpha_A$, i.e. $$ \alpha_A\left(\beta_A(B)\right)=\beta_A\left(\alpha_A(B)\right)\quad\forall B. $$

Answer 2

$$\alpha_A (\beta_A(B)) = \alpha_A(\frac{1}{2i}(AB-BA^*)) = \frac{1}{4i} (A(AB-BA^*) + (AB-BA^*)A^*)$$ $$\beta_A(\alpha_A(B)) = \beta_A(\frac{1}{2}(AB+BA^*)) = \frac{1}{4i}(A(AB+BA^*)-(AB+BA^*)A^*)$$ Distribute the expressions on the right-hand side to see that they are the same.