How to find adjoint?

Question

where

From this question I concluded that L^T = L*. But when I find the transpose and then adjoint of the matrix I get

as the adjoint L* of L. But it is wrong. What am I doing wrong?

Answer 1

First of all, you have incorrectly found the "transpose and then adjoint" of the matrix, under any reasonable interpretation of those words. It is difficult to know what you're doing wrong since it's not clear how you got that answer in the first place. Note that the answer to the question given is not a $2 \times 2$ matrix as you seem to believe; it is a formula for the function $L^*$. In this case, it happens to be a formula of the form $L^*(X) = BX$ for some matrix $B \in \Bbb R^{2 \times 2}$, though it is not generally true that linear maps over $\Bbb R^{2 \times 2}$ are of this form.

As for the correct approach, note that the definition of the adjoint states that for all $X,Y$, we have $$ \langle L(X),Y \rangle = \langle X, L^*(Y)\rangle. $$ The usual way to answer this kind of question is as follows. Using the properties of trace, rearrange the expression $\langle AX,Y \rangle$ into an expression of the form $\langle X, ?\rangle$ (where $A = \pmatrix{-4&-6\\-1&-8}$ and ? does not depend on $X$). What ever you end up with as "$?$" gives you a formula for $L^*(Y)$.