Let $A\in M_{n \times n}(\Bbb F)$, and let $\gamma$ be the basis for $\Bbb F^n$. Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.
I start by applying the theorem: $[L_A]_γ = Q^{-1}[L_A]_{α} Q$
And let $[L_A]_α=A$, $α$ is another unknown basis of $\Bbb F^n$.
Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n \times n$ matrix.
But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.
Hope someone can give me some hints. Thanks a lot.
Hint: Let $\alpha$ denote the standard basis (that is, $e_1 = (1,0,\dots,0)$ is the first vector, $e_2 = (0,1,0,\dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $\gamma$, we have $$ Q = [I_n]_{\gamma \to \alpha} $$ We can that this is the case as follows: if $v_1,\dots,v_n$ are the elements of $\gamma$, then we note that the $i$th column of $[I_n]_{\gamma \to \alpha}$ is given by $$ [I_n]_{\gamma \to \alpha} e_i = [I_n]_{\gamma \to \alpha} [v_i]_{\gamma} = [I_n v_i]_{\alpha} = [v_i]_{\alpha} = v_i $$