Htykuut

Let $Ain M_{n times n}(Bbb F)$, 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
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$