Prove that $[T]_{B_1}^{B_1}$ = $[I]_{B_1}^{B_2}$ Let $B_1 = \{u_1,\dots,u_n\}$ and $B_2 = \{v_1,\dots,v_n\}$ be ordered bases of the vector space $V$. Let $T: V \rightarrow V$ be the linear operator defined by $Tu_1 = v_1, Tu_2 = v_2, \dots,  T_{u_{n}} = v_n$. Prove that $[T]_{B_1}^{B_1} = [I]_{B_1}^{B_2}$. Hint: Compare the matrices column by column
So this is the problem basically, I really got no clue, I tried a theorem I knew which was
$[T]_{B_1}^{B_1} = [I]_{B_1}^{B_2}[T]_{B_2}^{B_2} [I]_{B_2}^{B_1}$, except I tried manipulating it and I didn't find the solution. I tried to do it based on the hint, I really couldn't get far since I didn't really understand 100%, for me this is $[I]_{B_1}^{B_2}$
$[I]_{B_1}^{B_2}$" />
but I got no clue the matrix of $[T]_{B_1}^{B_1}$ and even when I got it how do I compare it, sorry I'm kinda lost on this problem, I'm usually not this lost, linear algebra is a tuff one for me
 A: By definition, if $A = \{a_1, a_2,\dots,a_m\}$ and $B = \{b_1,b_2,\dots,b_n\}$ then
$$[T]^A_B = \left[\begin{array}{c|c|c|c}  \\ [Ta_1]_B & [Ta_2]_B &\cdots&[Ta_m]_B \\ \\ \end{array}\right].$$
Or maybe you have defined $[T]^A_B$ as the matrix such that $[T]^A_B [x]_A = [Tx]_B$, in which case the above is a theorem.
Anyways, you don't have to think, just calmly replace $A, B, T, m, n$ with what is given.
\begin{align*}
[I]^{B_2}_{B_1} &= \left[\begin{array}{c|c|c|c}  \\ [Iv_1]_{B_1} & [Iv_2]_{B_1} &\cdots&[Iv_n]_{B_1} \\ \\ \end{array}\right] \\
&= \left[\begin{array}{c|c|c|c}  \\ [v_1]_{B_1} & [v_2]_{B_1} &\cdots&[v_n]_{B_1} \\ \\ \end{array}\right]. \\
[T]_{B_1}^{B_1} &= \left[\begin{array}{c|c|c|c}  \\ [Tu_1]_{B_1} & [Tu_2]_{B_1} &\cdots&[Tu_n]_{B_1} \\ \\ \end{array}\right] \\
&= \left[\begin{array}{c|c|c|c}  \\ [v_1]_{B_1} & [v_2]_{B_1} &\cdots&[v_n]_{B_1} \\ \\ \end{array}\right].
\end{align*}
Or, you can do this using the $[T]^A_B [x]_A = [Tx]_B$ definition:
\begin{align*}
[T]^{B_1}_{B_1} [u_i]_{B_1} &= [Tu_i]_{B_1} = [v_i]_{B_1}, \\
[I]^{B_2}_{B_1} [v_i]_{B_2} &= [Iv_i]_{B_1} = [v_i]_{B_1}.
\end{align*}
Note that $[u_i]_{B_1} = [v_i]_{B_2} = e_i$, the $i$-th column of the identity matrix. So the two matrices $[T]^{B_1}_{B_1}$ and $[I]^{B_2}_{B_1}$ agree on a basis (the standard basis) and hence they must be equal.
