Find and prove an equation relating $[T^t]_{B^*}^{B^*}$ and $[(T^t)^t]_{B^{**}}^{B^{**}}$. Let $V$ be a finite dimensional vector space over the field $F$ and let $T:V \rightarrow V$ be  $F$-linear.  Suppose $B$ is an $F$-basis for $V$.  Find and prove an equation relating $[T^t]_{B^*}^{B^*}$ and $[(T^t)^t]_{B^{**}}^{B^{**}}$.
I am very confused how to even go about finding such an equation that relates these.  I believe that after finding the equation I will be able to prove that it is true.  Sorry if it seems like I haven't done any work on it, I have no idea how to even go about doing a problem like this.
 A: I don't get the point: why just simply ask for a relation between
$$
[T]^B_B \qquad \text{and} \qquad [T^t]^{B^*}_{B^*} \quad \text{?}
$$
(The one you're asking for would follow from that simpler one, wouldn't it?)
In that case, the relation is simple enough:
$$
[T^t]^{B^*}_{B^*} = \left( [T]^B_B  \right)^t
$$
Isn't it?
Remark. I'm assuming that your $T^t$ means the dual of $T$. That is, the linear map $T^t : V^* \longrightarrow V^*$ defined by $T^t\omega = \omega \circ T$. I also assume that your $B^*$ means the dual basis. That is, if $B = \left\{ u_1, \dots , u_n \right\}$ form a basis of $V$, its dual basis is $B^* = \left\{ u_1^*, \dots , u_n^* \right\} $, where the $u_i^* : V \longrightarrow F$ are the linear forms defined by $u_i^*(u_j) = 1 $ if $i =j$ and $0$ otherwise.
If my assumptions are true, I add another piece of standard notation: that $t$ exponent on the right means the transpose of the matrix.
So, in order to prove the stated relation, we can proceed as follows: let's call $A = [T]^B_B$ and display its entries
$$
A = 
\begin{pmatrix}
a^1_1   &  \dots & a^1_i  & \dots  & a^1_n  \\
\vdots  &        & \vdots &        & \vdots  \\
a^j_1   & \dots  & a^j_i  & \dots  & a^j_n  \\
\vdots  &        & \vdots &        & \vdots  \\
a^n_1   & \dots  & a^n_i  & \dots  & a^n_n
\end{pmatrix}
$$
This means that
$$
T(u_i) = a^1_i u_1 + \dots + a^j_iu_j + \dots + a^n_i u_n \ .
$$
And since we are saying that $[T^t]^{B^*}_{B^*} = A^t$, what we need to show is
$$
T^t(u^*_j) =  a^j_1 u^*_1 + \dots + a^j_iu^*_i + \dots + a^j_n u^*_n \ ,
$$
for all $j$. Right?
Ok, so now your turn: compute. On one hand, find out
$$
T^t(u^*_j) (u_i) 
$$
for every $i,j$. On the other hand, find out
$$
(  a^j_1 u^*_1 + \dots + a^j_iu^*_i + \dots + a^j_n u^*_n)(u_i) 
$$
too, for all $i,j$. Compare. Think.
