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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.

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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.

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    Yes. Indeed you have that equality and also $(B^*)^* = B$ and $(V^*)^* = V$. In fact, "equality" means "canonical isomorphism", to be precise. You can identify all those double duals with the original guys through the following isomorphism: $\phi : V \longrightarrow V^{**}$, $\phi (u) (\omega) = \omega (u)$.2012-10-18