I have an exam coming up and I have a whole bunch of questions to study from, and I am realizing that I really don't have a good grasp on basis matrices for linear transformations. One of the questions is:

For a linear map $f$ in $Hom_F( V,W)$ the (operator) adjoint $f^*:W^*\rightarrow V^*$ is defined by $f^*(g)=g\circ f$. If $f$ has matrix $A$ in some basis, what is the matrix of $f^*$ in the dual basis?

Similarly, here is another question that Im struggling a bit with that would help if I could get some intuition on...

A linear map between finite dimensional vector spaces $f:V\rightarrow W$ corresponds to a matrix when bases for $V$ and $W$ are chosen. HOw does this matrix change if we change bases in $V$ and $W$.

Let $(w_j)_{j=0}^{m-1},(g_j)_{j=0}^{m-1}$ be a basis-dual basis pair of $W$ and $W^*$. Let $(v_k)_{k=0}^{n-1},(f_k)_{k=0}^{n-1}$ be a basis-dual basis pair of $V$ and $V^*$. Suppose that $\Phi \in \operatorname{Hom}(V,W)$. Let $\mathbf M$ be the matrix of $\Phi$ w.r.t. $(w_j)_{j=0}^{m-1},(v_k)_{k=0}^{n-1}$ and $\mathbf N$ be the matrix of $\Phi^*$. By definition, $$\Phi^* g_j = \sum_{r=0}^{n-1} \mathit N_{j,r} f_r$$ Hence $$(\Phi^* g_j)(v_k) = \sum_{r=0}^{n-1} \mathit N_{j,r} f_r(v_k) = \mathit N_{j,k}$$ On the other hand,
$$(\Phi^* g_j)(v_k) = g_j(\Phi v_k) = g_j \left(\sum_{r=0}^{m-1} \mathit{M}_{k,r} w_r\right) = \mathit{M}_{k,j}$$ Hence $\mathbf M = \mathbf N^\intercal$.
• First question: why are your indices going to $(m-1)$ and $(n-1)$ ? And thank you for your quick response!! – jgcello Dec 18 '16 at 16:46
• @JennettaAlden $m$ and $n$ are the dimensions of $W$ and $V$, respectively. My indices are in the range $[0\ldotp\ldotp m - 1]$ and $[0\ldotp\ldotp n - 1]$ due to convention. – Henricus V. Dec 18 '16 at 16:47
• Great and one last question... I see how your algebra shows that $M=N^T$ But I dont really understand why. Is there anything you can say to help me see what is actually happening here? – jgcello Dec 18 '16 at 16:58
• @JennettaAlden I invoked these two facts: 1. $\Phi v_k = \sum_{j=0}^{m-1} \mathit M_{j,k} w_j$. 2. For any basis-dual basis pair $(v_k)_{k=0}^{n-1},(f_k)_{k=0}^{n-1}$, $f_k(v_j) = 1$ if and only if $j = k$, and $f_k(v_j) = 0$ otherwise. This allows the cancellation of many terms in the matrix equations. – Henricus V. Dec 18 '16 at 17:00