Let $f:V \times W \to \mathbb F$ and $s:V \to W^*, r:W \to V^*$. Find $[s]^B_{C^*}, [r]^C_{B^*}$.

Question

Let $f:V \times W \to \mathbb F$ bilinear map and $s:V \to W^*, r:W \to V^*$ linear transformations s.t: $s(v)(w) = f(v,w)$ and $r(w)(v) = f(v,w), \forall v \in V$ and $\forall w \in W$. Let $B$ an ordered basis of $V$ and $C$ an ordered basis of $W$. Let $A$ be the transformation matrix of $f$ in the bases $B,C$ whilst $B^*,C^*$ are the dual bases for $V^*,W^*$ respectively. Find: $[s]^B_{C^*}, [r]^C_{B^*}$

So we haven't really seen dual spaces in Linear Algebra #1 and so this question really got me confused.

Basically I know that $V^*$ for example is $Hom(V,\mathbb F)$ and given some basis $B$ of $V$ we have $B^*$ a basis for $V^*$ where $b_i \in B^*$ is the linear functional $\phi_i$ which satsfies:

$\phi_i = \begin{cases} 1 & i = j \\[2ex] 0 & i \neq j \end{cases}$

So past this I don't even know where to start.

Answer 1

To find a matrix means to find all its entries.

In general, the matrix $M$ of a linear transformation $\varphi:X\to Y$ in the bases $\{x_i\},\{y_j\}$ of $X,Y$, respectively, is defined by the following equations on its entries $M_{ij}$: $$\varphi(x_i)=\sum_j M_{ij}y_j.$$ I'd suggest you note the types of objects on both sides - we have an equality of what? Answer: of vectors in $Y$.

In your case, carefully instantiate all variables with the specific ones in the problem:

• $\varphi$ is $s$
• $X$ is $V$, and $Y$ is $W^*$
• the basis vectors $x_i$ are $b_i$, and the basis vectors $y_j$ are $c^*_j$
  I denoted here by $\{c^*_j\}$ the basis of $W^*$ which is dual to the basis $\{c_i\}$ of $W$. As you noted in your question, we know that $c_i^*(c_i)=1$ and if $i\neq j,c_j^*(c_i)=0$.
• $M$ is $[s]^B_{C^*}$

and compute the same equation:

$$s(b_i)=\sum_j \left([s]^B_{C^*}\right)_{ij}c^*_j$$

Again, what do we equate here? Answer: elements of $W^*$. How do we get hold of the matrix entries $\left([s]^B_{C^*}\right)_{ij}$? By getting rid of the (co)vectors (functionals) $c^*_j$'s. How do we get rid of them?

Well, it's in their nature to eat vectors (they are functions from $\text{Hom}(W,\mathbb R)$, that is $c_j^*:W\to\mathbb R$, after all), and as you said, you know how the dual basis acts on (eats) the basis it is dual of - $c_i^*(c_i)=1$ and if $i\neq j, c_j^*(c_i)=0.$

Therefore it makes sense to act on both sides with $c_k$ (we have to use a new index so that it does not clash with the $i$ and $j$ we already use):

$$ \begin{align*} s(b_{i})(c_{k}) &=\left(\sum_{j}\left([s]_{C^{*}}^{B}\right)_{ij}c_{j}^{*}\right)(c_{k})\\ &=\sum_{j}\left([s]_{C^{*}}^{B}\right)_{ij}c_{j}^{*}(c_{k}) \end{align*}$$

Ask yourself again, what type of objects are we equating here?

Now if you apply your knowledge about the action $c_{j}^{*}(c_{k})$ in the cases where $j=k$ and $j\neq k$, you'll eliminate all terms in the sum but one and will find $\left([s]_{C^{*}}^{B}\right)_{ik}$.