Dual space and Transposed matrix

Question

Transpose of matrix $T$ denoted by $T'$ is a matrix that holds this: $(T' l , x) = (l, Tx)$ where $$ T: U \mapsto V , x \in U, $$ and $l$ is in dual of $V$, which is $V'$.

I kinda understand that $(T')^{-1}$ is the change of basis from dual $U$, $U'$ to dual $V$, $V'$ such that the pairwise multiplication of transformed members of $U$ and $U'$ stays same.

But still I don't understand why such a transformation $T'$ should exist and is unique.

Answer 1

You don't need bases for this. For fixed $l\in V'$, the map $x\mapsto (l, Tx)$ is a linear functional on $U$, because it's a composition of two linear maps $x \mapsto Tx$ followed by $Tx \mapsto (l, Tx)$. So by definition of dual, there is a unique element of $U'$ which is called $T'l$ that acts on $x\in U$ like $(l, Tx)$, in other words $(T'l, x) = (l, Tx)$ for all $x \in U$. It remains to show that $l \mapsto T'l$ is linear in $l$. This holds because $(,)$ is bilinear - which is proven using the definition of scalar multiplication and addition in the dual. Note that $T'$ is a linear map from $V'$ to $U'$. The basis-free formulation is more helpful in infinite dimension.

Answer 2

Let $V$ and $W$ be vector spaces over the field $F$, and let $T:V \rightarrow W$ be a linear transformation. Denote the dual of these spaces $V^*$ and $W^*$ respectively. Fix ordered bases $\beta=\left\{v_1, \dots, v_n\right\}$ for $V$, $\beta^*=\left\{f_1, \dots, f_n\right\}$ for $V^*$, $\beta'=\left\{w_1, \dots, w_m\right\}$ for $W$, and $\beta^{'*}=\left\{g_1, \dots, g_n\right\}$ for $W^*$ (where $\beta^*$ and $\beta^{'*}$ are the natural dual bases).

The matrix representation of $T$ relative to the ordered bases $\beta$ and $\beta'$ is the $m \times n$ matrix $A=\left[T\right]_{\beta}^{\beta'}$ where $A=\left[\left[T(v_1)\right]_{\beta'} \mid\cdots\mid \left[T(v_n)\right]_{\beta'}\right]$. We'd like to show the existence of a transformation $T^t$ whose matrix representation $B$ has rows equal to the columns of $A$. Clearly if such a transformation exists then its matrix is unique.

Consider the transformation $T^t: W^* \rightarrow V^*$ given by $T^t(g(v))=g(T(v))$. Observe $\left[T^t\right]_{\beta^{'*}}^{\beta^*}=\left[\left[T^t(g_1)\right]_{\beta^{'*}} \mid\cdots\mid \left[T^t(g_m)\right]_{\beta^{'*}}\right]$.

Now see that if the $i^{\text{th}}$ column of $A$ is $\left[T(v_i)\right]_{\beta'}=\begin{bmatrix} a_{1,i} \\ a_{2,i}\\ \vdots\\a_{n,i}\end{bmatrix}$, then $T(v_i)=\sum\limits_{k=1}^{n}a_{k,i}w_k$ and thus $g_j(T(v_i))=\sum\limits_{k=1}^{n}a_{k,i}g_j(w_k)=a_{j,i}$. So $A_{j,i}=B_{i,j}$ and we have proved existence.