Show that $(T\circ S)'=S'\circ T'$

Question

Question: Let $S:U\rightarrow V$ and $T:V\rightarrow W$ be linear mappings with S' and T' their transposes. Show that $(T\circ S)'=S'\circ T'$

My approach: Let dimensions of $U,V$ and $W$ are $m,n$ and $p$ and their basis are $\alpha=\{ \alpha_1,\alpha_2,\cdots \alpha_m \},\beta=\{ \beta_1,\beta_2,\cdots \beta_n \}$ and $\gamma=\{ \gamma_1,\gamma_2,\cdots \gamma_p \}$ respectively. Then find matrix representation of $S$ and $T$ separately. Matrix representation of the composition of linear transformations $T\circ S$ is as same as the multiplication of their matrix representation. Hence show that transpose of $(TS)$ matrix is as same as $S'T'$. But I think it is not enough to proof the statement in general.
This Question is similar but the answer is not clear to me. Because how transpose is related with dual map as I know that dual map is a linear functional which map a vector space to it's field and how they get \begin{eqnarray} (UT)^*(h)=h(UT),~h\in Z^* \end{eqnarray} Where \begin{array}{lcl} \\ V\overset{T}{\rightarrow} W & & \color{red}{Z^*}\overset{U^*}{\rightarrow}W^* \\ {\tiny{UT}}\searrow~\downarrow {\tiny{U}}& & \color{red}{{\tiny{T^*U^*}}\searrow}~\downarrow {\tiny{T^*}} \\ ~~~~~~~~~~Z & & ~~~~~~~~~~~~\color{red}{V^*} \end{array}
Now I am confused between dual map and transpose. Any explanation or solution will be appreciated.
Thanks in advance .

Answer 1

Recall that the dual space of a vector space $\textsf{V}$ (over a field $F$ ) is the set of all their linear functionals (all the linear transformations from $\textsf{V}$ to the field). That is $$\textsf{V}^*=\{f:\textsf{V}\to F\, : \, f \textrm{ is linear}\}$$

Recall that also, given a linear map $\textsf{T}:\textsf{V}\to \textsf{W}$, their transpose is defined by the linear transformation $\textsf{T}^t :\textsf{W}^* \to \textsf{V}^*$ with correspondence rule

$$\textsf{T}^t (f) = f\circ \textsf{T} \quad \textrm{for all } f\in \textsf{W}^*$$

Now, the proof is simple, it would go something like this :

For any linear functional $f\in \textsf{W}^*$ we have $$( \textsf{S}^t \circ \textsf{T}^t )(f)= \textsf{S}^t(\textsf{T}^t (f))= \textsf{S}^t (f\circ \textsf{T}) = (f\circ \textsf{T})\circ \textsf{S}=f\circ (\textsf{T} \circ \textsf{S}) = (\textsf{T}\circ \textsf{S})^t (f)$$ It is enough to show that $(\textsf{T}\circ \textsf{S})^t =\textsf{S}^t \circ \textsf{T}^t$.

Answer 2

For a more intuitive and less notational reason, I think it is best to think about the 4 spaces of a matrix (As a linear transformation). Every linear transformation has 4 basic spaces associated with it, Column Space, kernel, row space and co-kernel. Column Space is defined as the vector space spanned by the column vectors of the matrix, and the row space is defined in the same way. Co-Kernel is the Kernel of the Transpose matrix.

The kernel is the vector space of all inputs that are mapped to the zero vector by the matrix. An intuitive way to think about a matrix is as a mapping from its Column Space to its Row Space and its kernel. The transpose of a matrix then maps from its Row Space to its Column Space (Since the transpose switches the two spaces) and whats called the co-kernel. The kernel is important in general linear algebra but can be ignored for this result.

Now if you take a composition of matrices $A*B$ you can think of it as mapping from the Column Space of A to the Column Space of B to the Row Space of B. Now if you want to take the transpose of this composition you want to map from the Row Space of B to the Row Space of A to the Column Space of A. Thus $(AB)' = B'A'$

Dual Space can be thought of in a similar way which I won't explain here since it follows from similar logic as above and I think others have better answers about the dual vector space specifically.