Why is $A^T : W^* \to V^*$ if $A : V \to W$?

Question

I was wondering what the meaning of a transpose matrix was. I found an answer on this forum, but there's something I don't understand.

It was said that if $A$ transforms $\mathbf{x}$ into $\mathbf{y}$, then $A^T$ transforms $\mathbf{y}^T$ into $\mathbf{x}^T$.

In other "words", I can write $A\mathbf{x}=\mathbf{y} $. But I could also write \begin{aligned}(A\mathbf{x})^T&=\mathbf{y}^T \\ \mathbf{x}^TA^T&=\mathbf{y}^T.\end{aligned}

In this context, $A^T$ let me go from $V^*$ to $W^*$ and not the reverse.

In addition, I couldn't see the link well with the other definition: \begin{aligned}^tf : W^* &\to V^* \\ \phi &\mapsto \phi \circ f \end{aligned}

EDIT: Since I solved my problem, I didn't want to let some of the bad things I wrote.

From how I defined $^tf$, we can see that $\mathbf{y}^T$ must become $\mathbf{y}^TA$, but the dual vector in the dual basis is not represented by $\mathbf{y}^T$ (which is useful to apply the dual vector on vectors) but by $\mathbf{y}$ itself, so the components of this dual vector become $(\mathbf{y}^TA)^T=A^T\mathbf{y}$.

Those explanations may sound a bit clunky, so see egreg's answer to my question to have better foundations that I didn't want to rewrite.

Answer 1

You have discovered that confusing linear maps with their matrix representation is dangerous.

Let's stick to finite dimensional vector spaces over the field $F$ (which may well be $\mathbb{R}$, if this is where your vector spaces are based on).

If $\mathscr{B}=\{v_1,\dots,v_n\}$ is an (ordered) basis of the vector space $V$, let's denote by $C_{\mathscr{B}}\colon V\to F^n$ the map implicitly defined by $$ C_{\mathscr{B}}(v)=\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \quad\text{if and only if}\quad v=a_1v_1+a_2v_2+\dots+a_nv_n $$ If $f\colon V\to W$ is a linear map, with $\mathscr{B}$ and $\mathscr{D}$ bases for $V$ and $W$, there exists a unique matrix $A$ such that, for every $v\in V$, $$ C_{\mathscr{D}}(f(v))=AC_{\mathscr{B}}(v) $$ This is the matrix representation of the linear map $f$, with respect to the bases $\mathscr{B}$ and $\mathscr{D}$.

When you consider dual spaces, you know that to any basis $\mathscr{B}=\{v_1,\dots,v_n\}$ of $V$ there corresponds a dual basis $\mathscr{B}^*=\{v_1^*,v_2^*,\dots,v_n^*\}$ such that $$ v_i^*(v_j)=\begin{cases} 1 & \text{if $i=j$} \\[4px] 0 & \text{if $i\ne j$} \end{cases} $$

You also know that to a linear map $f\colon V\to W$ there corresponds a linear map $f^*\colon W^*\to V^*$, defined by $$ f^*(\xi)\colon v\mapsto\xi(f(v)) $$ for $\xi\in W^*$.

What happens is that

if $\mathscr{B}$ and $\mathscr{D}$ are bases of $V$ and $W$ and $A$ is the matrix representation of $f\colon V\to W$ with respect to these bases, then $A^T$ is the matrix representation of $f^*\colon W^*\to V^*$ with respect to the bases $\mathscr{D}^*$ and $\mathscr{B}^*$

which means: for every $\xi\in W^*$, $$ C_{\mathscr{B}^*}(f^*(\xi))=A^TC_{\mathscr{D}^*}(\xi) $$ according to the definition of matrix representation. Note that the matrix is always on the left of the coordinate vector.

There is a special case for this, which might be where you lost your steps. If $V=F^n$ and $W=F^m$, then every linear map is the multiplication by a matrix, precisely the matrix representation of the linear map with respect to the standard bases.

You can see the dual space of $F^n$ as $F^n$ itself, via identification of the dual basis with the standard basis; but observe that if $f\colon F^n\to F^m$ is a linear map, then its matrix representation $A$ is an $m\times n$ matrix. The transpose matrix $A^T$ is $n\times m$ and indeed it induces a linear map $w\mapsto A^Tw$ from $F^m$ to $F^n$.

Answer 2

For a finite-dimensional $\mathbb{R}$-vector space $V \cong \mathbb{R}^n$ one can show, that for any linear functional $\alpha \in V^*$ there is a unique vector $x_\alpha \in V$ such that $\alpha ( v ) = \langle v , x_\alpha \rangle = v^T \cdot x_\alpha$. In that sense, we can identify $V$ and $V^*$.
For a linear operator $T \colon V \to W$ between finite dimensional $\mathbb{R}$-vector spaces, we define the adjoint map as the uniquely determined linear map $$T^* \colon W^* \to V^* \colon \alpha \longmapsto T^* \alpha,$$ where $T^* \alpha (v) := \alpha \big(T(v) \big)$.
If we now consider the map $T \colon V \to W \colon v \mapsto A\cdot v$, what is the adjoint map for $T$? With our above identification of $V$ and $V^*$ we just want to know what for given $\alpha \in W^*$ and corresponding $x_\alpha \in W$ the vector $x_{T^* \alpha} \in V$, corresponding to $T^* \alpha \in V^*$ is.
We get that by calculating $$T^* \alpha (v) = \alpha \big(T(v)\big) = \alpha (A v) = \langle A v , x_\alpha \rangle = (Av)^T x_\alpha = v^T A^T x_\alpha = \langle v , A^T x_\alpha \rangle .$$

Answer 3

Here's another point of view. $V,W$ are finite dimensional vector spaces

Let $f:V\to W$ be any linear operator, fix bases $(e_i),(f_j)$ of $V,W$ respectively, so that the matrix of $f$ with respect to those bases is $A$.

Then you have natural bases of $V^*,W^*$ that are $(e_i^*), (f_j^*)$, where $e_i^*(\displaystyle\sum_k \lambda_k e_k) = \lambda_i$ and similarly for $f_j$.

Note that if $l\in V^*$, then $e_i^{**}(l) = l(e_i)$ : this is easily checked by noting that for all $k$, $e_i^{**}(e_k^*) = \delta_{ik} = e_k^*(e_i)$

Then let $f^* : W^*\to V^*$ denote $l\mapsto l\circ f$, and let's compute the matrix of $f^*$ in those bases.

The $(i,j) $-coefficient is $e_i^{**}(f^*(f_j^*))= f^*(f_j^*)(e_i)=(f_j^*\circ f)(e_i)= f_j^*(\displaystyle\sum_k A_{ki}f_k)= A_{ji} = (A^T)_{ij}$