I'm reading Axler's Linear Algebra Done Right and trying to understand
exactly what the transpose of a matrix represents. To get some intuition, I
constructed the following example:
Let's say we're working with the two vector spaces $\mathbb{R}^3$ and
$\mathbb{R}^2$.
We have the linear map $T$:
$$ T : \mathbb{R}^3 \rightarrow \mathbb{R}^2$$
$$ T(x,y,z) = (2x + 3y, 4z) $$
We also have the linear functional $\phi$:
$$ \phi : \mathbb{R}^2 \rightarrow \mathbb{R}$$
$$ \phi(x,y) = (x + y) $$
We can compose these together:
$$ \phi \circ T : \mathbb{R}^3 \rightarrow \mathbb{R} $$
$$ [\phi \circ T](x,y,z) = (2x + 3y – 4z) $$
We can also represent $T$ and $\phi$ with matrices:
$$ \mathcal{M}(T) = \left( \begin{array} & 2 & 3 & 0 \\ 0 & 0 & -4 \end{array} \right) $$
$$ \mathcal{M}(\phi) = \left( \begin{array} & 1 & 1 \end{array} \right)$$
So the matrix representation of $\phi \circ T$ is:
$$\mathcal{M}(\phi)\mathcal{M}(T) = \left( \begin{array} & 2 & 3 & -4 \end{array} \right) $$
The dual space of $\mathbb{R}^3$, $\mathbb{R}^{3\prime}$, is the set of linear
functionals $\mathbb{R}^3 \rightarrow \mathbb{R}$ and likewise for
$\mathbb{R}^2$
Now, let $T\prime$ be the dual map of $T$:
$$
T^\prime : \mathbb{R}^{2\prime} \rightarrow \mathbb{R}^{3\prime}
$$
$$
T^\prime(\phi) = \phi \circ T
$$
So $T^\prime$ is a "higher-order function" that maps one type of linear
functionals to another type of linear functional.
Supposedly, the matrix representation of $T^\prime$ is $\mathcal{M}(T)^T$, so
$$\mathcal{M}(T^\prime) =
\left(
\begin{array}
& 2 & 0 \\
3 & 0 \\
0 & -4
\end{array}
\right)
$$
If everything is correct so far, the question I have is how to get from this
matrix to $ \left( \begin{array} & 2 & 3 & -4 \end{array} \right)$, which I
would expect to, again, be the final result. I would have expected that I could
simply multiply $\mathcal{M}(T^\prime)$ by $\mathcal{M}(\phi)$, since $\phi$ is
already an element of $\mathbb{R}^{2\prime}$, but that matrix is 1×2. So it seems
I would have to take its transpose as well, multiply it to the right of
$T^\prime$ to get a 3×1 matrix, and then transpose that to get the expected 1×3
matrix.
I'm guessing this is related to the algebraic property of dual maps that
$$ (ST)^\prime = T^\prime S^\prime $$
for all linear maps $S$ and $T$, but I don't understand why.
Best Answer
Simply do$$\begin{pmatrix}2&0\\3&0\\0&-4\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}2\\3\\-4\end{pmatrix}.$$That's because the coordinates of $\phi$ in the dual of the canonical base are $1$ and $1$, that is, if the canonical basis is $(e_1,e_2)$, then $\phi={e_1}^*+{e_2}^*$.