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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 1x2. So it seems I would have to take its transpose as well, multiply it to the right of $T^\prime$ to get a 3x1 matrix, and then transpose that to get the expected 1x3 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.

Related: Intuition behind definition of transpose map

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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 $\phi={e_1}^*+{e_2}^*$.

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  • $\begingroup$ I don't doubt that this is correct, but I don't understand it. Would you mind expanding on it? If not, I will just mark as answered. $\endgroup$ – user2987808 Aug 1 '17 at 13:31
  • $\begingroup$ @user2987808 The matrix of $T'$ with respect to the dual of the canonical basis is $\left(\begin{smallmatrix}2&0\\3&0\\0&-4\end{smallmatrix}\right)$ and both coordinates of $\phi$ in that basis are equal to $1$. Therefore the coordinates of $T'(\phi)$ are given by the product $\left(\begin{smallmatrix}2&0\\3&0\\0&-4\end{smallmatrix}\right)\left(\begin{smallmatrix}1\\1\end{smallmatrix}\right)$. So, the coordinates are $2$, $3$, and$-4$, as you wanted them to be. $\endgroup$ – José Carlos Santos Aug 1 '17 at 15:58

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