Transposes

A good background video

Transposes are inextricably linked to the concept of duality.

The transpose of a vector is its dual vector: $v^T$ essentially represents the operation “project onto $v$”, and it can be applied to some vector $x$ simply by left-multiplying, i.e. $v^{T}x$. Therefore, $v^T$ should primarily be thought of as a function.

One way to visualize a row vector $v^T$ is by drawing its level curves. That is, for each integer $k$, we draw a single line representing all vectors $x$ such that $v^{T}x=k$. For instance, here’s how we’d represent the row vector $[1,1]$:

The line going through the origin corresponds to $k=0$, and the line going through $(0,1)$ and $(1,0)$ corresponds to $k=1$.

Note that scaling up the vector will simply make the level curves more dense. Shown below are the level curves for $v^T=[2,2]$:

Lemma

The distance between the level curves of $v$ is $\frac{1}{|v|}$.

Now, we’ll use this visualization to prove the following identity geometrically:

Theorem

For any 2x2 matrix $A$, $\det (A)=\det (A^T)$.

As a quick refresher, $\det ([x_1{x}_2])$ is the signed area of the parallelogram defined by the column vectors $x_1$ and $x_2$. For instance, $\det (\left(\begin{array}{cc}1& 4\\ 2& 1\end{array}\right))$ represents the following area:

Equivalently, the determinant is the scale factor by which $A$ transforms the area of any shape.

Now, how do we visualize the scaling induced by $A^T$? It will actually be easier to visualize this in the backwards direction---that is, to find the area of the shape that $A^T$ maps to the unit square. If this area is $x$, then we hope to show that $\det (A)=\frac{1}{x}$.

Let’s draw in level curves for $A=\left(\begin{array}{cc}1& 4\\ 2& 1\end{array}\right)$. Then, by definition of the level curves, the shape which $A^T$ maps to the unit square is precisely the shaded region:

Now, how can we relate the area of this parallelogram to our first one? Well, according to our lemma, the two altitudes of the new parallelogram are $\frac{1}{|x_1|}$ and $\frac{1}{|x_2|}$, respectively. We also know that the two altitudes of the original parallelogram are $\frac{\det (A)}{|x_1|}$ and $\frac{\det (A)}{|x_2|}$, respectively. And since the two parallelograms have the same angles, this means that they must be similar: the new one is just the original one scaled by a factor of $\frac{1}{\det (A)}$! This means the area of our new parallelogram is simply

$$\det (A){\left(\frac{1}{\det (A)}\right)}^2=\frac{1}{\det (A)}$$

as desired.

By the way, the basis which defines the second parallelogram has a special name: it is the dual basis of $A$.

Also unfortunately, this similarity argument doesn’t seem to extend to higher dimensions. However, I hope it still provided a more visceral feel for what exactly the transpose does, and how it’s visually related to the original basis.