Groking The Determinant

grok | Verb

“to understand profoundly and intuitively” - Merriam-Webster

$\det$ is a function of all time. However, I think most students using it don’t grok it, despite their understanding of one of its “less abstract” definitions. In this post I will try to illustrate my journey with this function, and the path I took to grok it.

In high school, the following formula occasionally showed its face in math competitions or a precalculus class: $\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$, accompanied by the usual “area” argument as follows:

Given a matrix $A \in \R^{2 \times 2}$, we have that $| \det A |$ is the area of the parallelogram created when applying it to $\vec{e}_1$ and $\vec{e}_2$. For example consider the matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$ with $|\det A| = |3 \cdot 2 - 1 \cdot 1| = 5$.

Of course this definition can be extended to higher dimensions with the mythical, hideously named, and potentially hard to imagine parallelepiped. [I didn’t know there was a third e until writing this].

Now what happens to our nice looking ad - bc when we go to higher dimensions.

$\det \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = a(ei-fh) - b(di-fg) + c(dh-eg)$

GAH! What murderous manner of mechanism has manifested such a miraculously malicious formula. Well it turns ot that:

$\det \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = a \cdot \det \begin{bmatrix} e & f \\ h & i \end{bmatrix} - b \cdot \det \begin{bmatrix} d & f \\ g & i \end{bmatrix} + c \cdot \begin{bmatrix} d & e \\ g & h \end{bmatrix}$

Who would’ve guessed? [Certainly not me.] This all leads to the frankly tragic recursive definition of the determinant which came in college:

$\det A = \sum_{i=1}^n (-1)^i \det A_{i,j}$

Where $A_{i,j}$ is a minor of the matrix and $1 \leq j \leq n$. [This definition also works iterating on rows the columns.] This definition is insanely useful and helps us do lots of nice things with the list of identities involving $\det$.

But can we do better? Yes. Let’s start from the beginning. Lets say we want a function that gives us the volume of the space that we get when mapping the box defined by $e_1, \dots, e_n$ through a matrix $A$.

What kinds of properties might we want such a function $f$ to have? [We will defined $f$ as a function on $\R^n \times \dots \times \R^n$, where each entry represents a column in $A$] It turns out there are three:

• - $f(I) = 1$
• - $f$ is multilinear
• - $f$ is alternating

Clearly we want $f(I) = 1$ as the mapping our box through the identity produces the same unit box.

Now what about multilinear. Suppose we have $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$ as before and $B = \begin{bmatrix} 3 & 3 \\ 1 & 5 \end{bmatrix}$. Now how can we add the areas? Lets go about this geometrically:

With $C = \begin{bmatrix} 3 & 7 \\ 1 & 6 \end{bmatrix}$. Note here second column of $C$ is the sum of the second columns of $A$ and $B$, and the first column is their first column. I claim that $f(A) + f(B) = f(C)$. Hopefully you can see this by sliding the shape formed by the blue and red bits into place.

Thus for $f$ we want that $f(\vec{v}_1, \dots, \vec{a}, \dots, \vec{v}_n) + f(\vec{v}_1, \dots, \vec{b}, \dots, \vec{v}_n) = f(\vec{v_1}, \dots, \vec{a} + \vec{b}, \dots, \vec{v}_n)$. A similar argument exists for the scalar multiple property of being multilinear but I will leave that as an exercise for the reader [There is already enough tikzpicture in this post].

Now what about alternating? This induces the discussion of signed volume. Which initially seems a little unclear if we are looking for a function that gets the area of some space. But lets suspend our disbelief for a little bit and assume we want signed volume. For example if we transformed the unit square across the $x$-axis it is a fair assumption to want its volume to negative.

Lets consider this question: What happens if we permute two vectors in the input to $f$? We can think of this as permuting the basis vectors of the original space. Consider the following geometric argument of permuting the $x$ and $y$ axis.

By flipping these two axis we have essentially inverted the same across the line of $y = x$! Thus we will say that this inversion negates the volume. So we conclude that $f$ should be alternating.

Now lets examine the class of function that $f$ is. To recap $f$ is an $n$ alternating multilinear map. You can verify for yourself that this is a vector space. So a natural question arises. What is the dimension of this vector space?

TO BE CONTINUED