Volumes and determinants

Determinants also relate to volumes of geometric objects. It may seem silly but let’s look at the one-dimensional setting first. (In general, if you don’t understand the 1-d setting, then you should start your learning there.)

In $\mathbb{R}^1 = \mathbb{R}$, a vector is just a scalar or a element of $\mathbb{R}$. The length of the vector $a$ is $|a|$. In 1-d, length is volume. So we have the following relation between the volume of the line segment $P(a)$ spanned by $a$ is $$\operatorname{Vol}(P(a)) = |\det(a)|$$ The determinant is a signed volume in 1-d.

With some confidence with one dimension, we turn to 2-d. Let’s start with a $2 \times 2$ matrix $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ The columns give up two vectors $$\mathbf{v}_1 = \begin{pmatrix} a \\ c \end{pmatrix}, \ \mathbf{v}_2 = \begin{pmatrix} b \\ d \end{pmatrix}$$ The vectors $\mathbf{v}_1, \mathbf{v}_2$ span a parallelogram $P = P(A)$.

Let’s check that $$\operatorname{Vol}(P(A)) = |\det(A)|$$

Neither the volume nor the determinant depend on where we place the parallelogram in the plane so assume that the unlabelled vertex is at the origin.

Let’s rotate $P$ so that $\mathbf{v}_1$ lies along the $x$-axis.

Notice that $$\mathbf{v}_1^\prime = \begin{pmatrix} a^\prime \\ 0 \end{pmatrix}, \ \mathbf{v}_2^\prime = \begin{pmatrix} b^\prime \\ d^\prime \end{pmatrix}$$ so for the rotated matrix $$\det A^\prime = a^\prime d^\prime$$ which is exactly the volume of the rotated parallelogram $P^\prime$ (base $\times$ height).

To finish, we want to see that the determinant also does not change up when rotating in $\mathbb{R}^2$. However, a rotation is a linear transformation! So it can be represented by a matrix when using the standard basis. For rotating counter-clockwise through an angle $\theta$ from $x$-axis, we have $$R_\theta := \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$ as the matrix. Since $$A^\prime = R_\theta A$$ we get $$\det A^\prime = \det(R_\theta) \det(A)$$ and $$\det R_\theta = \cos^2 \theta + \sin^2 \theta = 1.$$ So the determinant is unchanged also.

We can conclude that $$\operatorname{Vol} P = |\det(A)|$$ in 2-d, like 1-d.

We feel pretty good about making the following statement.

For any $n \times n$ matrix $A$ we have $$\operatorname{Vol} P(A) = |\det(A)|$$ where $P(A)$ is the parallelopiped spanned by the column vectors of $A$.

In three dimensions, the parallelopiped looks like

For higher dimensions, we rely on our intuition from dimensions 1 through 3.

The proof of $$\operatorname{Vol} P(A) = |\det(A)|$$ can go one of two ways:

• We can brute force things along lines of two dimensions above. Use explicit formulas and manipulations to match up the results.
• Or we can reach for the power of abstraction and use the properties that completely capture the behavior $\operatorname{Vol}$ and $|\det(A)|$.

Proposition. If $$f,g : \operatorname{Mat}_{n \times n}(\mathbb{R}) \to \mathbb{R}$$ are two functions which both satisfy

• $$\phi(L(i,j,c)A) = \phi(A)$$
• $$\phi(P(i,j)A) = \phi(A)$$
• $$\phi(S(i,c)A) = |c|\phi(A)$$
• $$\phi(I_n) = 1.$$
• and $$\phi(A) = 0$$ for singular $A$.

Then, $f = g$.

Corollary. For any $n \times n$ matrix $A$ we have $$\operatorname{Vol} P(A) = |\det(A)|$$ where $P(A)$ is the parallelopiped spanned by the column vectors of $A$.

This corollary tells us that the determinant can also be viewed as a way to provide signed volumes in all dimensions. This is useful for multi-variable integration, for example, which is why you find determinants showing up in change of variable formulae for multi-variable integrals.