Volumes

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.

Rotated parallelogram spanned by two vectors

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, $f = g$.

Proof. (Expand to view)

From Gaussian Elimination, we know that any matrix can be written as $A = C U$ where $C$ is a product of $L(i,j,c), S(i,c)$, and $P(i,j)$ and $U$ is in reduced row echelon form. Thus, $\phi(A) = \phi(U)$ and $\phi$ only depends on the reduced row echelon form. We have seen that $A$ is invertible if and only $U = I_n$. Thus, we have specified $\phi(A)$ for invertible $A$ completely from the properties.

For non-invertible $A$, we know that $\phi(A) = 0$. ■

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 $A$.

Proof. (Expand to view)

Both $\operatorname{Vol}(P(-))$ and $|\det(-)|$ satisfy the conditions of the previous proposition.

We have already seen that $\det(A)$ satisfies all the conditions with the exception that $\det(P(i,j)A) = -\det(A)$ But the absolute value doesn’t change.

For the volume, we know it doesn’t change with rigid motions. Let’s write $\mathbf{v}1,\ldots,\mathbf{v}_n$ be the _row vectors of $A$. Then, we have an intuitively clear formula $\operatorname{Vol} (P(A)) = \operatorname{ht}(\mathbf{v}_i}) \operatorname{Vol} (P_i)$ where $P_i$ is the parallelopiped spanned by all the vectors $\mathbf{v}1,\ldots,\mathbf{v}_n$ _except $\mathbf{v}_i$ and $\operatorname{ht}(\mathbf{v}_i})$ is height of $\mathbf{v}_i$ above that parallelopiped. (We should really prove this too but we will rely a bit on “geometric intuition” here and below.)

From this formula, we see that $\operatorname{Vol}(P(P(i,j)A)) = \operatorname{Vol}(P(A))$ and $\operatorname{Vol}(P(S(i,c)A)) = |c|\operatorname{Vol}(P(A))$ using induction on $n$.

For $\operatorname{Vol}(P(L(i,j,c)A)) = \operatorname{Vol}(P(A))$ we note that the height of $\mathbf{v}_j + c\mathbf{v}_i$ has the same height above $P_j$ as $\mathbf{v}_j$.

Finally, the volume of $P(I_n)$ is the volume of the unit cube in $n$-dimensions which is $1$. And if $A$ is singular, then it spans at most a $(n-1)$-dimensional subspace and must have $0$ volume. ■

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.