# Math 31AH: Lecture 24

Let us begin with a brief recap of Lecture 23, which introduced a number of new concepts which may seem complex at first, but are in fact rather simple. Let $\mathbf{V}$ be an $n$-dimensional vector space; the dimension $n$ could be arbitrarily large, so visualizing geometry in $\mathbf{V}$ can be arbitrarily hard. However, no matter what $n$ is, we have a naturally associated $1$-dimensional vector space $\mathbf{V}^{\wedge n}.$ We can visualize any $1$-dimensional vector space easily, by picturing it as a line. Algebraically, we associate to a given set of $n$ vectors $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ of $n$ vectors in $\mathbf{V}$ a point on the line $\mathbf{V}^{\wedge n},$ namely the point $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n.$ The set $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ is linearly independent if and only if the point $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$ is not the zero point on the line $\mathbf{V}^{\wedge n}.$ This is pure algebra. If $\mathbf{V}$ is a Euclidean space, meaning that we can measure lengths and angles in $\mathbf{V}$ using a scalar product, then the line $\mathbf{V}^{\wedge n}$ is a Euclidean line, i.e. a line on which we have a notion of distance defined. The volume of the parallelepiped $\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n) \subset \mathbf{V}$ is then the distance from the point $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$ to zero on the line $\mathbf{V}^{\wedge n}.$ There are only two points at distance one from the origin on the line $\mathbf{V}^{\wedge n},$ which we call $\omega$ and $-\omega.$ These are analogous to the numbers $1$ and $-1$ on the number line $\mathbb{R}.$ Choosing one of these, say $\omega,$ is what it means to choose an orientation on $\mathbf{V}.$ The oriented volume of the parallelepiped $\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n)$ is then the unique number $a \in \mathbb{R}$ such that $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n = a\omega.$ For example, if $\mathbf{V}=\mathbb{R},$ choosing $\omega =1$ means the number line is oriented left to right, and choosing $\omega=-1$ means the number line is oriented right to left. In the former case, the oriented length of the line segment $\mathcal{P}(2)$ is $2,$ while in the latter case it is $-2.$ That’s it, more or less.

In this lecture, we use the above machinery to define a function

$\det \colon \mathrm{End} \mathbf{V} \to \mathbb{R}$

which tells us when a given linear operators $A \in \mathrm{End}\mathbf{V}$ is invertible. This function is called the determinant. All the complexity (i.e. the hard stuff) is concentrated in the definitions from the past two lectures, and if you understand these, then the definition of the determinant is exceedingly simple.

Let $A \in \mathrm{End}\mathbf{V}$ be a given linear operator. We associate to $A$ the linear operator $A^{\wedge n} \in \mathrm{End}\mathbf{V}^{\wedge n}$ defined by

$A^{\wedge n}\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n :=A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n.$

Definition 1: The determinant of $A$ is the unique eigenvalue of $A^{\wedge n}.$

Let us unpack this definition. The key point is that the vector space $\mathbf{V}^{\wedge n}$ is $1$-dimensional. This means that every operator in $\mathrm{End}\mathbf{V}^{\wedge n}$ is simply multiplication by a number (see Problem 1 on Assignment 6). We are defining $\mathrm{det}(A)$ to be the number by which the operator $A^{\wedge n} \in \mathbf{V}^{\wedge n}$ scales its argument. That is, $\det(A)$ is the number such that

$A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n = \det(A)\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$

for all $\mathbf{v}_1,\dots,\mathbf{v}_n \in \mathbf{V}.$

Theorem 1: An operator $A \in \mathrm{End}\mathbf{V}$ is invertible if and only if $\det(A) \neq 0.$

Proof: Suppose first that $A$ is invertible. Let $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ be a linearly independent set in the $n$-dimensional vector space $\mathbf{V}.$ Since $A$ is invertible, $\{A\mathbf{v}_1,\dots,A\mathbf{v}_n\}$ is also a linearly independent set in $\mathbf{V}$, and the linear independence of this set is equivalent to the statement that

$A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n$

is not the zero tensor. Since the above nonzero tensor is equal to

$\det(A)\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n,$

we have $\det(A) \neq 0.$

Conversely, suppose that $\det(A) \neq 0.$ Let $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ be a linearly independent set in $\mathbf{V}.$ The statement that $A$ is invertible is equivalent to the statement that $\{A\mathbf{v}_1,\dots,A\mathbf{v}_n\}$ is linearly independent, which is in turn equivalent to the statement that

$A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n$

is not the zero tensor. But

$A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n = \det(A) \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$

is the nonzero tensor $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$ scaled by the nonzero number $\det(A)$, hence it is nonzero.

— Q.E.D.

If you already know something about determinants of matrices, you are probably wondering how Definition 1 relates to this prior knowledge; let us explain this now. Let $E=\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ be an orthonormal basis in $\mathbf{V},$ and let

$\omega = \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n$

be the corresponding unit volume tensor in $\mathbf{V}^{\wedge n}.$ We then have that

$\det(A) = \langle \omega,A^{\wedge n}\omega\rangle;$

this follows from the fact that $\det(A)$ is by definition of the eigenvalue of the operator $\mathbf{A}^{\wedge n}$ together with the fact that $\{\omega\}$ is an orthonormal basis of $\mathbf{V}^{\wedge n}.$ Now, we can explicitly evaluate the above inner product in terms of the matrix elements of $A$ relative to the basis $E$ of $\mathbf{V}.$ Here is the computation:

$\det(A) = \langle \omega,A^{\wedge n}\omega\rangle = \langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,A\mathbf{e}_1 \wedge \dots \wedge A\mathbf{e}_n\rangle = \sum\limits_{\pi \in \mathrm{S}(n)} \mathrm{sgn}(\pi) \langle \mathbf{e}_1,A\mathbf{e}_{\pi(1)} \rangle \dots \langle \mathbf{e}_n,A\mathbf{e}_{\pi(n)} \rangle.$

We recall that $\mathrm{S}(n)$ is the set of permutations $\pi$ of the numbers $1,\dots,n$, i.e. bijections $\pi \colon \{1,\dots,n\} \rightarrow \{1,\dots,n\}.$ For example, in the case $n=2,$ this computation reads

$\det(A) = \langle \mathbf{e}_1 \wedge \mathbf{e}_2, A\mathbf{e}_1 \wedge A\mathbf{e}_2 \rangle = \langle \mathbf{e}_1,A\mathbf{e}_1\rangle \langle \mathbf{e}_2,A\mathbf{e}_2\rangle - \langle \mathbf{e}_1,A\mathbf{e}_2\rangle \langle \mathbf{e}_2,A\mathbf{e}_1\rangle.$

This is the formula for the determinant of the $2 \times 2$ matrix

$[A]_E = \begin{bmatrix} \langle \mathbf{e}_1,A\mathbf{e}_1\rangle & \langle \mathbf{e}_1,A\mathbf{e}_2\rangle \\ \langle \mathbf{e}_2,A\mathbf{e}_1\rangle & \langle \mathbf{e}_2,A\mathbf{e}_2\rangle \end{bmatrix}$

which you likely learned in high school algebra (if you didn’t learn it then, you just learned it now).

To reiterate, for any operator $A \in \mathrm{End}V,$ the determinant $\det(A)$ is the constant such that

$A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n = \det(A)\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$

for all $\mathbf{v}_1,\dots,\mathbf{v}_n.$ Taking the scalar product with a unit volume tensor $\omega$ on either side, this becomes

$\langle \omega, A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n \rangle= \det(A)\langle \omega,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\rangle.$

The scalar product on the LHS is the oriented volume of the parallelepiped $\mathcal{P}(A\mathbf{v}_1,\dots,A\mathbf{v}_n),$ while the scalar product on the right is the oriented volume of the parallelepiped $\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n).$ Assuming that the latter volume is nonzero — i.e. that $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ is a linearly independent set in $\mathbf{V}$ — we thus have

$\det(A) = \frac{\langle \omega, A\mathbf{v}_1 \wedge \dots \wedge A\mathbf{v}_n \rangle}{\langle \omega,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\rangle},$

which says that $\det(A)$ is the scalar factor by which the oriented volume of $\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n)$ transforms when each of the vectors $\mathbf{v}_1,\dots,\mathbf{v}_n$ transforms by $A.$ As an application of this geometric interpretation of the determinant, let us calculate the area of an ellipse whose semimajor and semiminor axes have lengths $a$ and $b$, respectively. The basic observation is that such an ellipse is the image of the unit disc in the Euclidean plane $\mathbf{V}= \mathbb{R}^2$ under the linear transformation $A \in \mathrm{End}\mathbb{R}^2$ defined by

$A\mathbf{e}_1=a\mathbf{e}_1,\ A\mathbf{e}_2=b\mathbf{e}_2,$

where $\mathbf{e}_1=(1,0)$ and $\mathbf{e}_2=(0,1)$ is the standard basis. The determinant of the operator $A$ is

$\det(A) = \frac{\langle \mathbf{e}_1 \wedge \mathbf{e}_2, A\mathbf{e}_1 \wedge A\mathbf{e}_2\rangle}{\langle \mathbf{e}_1 \wedge \mathbf{e}_2, \mathbf{e}_1 \wedge \mathbf{e}_2\rangle} = ab.$

Let $\epsilon > 0$ be an extremely small positive number, and tile the unit disc with translated copies of the tiny square $P(\varepsilon \mathbf{e}_1,\varepsilon \mathbf{e}_2),$ so that the tiling approximates the area of the disc up to an exceedingly small error. The oriented area of this tiny square is

$\langle \mathbf{e}_1 \wedge \mathbf{e}_2, \varepsilon\mathbf{e}_1 \wedge \varepsilon \mathbf{e}_2 \rangle = \varepsilon^2 \langle \mathbf{e}_1 \wedge \mathbf{e}_2,\mathbf{e}_1 \wedge \mathbf{e}_2 \rangle = \varepsilon^2.$

The image of the tiny square under the transformation $A$ is the tiny parallelogram $\mathcal{P}(A\varepsilon \mathbf{e}_1,A\varepsilon \mathbf{e}_2),$ and the oriented area of this tiny parallelogram is

$\det(A) \langle \mathbf{e}_1 \wedge \mathbf{e}_2, \varepsilon\mathbf{e}_1 \wedge \varepsilon \mathbf{e}_2 \rangle = \varepsilon^2 ab.$

We conclude that the area of the parallelogram tiling which approximates the ellipse is the area of the square tiling which approximates the circle scaled by $ab,$ and hence that the area of the ellipse equals the area of the disc scaled by $ab.$ Since the area of the unit disc is $\pi,$ the area of the ellipse in question is $\pi a b.$