Properties of Determinants

One magic number per square matrix

This lecture is “like the beginning of the second half” of the course. Up to now the attention went to rectangular matrices; from here on it’s square matrices and two big topics: determinants and eigenvalues. The big reason we need determinants is for the eigenvalues — that’s the next big chunk of 18.06.

The one fact you’ve seen before, and will see again: $A$ is invertible exactly when $\det A \neq 0$; $A$ is singular exactly when $\det A = 0$. The determinant is a test for invertibility. But it has much more to it than that.

How to start? Strang could give the big formula for the determinant all in one gulp — “I don’t think so!” That formula has too much packed into it. Instead he starts with three properties, and those three properties define the determinant. Everything else, formula included, has to come out of them.

The three defining properties

Property 1. $\det I = 1$. The determinant of the identity is one.

Property 2. Exchanging two rows reverses the sign of the determinant. A lot of determinant work is exactly this: paying attention to plus and minus signs.

Notice what these two already give us. We know $\det I$, and every permutation matrix comes from the identity by row exchanges, so

$$\det P = \pm 1$$

— plus one if the number of exchanges was even, minus one if it was odd. In the two-by-two world:

$$\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1, \qquad \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1.$$

(Off to the side, Strang writes the formula we all know, $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$, and leaves it up — not as an assumption, but as a check. Every property should agree with it, and by the end of the lecture the properties will produce it.)

Property 3 is the key property, in two halves. Both are about the first row alone, with the other $n-1$ rows left unchanged:

$$\textbf{3A:}\quad \begin{vmatrix} ta & tb \\ c & d \end{vmatrix} = t \begin{vmatrix} a & b \\ c & d \end{vmatrix} \qquad\qquad \textbf{3B:}\quad \begin{vmatrix} a + a' & b + b' \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a' & b' \\ c & d \end{vmatrix}$$

In words: the determinant is a linear function of each row separately, when all the other rows stay the same.

Those are the three properties. Everything from here on must come from properties 1, 2, 3 — that’s all we have to work with.

Properties 4, 5, 6: what elimination does to a determinant

Property 4: two equal rows give determinant zero. Sure, in the two-by-two case $\begin{vmatrix} a & b \\ a & b \end{vmatrix} = ab - ab = 0$, but the argument has to work for a seven-by-seven with row one equal to row three. Here it is: exchange the two equal rows. The matrix doesn’t change — the rows were equal — so the determinant doesn’t change. But property 2 says the sign did change. A number whose sign both changes and doesn’t change must be zero. And this matches what we know: two equal rows mean rank less than $n$, a matrix that isn’t invertible.

Property 5: elimination doesn’t change the determinant. Subtracting a multiple $\ell$ of one row from another row leaves $\det A$ alone. There’s nothing special about row one here — any row $i$ subtracted from any row $k$. This is the property we really want, because it means we can take a full matrix whose determinant we don’t know, run elimination to clean it out below the diagonal, and

$$\det A = \det U .$$

The proof, in the two-by-two case, uses exactly the properties in order:

Property 6: a row of zeroes gives determinant zero. This is the singular case showing up: elimination might produce a complete row of zeroes, our old test for rank less than $n$. Why is the determinant zero? A student in the lecture has the slick answer — “you had the idea better” — take $t = 0$ in rule 3A. Multiplying the zero row by $t = 0$ changes nothing, but pulls a factor $0$ out front. So the determinant equals $0$ times itself: zero.

Property 7: triangular matrices, and the formula that software uses

Now the payoff. Suppose elimination has arrived at an upper triangular $U$ with pivots $d_1, d_2, \ldots, d_n$ on the diagonal, unknown stuff above, zeroes below.

Why don’t they matter? Run elimination upward: with nonzero pivots, multiples of lower rows kill every entry above the diagonal (property 5, so no change to the determinant), leaving a diagonal matrix. Then use 3A row by row: factor $d_1$ out of row one, $d_2$ out of row two, and so on. What’s left after all the $d$‘s come out? The identity — and finally rule 1 chips in: $\det I = 1$. So the determinant is the product of the $d$‘s. The proof used rule 5, rule 3A, and “our best friend, rule 1.” And if some pivot $d_i$ is zero, elimination produces a row of zeroes and rule 6 says the determinant is zero — the singular case checks too.

This is how MATLAB, or any reasonable software, actually computes a determinant of a matrix of size a hundred: eliminate to triangular, multiply the pivots.

$$\det A = \pm\, (\text{product of the pivots})$$

The plus or minus is there because row exchanges may have been needed along the way — each one flips the sign. If the simple LU code goes straight through with no exchanges, it’s exactly the product of the pivots.

Property 8 falls out: $\det A = 0$ exactly when $A$ is singular, and $\det A \neq 0$ exactly when $A$ is invertible. If $A$ is singular, elimination reaches a row of zeroes — determinant zero. If $A$ is invertible, elimination produces a full set of nonzero pivots — determinant is their (nonzero) product.

Properties 9 and 10: products and transposes

Two more, quick to write down and very, very useful.

This one comes partly out of the blue — we haven’t been multiplying matrices, and suddenly the determinant has this multiplying property. Look what it gives immediately. Since $A^{-1}A = I$, take determinants of both sides: $\det(A^{-1}) \det A = \det I = 1$, so

$$\det(A^{-1}) = \frac{1}{\det A}.$$

Check it on a diagonal matrix: if $A$ has diagonal $2, 3$ then $A^{-1}$ has diagonal $\frac{1}{2}, \frac{1}{3}$, and sure enough $6$ and $\frac{1}{6}$ are reciprocals. And notice how the rule agrees with singular/non-singular: if $\det A = 0$, then $1/\det A$ is ridiculous — and $A^{-1}$ indeed doesn’t exist.

Two more consequences: $\det(A^2) = (\det A)^2$ — square the matrix, square the determinant. But $\det(2A)$ is not $2 \det A$. Doubling the matrix doubles every row, and 3A factors a $2$ out of each of the $n$ rows:

$$\det(2A) = 2^n \det A .$$

Check on the two-by-two: both come out $ad - bc$. For a quick proof (covering almost every case), factor $A = LU$, so $A^T = U^T L^T$, and use rule 9 on both sides. It remains to compare $\det L \cdot \det U$ with $\det U^T \cdot \det L^T$. But $L$ and $L^T$ are triangular with ones on the diagonal — determinant one each — and $U$ and $U^T$ are triangular with the same diagonal, so they have the same determinant, the product of those diagonal entries. Both sides agree.

The consequence is bigger than it looks: everything said about rows now holds for columns. A column of zeroes gives determinant zero. Exchanging two columns reverses the sign — transpose, exchange the rows, transpose back. Nothing was special about rows.

One hidden fact closes the lecture. Property 2 would be inconsistent if some permutation could be reached by both seven exchanges and ten exchanges — the sign would have to be both $-1$ and $+1$, and “my whole lecture would die.” The rescue: every permutation is either odd or even. An odd permutation might take $7$, $21$, or $101$ exchanges, but never an even number. That little algebraic fact is what makes the determinant well-defined by properties 1, 2, 3 — with properties 4 through 10 following.

Problems

Everything here follows from the ten properties — no big formula needed.

Problem 18.1 Using the defining properties

Using only properties 1–3, find

$$\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} \qquad \text{and} \qquad \begin{vmatrix} 3 & 6 \\ 1 & 2 \end{vmatrix}.$$

Show solution

First matrix: it is a permutation matrix. Exchange rows 1 and 3 to reach the identity — one exchange, so the sign flips once:

$$\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} = -\det I = -1.$$

Second matrix: by 3A, factor $3$ out of row one:

$$\begin{vmatrix} 3 & 6 \\ 1 & 2 \end{vmatrix} = 3 \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 3 \cdot 0 = 0$$

using property 4 (two equal rows). The matrix is singular.

Problem 18.2 Determinant via pivots

Compute the determinant of

$$A = \begin{bmatrix} 2 & 1 & 0 \\ 4 & 5 & 3 \\ -2 & 2 & 7 \end{bmatrix}$$

by elimination: reduce to $U$ and multiply the pivots.

Show solution

Row 2 $-\,2\,\cdot$ Row 1 and Row 3 $+\,$ Row 1 (property 5, no change to the determinant):

$$\begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 3 \\ 0 & 3 & 7 \end{bmatrix}$$

Then Row 3 $-$ Row 2:

$$U = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 3 \\ 0 & 0 & 4 \end{bmatrix}$$

No row exchanges were needed, so

$$\det A = \det U = 2 \cdot 3 \cdot 4 = 24.$$

Problem 18.3 Product and inverse rules

Suppose $A$ is $4 \times 4$ with $\det A = 3$. Find $\det(A^2)$, $\det(A^{-1})$, $\det(2A)$, and $\det(A^T A)$.

Show solution

By property 9, $\det(A^2) = (\det A)^2 = 9$.

$\det(A^{-1}) = 1/\det A = \dfrac{1}{3}$, from taking determinants of $A^{-1}A = I$.

$\det(2A) = 2^4 \det A = 16 \cdot 3 = 48$: doubling the matrix doubles all four rows, and 3A pulls a factor $2$ out of each row.

By properties 9 and 10, $\det(A^T A) = \det(A^T)\det A = (\det A)^2 = 9$.

Problem 18.4 Sign of a permutation

A $5 \times 5$ permutation matrix $P$ reverses the order of the rows of the identity (row 1 of $P$ is row 5 of $I$, row 2 is row 4, and so on). What is $\det P$? Could this same permutation ever be produced by an even number of exchanges?

Show solution

Undo $P$ by exchanges: swap rows 1 and 5, then rows 2 and 4; row 3 is already in place. Two exchanges reach the identity, so

$$\det P = (-1)^2 \det I = +1.$$

It is an even permutation. Since every permutation is either odd or even — the hidden fact that makes property 2 consistent — this permutation can only be reached by an even number of exchanges: 2, or 4, or 100, but never 7. Otherwise the determinant would have to be $+1$ and $-1$ at once.

Problem 18.5 Reasoning from the properties

(a) Explain, using only the ten properties, why a matrix with a column of zeroes has determinant zero. (b) True or false, with a reason: if $\det A = 0$, then at least one row of $A$ is a combination of the other rows.

Show solution

(a) By property 10, $\det A = \det(A^T)$. Transposing turns the zero column into a zero row, and property 6 says a row of zeroes gives determinant zero. So $\det A = \det(A^T) = 0$.

(b) True. $\det A = 0$ means $A$ is singular (property 8), so elimination produces a row of zeroes: some row was eliminated entirely by subtracting multiples of the rows above it. That row was a linear combination of the others. (Equivalently: rank less than $n$ means the rows are dependent.)