Determinants - Existence

Determinants are functions which take matrices as inputs and produce numbers. They are of enormous importance in linear algebra, but perhaps you've also seen them in other courses. They're used to define the cross product of two 3-dimensional vectors. They appear in Jacobians which occur in the change-of-variables formula for multiple integrals.

In this section, I'll define determinants as functions satisfying three axioms. I'll show that there is at least one determinant function --- namely, one defined by cofactor expansion. I'll show how you can use row operations and cofactor expansion to compute determinants.

Later, I'll give a second formula for determinants based on permutations. This will allow me to show that there is only one determinant function satisfying the axioms. I'll also prove many important properties of determinants, such as the multiplication rule.

Determinants take as inputs $n \times
   n$ matrices with entries in R, where R is a commutative ring with identity. The set of such matrices is denoted $M(n, R)$ .

Definition. A determinant function is a function $D: M(n,R)
   \rightarrow R$ such that:

  1. D is a linear function in each row. That is, if $a \in R$ and $x, y \in R^n$ ,

$$D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & ax + y & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] = a\cdot D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] + D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right].$$

  1. A matrix with two equal rows has determinant 0:

$$D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] = 0.$$

  1. $D(I) = 1$ , where I is the $n
   \times n$ identity matrix.

Example. ( Linearity) The first axiom allows you to add or subtract, or move constants in and out, in a single row, assuming that all the other rows stay the same. Here's a $3 \times 3$ addition example, with all the action taking place in row 3:

$$D\left[\matrix{a & b & c \cr d & e & f \cr x_1 & x_2 & x_3 \cr}\right] + D\left[\matrix{a & b & c \cr d & e & f \cr y_1 & y_2 & y_3 \cr}\right] = D\left[\matrix{a & b & c \cr d & e & f \cr x_1 + y_1 & x_2 + y_2 & x_3 + y_3 \cr}\right].$$

Here's a $3 \times 3$ subtraction example, with all the action taking place in row 2:

$$D\left[\matrix{a & b & c \cr x_1 & x_2 & x_3 \cr d & e & f \cr}\right] - D\left[\matrix{a & b & c \cr y_1 & y_2 & y_3 \cr d & e & f \cr}\right] = D\left[\matrix{a & b & c \cr x_1 - y_1 & x_2 - y_2 & x_3 - y_3 \cr d & e & f \cr}\right].$$

Here's a $3 \times 3$ multiplication example, with all the action taking place in row 1:

$$k\cdot D\left[\matrix{x_1 & x_2 & x_3 \cr a & b & c \cr d & e & f \cr}\right] = D\left[\matrix{kx_1 & kx_2 & kx_3 \cr a & b & c \cr d & e & f \cr}\right].$$

Here's an example where you "take apart" a determinant using linearity. Notice that the first two rows are the same in all the matrices; all the action takes place in the third row:

$$D\left[\matrix{ a & b & c \cr d & e & f \cr x_1 + ky_1 & x_2 + ky_2 & x_3 + ky_3\cr}\right] = D\left[\matrix{ a & b & c \cr d & e & f \cr x_1 & x_2 & x_3 \cr}\right] + D\left[\matrix{ a & b & c \cr d & e & f \cr ky_1 & ky_2 & ky_3\cr}\right] =$$

$$D\left[\matrix{ a & b & c \cr d & e & f \cr x_1 & x_2 & x_3 \cr}\right] + k\cdot D\left[\matrix{ a & b & c \cr d & e & f \cr y_1 & y_2 & y_3\cr}\right].$$

Finally, here's a numerical example with entries in $\real$ :

$$D\left[\matrix{ 1 & -3 & 4 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] + D\left[\matrix{ -1 & 3 & -4 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] = D\left[\matrix{ 1 + (-1) & -3 + 3 & 4 + (-4) \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] = D\left[\matrix{ 0 & 0 & 0 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right].$$

You might suspect that a matrix with an all-zero row has determinant 0, and it's easy to prove using linearity:

$$D\left[\matrix{ 0 & 0 & 0 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] = D\left[\matrix{ 0 + 0 & 0 + 0 & 0 + 0 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] = D\left[\matrix{ 0 & 0 & 0 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] + D\left[\matrix{ 0 & 0 & 0 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right],$$

$$D\left[\matrix{ 0 & 0 & 0 \cr 2 & 0 & 17 \cr 7 & -6 & 1 \cr}\right] = 0.\quad\halmos$$


Example. ( Equal rows) The second axiom says that if a matrix has two equal rows, then its determinant is 0. Here's a $3 \times
   3$ example:

$$D\left[\matrix{a & b & c \cr d & e & f \cr a & b & c \cr}\right] = 0. \quad\halmos$$


Lemma. If $D: M(n,R) \to R)$ is a function which is linear in the rows (Axiom 1) and is 0 when a matrix has equal rows (Axiom 2), then swapping two rows multiplies the value of D by -1:

$$D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr & \vdots & \cr}\right] = -D\left[\matrix{& \vdots & \cr \leftarrow & r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr}\right].$$

Proof. The proof will use the first and second axioms repeatedly. The idea is to swap rows i and j by adding or subtracting rows.

In the diagrams below, all the rows except the $i^{\rm th}$ and the $j^{\rm th}$ remain unchanged.

$$\matrix{D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr & \vdots & \cr}\right] & = & D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr & \vdots & \cr}\right] + D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 2)} \cr & = & D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_i + r_j & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 1)} \cr & = & D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_i + r_j & \rightarrow \cr & \vdots & \cr}\right] - D\left[\matrix{& \vdots & \cr \leftarrow & r_i + r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_i + r_j & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 2)} \cr & = & D\left[\matrix{& \vdots & \cr \leftarrow & -r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_i + r_j & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 1)} \cr & = & D\left[\matrix{& \vdots & \cr \leftarrow & -r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_i + r_j & \rightarrow \cr & \vdots & \cr}\right] + D\left[\matrix{& \vdots & \cr \leftarrow & -r_j & \rightarrow \cr & \vdots & \cr \leftarrow & -r_j & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 2)} \cr & = & D\left[\matrix{& \vdots & \cr \leftarrow & -r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 1)} \cr & = & -D\left[\matrix{& \vdots & \cr \leftarrow & r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr}\right] & \hbox{(Axiom 1)} \cr}$$

Notice that in each addition or subtraction step, only one row changes at a time.

Remarks. (a) I'll show later that it's enough to assume (instead of Axiom 2) that $D(A) = 0$ vanishes whenever two {\it adjacent} rows of A are equal.

(b) Suppose that $D: M(n,R) \to R$ is a function satisfying Axioms 1 and 3, and suppose that swapping two rows multiplies the value of D by -1. Must D satisfy Axiom 2? In other words, is "swapping multiplies the value by -1" equivalent to "equal rows means determinant 0"?

Assuming that swapping two rows multiplies the value of D by -1, I have

$$D(A) = D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] = -D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] = -D(A).$$

(I swapped the two equal x-rows, which is why the matrix didn't change. But by assumption, this useless swap multiplies D by -1.)

Hence, $2\cdot D(A) = 0$ .

If R is $\real$ , $\rational$ , $\complex$ , or $\integer_n$ for n prime and not equal to 2, then $2\cdot D(A) = 0$ implies $D(A) = 0$ . However, if $R = \integer_2$ , then $2x = 0$ for all x. Hence, $2\cdot D(A) = 0$ , no matter what $D(A)$ is. Therefore, Axiom 2 need not hold. You can see, however, that it will hold if R is a field of characteristic other than 2.

Fortunately, since I take "equal rows means determinant 0" as an axiom for determinants, and since the lemma shows that this implies that "swapping rows multiplies the determinant by -1", I know that both of these properties will hold for determinant functions.


Example. ( Computing determinants using the axioms) Suppose that $D: M(3,\real) \to \real$ is a determinant function and

$$D\left[\matrix{\leftarrow & a_1 & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & c & \rightarrow \cr}\right] = 5 \quad\hbox{and}\quad D\left[\matrix{\leftarrow & a_2 & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & c & \rightarrow \cr}\right] = -3.$$

Compute

$$D\left[\matrix{\leftarrow & c & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & a_1 + 4a_2 & \rightarrow \cr}\right].$$

$$D\left[\matrix{\leftarrow & c & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & a_1 + 4a_2 & \rightarrow \cr}\right] = -D\left[\matrix{\leftarrow & a_1 + 4a_2 & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & c & \rightarrow \cr}\right] =$$

$$-\left(D\left[\matrix{\leftarrow & a_1 & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & c & \rightarrow \cr}\right] + 4D\left[\matrix{\leftarrow & a_2 & \rightarrow \cr \leftarrow & b & \rightarrow \cr \leftarrow & c & \rightarrow \cr}\right]\right) = -(5 + 4\cdot (-3)) = 7.\quad\halmos$$


I want to contruct a determinant function for $n \times n$ matrices. I'll begin by constructing one for $2 \times 2$ matrices; you may have seen this in (say) a multivariable calculus course.

Proposition. Define $|\cdot|: M(2,R) \rightarrow R$ by

$$\left|\matrix{a & b \cr c & d \cr}\right| = ad - bc.$$

Then $|\cdot|$ is a determinant function.

Proof. First,

$$\left|\matrix{a & b \cr a & b \cr}\right| = ab - ab = 0.$$

Therefore, Axiom 2 holds.

Next,

$$\left|\matrix{ka + a' & kb + b' \cr c & d \cr}\right| = kad + a'd - kbc - b'c = k(ad - bc) + (a'd - b'c) = k \left|\matrix{a & b \cr c & d \cr}\right| + \left|\matrix{a' & b' \cr c & d \cr}\right|.$$

$$\left|\matrix{a & b \cr kc + c' & kd + d' \cr}\right| = a(kd + d') - b(kc + c') = k(ad - bc) + (ad' - bc') = k\left|\matrix{a & b \cr c & d \cr}\right| + \left|\matrix{a & b \cr c' & d' \cr}\right|.$$ This proves linearity.

Finally,

$$\left|\matrix{1 & 0 \cr 0 & 1 \cr}\right| = 1.\quad\halmos$$


Example. On $M(2,\real)$ ,

$$\left|\matrix{1 & 2 \cr 3 & 4 \cr}\right| = 4 - 6 = -2.\quad\halmos$$


Example. ( Determinants and elementary row operations) How are determinants affected by elementary row operations?

The lemma I proved earlier shows that swapping two rows multiplies the determinant by -1. By linearity, multiplying a row by a multiplies the determinant by a.

Consider the operation of adding a multiple of a row to another row. Suppose, for example, I'm performing the operation $r_i \to r_i + a\cdot r_j$ . Let

$$A = \left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr}\right].$$

Then

$$D\left[\matrix{& \vdots & \cr \leftarrow & r_i + a\cdot r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr}\right] = D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr}\right] + a\cdot D\left[\matrix{& \vdots & \cr \leftarrow & r_j & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr}\right] = D\left[\matrix{& \vdots & \cr \leftarrow & r_i & \rightarrow \cr & \vdots & \cr \leftarrow & r_j & \rightarrow \cr}\right] = D(A).$$

Therefore, this kind of row operation leaves the determinant unchanged.

For example,

$$\left|\matrix{1 & 2 \cr 3 & 4 \cr}\right| = 4 - 6 = -2.$$

If I swap the rows, I get

$$\left|\matrix{3 & 4 \cr 1 & 2 \cr}\right| = 6 - 4 = 2.$$

The determinant was multiplied by -1.

If I multiply the second row of the original matrix by 5, I get

$$\left|\matrix{1 & 2 \cr 15 & 20 \cr}\right| = 20 - 30 = -10 = 5\cdot (-2).$$

This is 5 times the original determinant.

Finally, suppose I subtract 3 times row 1 from row 2. I get

$$\left|\matrix{1 & 2 \cr 0 & -2 \cr}\right| = -2 - 0 = -2.$$

The determinant was unchanged.


Example. ( Computing a determinant using row operations) Show that

$$\left|\matrix{x_1 - y_2 & x_2 - y_2 & x_3 - y_3 \cr y_1 & y_2 & y_3 \cr x_1 + z_1 & x_2 + z_2 & x_3 + z_3 \cr}\right| = \left|\matrix{x_1 & x_2 & x_3 \cr y_1 & y_2 & y_3 \cr z_1 & z_2 & z_3 \cr}\right|.$$

I'll prove the result by performing elementary row operations. Recall that adding a multiple of a row to another row doesn't chance the determinant. So

$$\left|\matrix{x_1 - y_2 & x_2 - y_2 & x_3 - y_3 \cr y_1 & y_2 & y_3 \cr x_1 + z_1 & x_2 + z_2 & x_3 + z_3 \cr}\right| \matrix{\vphantom{x} \cr = \cr r_1 \to r_1 + r_2 \cr} \left|\matrix{x_1 & x_2 & x_3 \cr y_1 & y_2 & y_3 \cr x_1 + z_1 & x_2 + z_2 & x_3 + z_3 \cr}\right| \matrix{\vphantom{x} \cr = \cr r_3 \to r_3 - r_1 \cr} \left|\matrix{x_1 & x_2 & x_3 \cr y_1 & y_2 & y_3 \cr z_1 & z_2 & z_3 \cr}\right|.\quad\halmos$$


When you define a mathematical object by axioms, two questions arise immediately:

1. ( Existence) Are there any objects which satisfy the axioms?

2. ( Uniqueness) Is there more than one object which satisfies the axioms?

I defined above a determinant function on $2 \times 2$ matrices. This solves the existence problem for determinant functions on $M(2, R)$ , but I want to show that there is a determinant function on $M(n, R)$ . I'll do this inductively, using Expansion by cofactors.

Later, I'll show that the determinant function defined in this way is the only determinant function on $M(n, R)$ . This solves the {\it uniqueness} problem for determinants on $M(n, R)$ .

Mathematicians often proceed in this way: Define an object by identifying properties which characterize it, rather than simply writing down a formula. I could have defined the determinant by cofactor expansion, but it says more about what determinants really are to characterize them by the axioms given above.

Moreover, the axiomatic approach will provide a clean proof of properties such as multiplicativity. Try proving that $D(AB) = D(A)D(B)$ from the cofactor expansion!

Earlier I discussed the connection between "swapping two rows multiplies the determinant by -1" and "when two rows are equal the determinant is 0". The next lemma is another piece of this picture. It says for a function which is linear in the rows, if "when two {\it adjacent} rows are equal the determinant is 0", then "swapping two rows multiplies the determinant by -1". (Axiom 2 does not require that the two equal rows be adjacent.)

Lemma. Let $f: M(n,R) \rightarrow R$ be a function which is linear in each row and satisfies $f(A) = 0$ whenever two adjacent rows are equal. Then swapping (any) two rows multiplies the value of f by -1.

Proof. First, I'll show that swapping two adjacent rows multiplies the value of f by -1. I'll show the required manipulations in schematic form:

$$f\left[\matrix{& \vdots & \cr \leftarrow & x & \rightarrow \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr}\right] = f\left[\matrix{& \vdots & \cr \leftarrow & y + (x - y) & \rightarrow \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr}\right] = f\left[\matrix{& \vdots & \cr \leftarrow & y & \rightarrow \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr}\right] + f\left[\matrix{& \vdots & \cr \leftarrow & x - y & \rightarrow \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr}\right] =$$

$$f\left[\matrix{& \vdots & \cr \leftarrow & x - y & \rightarrow \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr}\right] = f\left[\matrix{& \vdots & \cr \leftarrow & x - y & \rightarrow \cr \leftarrow & x + (y - x) & \rightarrow \cr & \vdots & \cr}\right] = f\left[\matrix{& \vdots & \cr \leftarrow & x - y & \rightarrow \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr}\right] + f\left[\matrix{& \vdots & \cr \leftarrow & x - y & \rightarrow \cr \leftarrow & y - x & \rightarrow \cr & \vdots & \cr}\right] =$$

$$f\left[\matrix{& \vdots & \cr \leftarrow & x & \rightarrow \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr}\right] - f\left[\matrix{& \vdots & \cr \leftarrow & y & \rightarrow \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr}\right] - f\left[\matrix{& \vdots & \cr \leftarrow & x - y & \rightarrow \cr \leftarrow & x - y & \rightarrow \cr & \vdots & \cr}\right] = -f\left[\matrix{& \vdots & \cr \leftarrow & y & \rightarrow \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr}\right].$$

To complete the proof, I must show that swapping two non-adjacent rows multiplies the value of f by -1. Since swapping adjacent rows multiplies the determinant by -1, it will suffice to show that swapping any two rows can be accomplished by an odd number of adjacent row swaps.

Without loss of generality, then, I may suppose the rows to be swapped are rows 1 and n. I'll indicate how to do the swaps by just displaying the row numbers. First, I do $n - 1$ adjacent row swaps to move row 1 into the n-th position:

$$\matrix{ 1 & 2 & 3 & 4 & \ldots & n - 2 & n - 1 & n \cr 2 & 1 & 3 & 4 & \ldots & n - 2 & n - 1 & n \cr & & \vdots & & \ldots & & \vdots & \cr 2 & 3 & 4 & 5 & \ldots & n - 1 & 1 & n \cr 2 & 3 & 4 & 5 & \ldots & n - 1 & n & 1 \cr}$$

Now I do $n - 2$ more swaps to move (the old) row n into the first position:

$$\matrix{ 2 & 3 & 4 & 5 & \ldots & n - 1 & n & 1 \cr 2 & 3 & 4 & 5 & \ldots & n & n - 1 & 1 \cr & & \vdots & & \ldots & & \vdots & \cr 2 & 1 & 3 & 4 & \ldots & n - 2 & n - 1 & 1 \cr 1 & 2 & 3 & 4 & \ldots & n - 2 & n - 1 & 1 \cr}$$

I've done a total of $(n - 1) + (n -
   2) = 2n - 3$ swaps, an odd number. This proves the result.

Definition. Let $A \in M(n,R)$ . Let $A(i\mid j)$ be the $(n - 1)
   \times (n - 1)$ matrix obtained by deleting the i-th row and j-th column of A. If D is a determinant function, then $D[A(i\mid j)]$ is called the $(i,j)^{\rm th}$ minor of A.

The $(i,j)^{\rm th}$ cofactor of A is $(-1)^{i+j}$ times the $(i,j)^{\rm
   th}$ minor, i.e. $(-1)^{i+j}\cdot D[A(i\mid j)]$ .


Example. Consider the real matrix

$$A = \left[\matrix{1 & 2 & 3 \cr 4 & 5 & 6 \cr 7 & 8 & 9 \cr}\right].$$

To find the $(2,3)^{\rm th}$ minor, strike out the $2^{\rm nd}$ row and the $3^{\rm
   rd}$ column (i.e. the row and column containing the $(2,3)^{\rm
   th}$ element):

$$\left[\matrix{1 & 2 & * \cr * & * & * \cr 7 & 8 & * \cr}\right].$$

Take the determinant of what's left:

$$\left|\matrix{1 & 2 \cr 7 & 8 \cr}\right| = 8 - 14 = -6.$$

To get the $(2,3)^{\rm th}$ cofactor, multiply this by $(-1)^{2+3} = (-1)^5 = -1$ . I get $(-1)\cdot (-6) = 6$ .

The easy way to remember whether to multiply by $+1$ or -1 is to make a checkboard pattern of +'s and -'s:

$$\left[\matrix{+ & - & + \cr - & + & - \cr + & - & + \cr}\right].$$

Use the sign in the $(i,j)^{\rm
   th}$ position. For example, there's a minus sign in the $(2,3)^{\rm th}$ position, which agrees with the sign I computed using $(-1)^{i+j}$ .


The next result says that I can use cofactors to extend a determinant function on $(n - 1) \times (n -
   1)$ matrices to a determinant function on $n \times n$ matrices.

Theorem. Let C be a determinant function on $M(n-1, F)$ . For any $j \in
   \{1, \ldots, n\}$ , define

$$D(A) = \sum_i (-1)^{i+j} A_{ij} C\left(A(i\mid j)\right).$$

Then D is a determinant function on $M(n, R)$ .

Notice that the summation is on i, which is the row index. That means you're moving down a column as you sum. Consequently, this is a cofactor expansion by columns.

Proof. I need to show that D is linear in each row, D is alternating, and $D(I) = 1$ .

Linearity: I'll prove linearity in row k. Let $a \in F$ , $x, y \in F^n$ . I want to prove that

$$D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & ax + y & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] = a\cdot D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right] + D\left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right].$$

(All the action is taking place in the k-th row --- the one with the x's and y's --- and the other rows are the same in the three matrices).

Label the three matrices above:

$$P = \left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & ax + y & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right], \quad Q = \left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & x & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right], \quad R = \left[\matrix{\leftarrow & r_1 & \rightarrow \cr & \vdots & \cr \leftarrow & y & \rightarrow \cr & \vdots & \cr \leftarrow & r_n & \rightarrow \cr}\right].$$

Expand each of the D's in the equation using the cofactor summation. Since C is a determinant function, C is linear. For $i \ne k$ , then, the terms of the summation on the two sides clearly agree.

Consider the terms generated on both sides when $i = k$ . On the left, I have

$$(-1)^{k+j} (ax_j + y_j) C\left(P(k\mid j)\right).$$

On the right, I have

$$(-1)^{k+j} ax_j C\left(Q(k\mid j)\right) + (-1)^{k+j} y_j C\left(R(k\mid j)\right).$$

However,

$$C\left(P(k\mid j)\right) = C\left(Q(k\mid j)\right) = C\left(R(k\mid j)\right),$$

because P, Q, and R only differ in row k, which is being deleted.

Therefore, the terms on the left and right are the same for all i, and D is linear.

Alternating: I have to show that if two rows of A are the same, then $D(A) = 0$ . First, I'll show that if two adjacent rows of A are the same, then $D(A) = 0$ .

Suppose without loss of generality that rows 1 and 2 are equal. ("Without loss of generality" means that the same argument will work for any two adjacent rows.) All the terms in the cofactor expansion are 0 except the first and second, since in all other cases the determinant function C will be applied to a matrix with two equal rows.

The first and second terms are

$$(-1)^{1+j} A_{1j} C\left(A(1\mid j)\right) + (-1)^{2+j} A_{2j} C\left(A(2\mid j)\right).$$

Now $A_{1j} = A_{2j}$ , since the first and second rows are equal. Likewise, since the rows are equal, I get the same matrix by deleting one or the other. Therefore, $C\left(A(1\mid j)\right) = C\left(A(2\mid j)\right)$ . The only things left are the signs, and $(-1)^{1+j}$ and $(-1)^{2+j}
   A_{2j}$ have opposite parity. Therefore, the cofactor expansion is equal to 0, and $D(A) = 0$ , as I wished to prove.

By applying the preceding lemma, I know that swapping two rows multiplies the value of D by -1.

Next, suppose two rows of A (which are not necessarily adjacent) are equal. I can swap rows until I get a matrix B which has two equal rows that are adjacent --- let's say that k swaps are needed to do this. Then

$$D(A) = (-1)^kD(B) = (-1)^k\cdot 0 = 0.$$

This completes the proof that $D(A)
   = 0$ if A has two equal rows.

The identity has determinant 1: Suppose $A = I$ . Notice that $A_{ij} = 0$ unless $i = j$ . Therefore, the cofactor expansion of $D(A)$ reduces to

$$(-1)^{j+j} A_{jj} C\left(A(j\mid j)\right) = 1\cdot 1\cdot C(I) = 1.\quad\halmos$$

Notation. If A is an $n \times n$ matrix, the determinant of A (as defined by the cofactor expansion) will be denoted $|A|$ or $\det A$ .


Example. ( Computing a determinant by cofactors) You can use cofactor expansion to compute determinants. I'll show later on that $\det A = \det A^T$ . Since transposing sends rows to columns, this means that you can expand by cofactors of rows as well as columns.

You can expand along any row or column you want, but it's usually good to pick one with lots of 0's. For example, consider the following real matrix. Expanding along the second column, I get

$$\left|\matrix{1 & 3 & -5 \cr 1 & 0 & -2 \cr 6 & 1 & 1 \cr}\right| = (-1)(3) \left|\matrix{1 & -2 \cr 6 & 1 \cr}\right| + (1)(0) \left|\matrix{1 & -5 \cr 6 & 1 \cr}\right| + (-1)(1) \left|\matrix{1 & -5 \cr 1 & -2 \cr}\right| = -42. \quad\halmos$$


Example. ( Computing a determinant using row operations and cofactors) You can use row operations to simplify a matrix before expanding by cofactors. For example, I'll compute the determinant of the following matrix in $M(3, \real)$ :

$$\matrix{ \left|\matrix{1 & 3 & -5 \cr 1 & 0 & -2 \cr 6 & 1 & 1 \cr}\right| & = & \left|\matrix{1 & 3 & -5 \cr 0 & -3 & 3 \cr 6 & 1 & 1 \cr}\right| & (r_2 \to r_2 - r_1) \cr & = & \left|\matrix{1 & 3 & -5 \cr 0 & -3 & 3 \cr 0 & -17 & 31 \cr}\right| & (r_3 \to r_3 - 6r_1) \cr & = & (1)(1)\left|\matrix{-3 & 3 \cr -17 & 31 \cr}\right| & (\hbox{Cofactors of column 1}) \cr & = & -42 & \cr}$$

Again anticipating the later result that $\det A = \det A^T$ , it follows that you can perform elementary column operations on a matrix when computing its determinant. But be careful! Column operations are {\it not} permitted in most other situations.


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