## Thursday, June 07, 2007

### Determinant of an n x n matrix

In a previous blog, I defined the determinant of a 2 x 2 matrix as:

In today's blog, I will offer a more general definition that is taken from Matrices and Linear Transformations by Charles G. Cullen.

Before offering this definition, I need to establish some notation for elementary operations of matrices. See here for review of these elementary operations.

Definition 1: A(kRi)

This is the matrix that results from multiplying a scalar k to the ith row of the matrix A.

Definition 2: A(Ri ↔ Rj)

This is the matrix that results from interchanging row i with row j in the matrix A.

Definition 3: A(kRj + Ri)

This is the matrix that results from adding (a scalar k * row j) to row i in the matrix A.

Now, I can use these definitions to offer a more general definition of the determinant of an n x n matrix:

Definition 4: Determinant of an n x n matrix A det(A)

A function is said to be the determinant of an n x n matrix if:
(I) det(AB) = det(A)*det(B)
(II) det(I(kR1)) = k

NOTE: I(kR1) is the identity matrix In with k multiplied to row 1 in I. [See here for details on how modifying the identity matrix and how this relates to elementary operations.]

For example, I(5R1) =

since for identity I2:

Now, from our previous result, we have shown that the above rule applies to 2 x 2 matrices.

Lemma 1: 2 x 2 matrices are determinants by Definition 4 above

Proof:

(1) Let A, B be 2 x 2 matrices.

(2) det(AB) = det(A)det(B) [See Lemma 1 here]

(3) I(kR1) =

(4) Using the definition of determinant for 2 x 2 matrices gives us:

k*1 - 0*0 = k

(5) Thus, we have shown that our proposed definition is consistent with the previous definition for 2 x 2 matrices.

QED

Lemma 2: I(kRj) = I(Rj ↔ R1)I(kR1)I(Rj ↔ R1)

Proof:

(1) I =

(2) I(Rj ↔ R1) =

(3) I(kR1)I(Rj ↔ R1)

(4) I(Rj ↔ R1)I(kR1)I(Rj ↔ R1)

QED

Lemma 3: I(Rj ↔ R1)I(Rj ↔ R1) = I.

Proof:

(1) I =

(2) I(Rj ↔ R1) =

(3) I(Rj ↔ R1)I(Rj ↔ R1) =

QED

Lemma 4: det(I(kRj)) = k

Proof:

(1) det(I(kRj)) = det( I(Rj ↔ R1)I(kR1)I(Rj ↔ R1) ) [See Lemma 2 above]

(2) det( I(Rj ↔ R1)I(kR1)I(Rj ↔ R1) ) = det( I(Rj ↔ R1) )det( I(kR1) )det( I(Rj ↔ R1) ) [See Definition 4 above]

(3) det( I(Rj ↔ R1) )det( I(kR1) )det( I(Rj ↔ R1) ) = det( I(kR1) )det( I(Rj ↔ R1) )det( I(Rj ↔ R1) ) [from the Commutativity of scalars]

(4) det( I(kR1) )det( I(Rj ↔ R1) )det( I(Rj ↔ R1) ) = det( I(kR1) )det( I(Rj ↔ R1)I(Rj ↔ R1) ) [ See Definition 4 above]

(5) det( I(kR1) )det( I(Rj ↔ R1)I(Rj ↔ R1) ) = det( I(kR1) )det(I) [See Lemma 3 above]

(6) det( I(kR1) )det(I) = det(I(kR1))det(I(1R1))

(7) det( I(kR1))det(I(1R1)) = k*1 = k [See Definition 4 above]

QED

Corollary 4.1: A = I(kRj)B → det(A) = k*det(B)

Proof:

(1) A = I(kRj)B

(2) det(A) = det(I(kRj)B) = det(I(kRj))det(B) = k*det(B) [See Lemma 4 above]

QED

Corollary 4.2: det(I(0Ri)) = 0

Proof:

(1) det(I(0Ri)) = 0 [This follows directly from Lemma 4 above for k=0]

QED

Corollary 4.3: det(I) = 1

Proof:

(1) det(I) = det(I(1Ri)) = 1. [This follows directly from Lemma 4 above for k = 1]

QED

Lemma 5: I(kRj + Ri) = I(k-1Rj)I(1Rj + Ri)I(kRj)

Proof:

(1) I(kRj + Ri) =

(2) I(kRj) =

(3) I(1Rj + Ri)I(kRj) =

(4) I(k-1Rj)I(1Rj + Ri)I(kRj) =

QED

Lemma 6: I(k-1Rj)I(kRj) = I

Proof:

(1) I(kRj) =

(2) I(k-1Rj)I(kRj) =

QED

Lemma 7: det(I(kRj + Ri)) = 1

Proof:

(1) Assume k = 0

(2) det(I(kRj + Ri)) = det(I(0Rj + Ri)) = det(I) = det(I(kR1)) = 1 [See Definition 4 above]

(3) Assume k ≠ 0

(4) det(I(kRj + Ri)) =det( I(k-1Rj)I(1Rj + Ri)I(kRj) ) [See Lemma 5 above]

(5) det( I(k-1Rj)I(1Rj + Ri)I(kRj) ) = det( I(k-1Rj) )det( I(1Rj + Ri) )det( I(kRj) ) [See Definition 4 above]

(6) det( I(k-1Rj) )det( I(1Rj + Ri) )det( I(kRj) ) = det( I(1Rj + Ri) )det( I(k-1Rj) )det( I(kRj) ) [from the Commutativity of scalars]

(7) det( I(1Rj + Ri) )det( I(k-1Rj) )det( I(kRj) ) = det( I(1Rj + Ri) )det( I(k-1Rj) I(kRj) ) [see Definition 4 above]

(8) det( I(1Rj + Ri) )det( I(k-1Rj) I(kRj) ) = det( I(1Rj + Ri) )det( I ) [See Lemma 6 above]

(9) det( I(1Rj + Ri) )det( I ) = det( I(1Rj + Ri) I) = det(I(1Rj + Ri)) [See Definition 4 above]

(10) So, we have shown that: det(I(kRj + Ri)) = det(I(1Rj + Ri))

(11) Now, (I(1Rj + Ri))2 = (I(1Rj + Ri))(I(1Rj + Ri)) = I(2Rj + Ri)

(12) But for k=2, we have:

det( (I(1Rj + Ri))(I(1Rj + Ri))) = det(I(2Rj + Ri) ) = det(I(1Rj + Ri))

(13) Now, this can only be true if det(I(1Rj + Ri)) = 0 or det(I(1Rj + Ri)) = 1.

(14) We note that I(1Rj + Ri)I(-1Rj + Ri) = I.

(15) Since det(I) = 1, it follows that det(I(1Rj + Ri)I(-1Rj + Ri)) = 1 so it is impossible for det(I(1Rj + Ri)) = 0.

(16) Therefore, det(I(1Rj + Ri)) = 1.

QED

Corollary 7.1: A = I(kRj + Ri)B → det(A) = det(B)

Proof:

(1) A = I(kRj + Ri)B

(2) det(A) = det(I(kRj + Ri)B) = det(I(kRj + Ri))det(B) = 1*det(B) = det(B) [See Lemma 7 above]

QED

Lemma 8: I(Ri ↔ Rj) = I(-Ri)I(1Ri + Rj)I(-Rj + Ri)I(1Ri + Rj)

Proof:

(1) I(Ri ↔ Rj) =

(2) I(1Ri + Rj) =

(3) I(-Rj + Ri)I(1Ri + Rj) =

(4) I(1Ri + Rj)I(-Rj + Ri)I(1Ri + Rj) =

(5) I(-Ri)I(1Ri + Rj)I(-Rj + Ri)I(1Ri + Rj) =

QED

Lemma 9: det(I(Ri ↔ Rj)) = -1.

Proof:

(1) det( I(Ri ↔ Rj) ) = det( I(-Ri)I(1Ri + Rj)I(-Rj + Ri)I(1Ri + Rj) ) [See Lemma 8 above]

(2) det( I(-Ri)I(1Ri + Rj)I(-Rj + Ri)I(1Ri + Rj) ) = det( I(-Ri) )det( I(1Ri + Rj) )det( I(-Rj + Ri) )det( I(1Ri + Rj) ) [See Definition 4 above]

(3) det( I(-Ri) ) = -1 [See Lemma 4 above]

(4) det( I(1Ri + Rj) ) = det( I(-Rj + Ri) ) = 1 [See Lemma 7 above]

(5) So, det( I(-Ri) )det( I(1Ri + Rj) )det( I(-Rj + Ri) )det( I(1Ri + Rj) ) = (-1)*(1)*(1)*(1) = -1.

QED

Corollary 9.1: A =I(Ri ↔ Rj)B → det(A) = -det(B).

Proof:

(1) A =I(Ri ↔ Rj)B

(2) det(A) = det(I(Ri ↔ Rj)B) = det(I(Ri ↔ Rj))det(B) = (-1)*det(B) = -det(B)

QED

Lemma 10:

Let Ei be an elementary matrix

Let A =

then:

det(A) = det(Ei)

Proof:

(1) There are only three types of elementary matrices so this proof is complete if I show it is true for each type.

(2) Type I: Ei = I(kRi)

(a) From Lemma 4 above, det(Ei) = k

(b) A, in this case, is equal to I(kRi+1)

(c) Using Lemma 4 above, det(A) = k

(3) Type II: Ei = I(kRj + Ri)

(a) From Lemma 7 above, det(Ei) = 1

(b) A, in this case, is equal to I(kRj+1 + Ri+1)

(c) Using Lemma 7 above, det(A) = 1.

(4) Type III: Ei = I(Ri ↔ Rj)

(a) From Lemma 9 above, det(Ei) = -1.

(b) A, in this case, is equal to I(Ri+1 ↔ Rj+1)

(c) Using Lemma 9 above, det(A) = -1

QED

References