Elementary Matrices

In a previous blog, I talked elementary matrices and showed that if two matrices A,B are row equivalent, then there exists a matrix P such that A = PB where P is a product of elementary matrices (see Theorem 5, here).

Today, I will extend this result and show that if a matrix A is invertible (see Definition 1, here), then it is equal to product of elementary matrices.

The content in today's blog is taken from Matrices and Linear Algebra by Hans Schneider and George Phillip Barker.

Lemma 1: Only square matrices are invertible.

Proof:

(1) Let A be an n x m matix where n ≠ m.

(2) To be invertible, there must exist a matrix B such that AB = BA = I (see Definition 1, here)

(3) Let's assume that B is a p x q matrix.

(4) In order for A to multiply with B, then p = m and the result AB is an n x q matrix. (See here for review of matrix multiplication)

(5) In order for B to multiply with A, q must = n and BA is then a p x m matrix.

(6) Since p must = m and q must = n, AB is an n x n matrix and BA is an m x m matrix.

(7) But AB ≠ BA since m ≠ n so there is no such B that satisfies the conditions in step #2.

(8) So, we reject our assumption in step #1 and conclude that n = m and A is therefore a square matrix.

QED

From a previous result, we know that each m x n matrix has a unique reduced echelon form (see Corollary 3.1, here).

Definition 1: A_r

I will denote the unique reduced echelon form of a given m x n matrix A as A_r.

[For review of reduced echelon form, see Definition 2, here]

Definition 2: Rank

The rank r of an m x n matrix A is the number of nonzero rows in its reduced echelon form A_r

Lemma 2: If A is invertible, then A_r is invertible.

Proof:

(1) A_r = E₁E₂...E_nA where E_i is an elementary matrix [See Theorem 5, here]

(2) Since each elementary matrix is invertible [see Lemma 4, here], we know that E₁E₂....E_n is invertible. [See Lemma 3, here]

(3) So, we can multiply the inverse of E₁*...*E_n to both sides to get:

(E₁*...*E_n)^-1A = I*A_r=A_r

(4) But since (E₁*...*E_n)^-1 is invertible and A is invertible, then A_r must also be invertible [See Lemma 3, here].

QED

Lemma 3: If a matrix A is invertible, then its rank is equal to its number of rows.

Proof:

(1) Assume that A is invertible.

(2) Since A is invertible, we know that A_r is invertible. [See Lemma 2 above]

(3) Let X = A_r^-1, B = A_r

(4) From Lemma 1 above, we know that A,A_r is an n x n matrix.

(5) Let e_*,n = B*x_*,n [NOTE: By e_*,n, I mean the vector of values including e_1,n ... e_n,n]

(6) Since BX = I_n, we know that e_n,n = 1 [See here for review of I_n]

(7) But we also know that e_n,n = ∑ (i=1,n) b_n,i*x_i,n [See here for review of matrix multiplication]

(8) So, therefore, it follows that the last row of A_r is nonzero.

(9) Since A_r is in reduced echelon form, it follows that all rows must be nonzero. [see Definition 2, here]

(10) Therefore, it follows that rank of A_r is n. [See Definition 2 above]

QED

Corollary 3.1: If a matrix A is invertible and has n rows, then A is row equivalent to I_n

Proof:

(1) Assume that A is invertible.

(2) Let n = the number of rows of A.

(3) By Lemma 3 above, its rank = n.

(4) Since its rank is n, we know that each row of A_r is nonzero and therefore contains a leading coefficient.

(5) But since A,A_r are square matricies (see Lemma 1 above), we know that they have n columns each since they have n rows each.

(6) Let l(i) = the column of leading nonzero element for A_r.

(7) l(i) = i for each row i since 1 ≤ l(1) ≤ l(2) is less than l(3) is less than ... is less than l(n) ≤ n.

(8) Thus a_i,i = 1 for i = 1 .. n

(9) Hence, every column of A_r contains a leading coefficient, and all other entries of A are 0.

(10) It therefore follows that A_r = I_n (See here for review of I_n, see Definition 2, here for review of reduced echelon form.)

QED

Lemma 4: If a matrix A is row equivalent to I_n, then it is equal to a product of elementary matrices.

Proof:

(1) Assume that A is row equivalent to I_n.

(2) Then there exists P such that A = PI_n where P is a product of elementary matrices. [See Theorem 5, here]

(3) But I_n = the elementary matrix = I_(1R1) [See Defintion 1, here]

(4) So A is a product of elementary matrices.

QED

Theorem 5: A matrix M is invertible if and only if it is a product of elementary matrices

Proof:

(1) Assume that a matrix M is invertible.

(2) Let n = the number of rows of M.

(3) Using Corollary 3.1 above, it follows that M is row equivalent to I_n.

(4) Using Lemma 4 above, this means that M is a product of elementary matrices.

(5) Assume that M is the product of elementary matrices so that A = E₁*E₂*...*E_n

(6) Since each elementary matrix is invertible [see Lemma 4, here], it follows that their product is also invertible [see Lemma 3, here].

QED

References

• Hans Schneider, George Philip Barker, Matrices and Linear Algebra, 1989.