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

Definition 1: Row-Column Equivalence

Two matrices A,B are row-column equivalent if and only if A can be obtained from B by a sequence of operations each of which is an elementary row operation (see Definition 4, here) or an elementary column operation (see Definition 4, here).

Definition 2: RC-Canonical Form

An n x m matrix C is in RC-Canonical Form if for some 0 ≤ r ≤ min(n,m):

if i ≤ r and i = j, then c

_{i,j}= 1

otherwise: c

_{i,j}= 0.

This is similar to reduced echelon form (see Definition 2, here) except that all values are either 0 or 1 and the only nonzero values are leading entries (see Definition 2, here). If 0

_{s}is an n x n matrix that consists entirely of 0's, then we can think of a matrix in RC-canonical form as:

Theorem 1:

If A is row-column equivalent to B, then there exists invertible matrices P,Q such that:

C = PAQ

Proof:

(1) Assume that A is row-column equivalent to B.

(2) Thus, B is obtained from A by a sequence of elementary row operations and elementary column operations.

(3) Let us separate out the row operations so that we have E

_{1}E

_{2}*...*E

_{n}

[See Theorem 5, here for proof that applying elementary row operations is equivalent to a product of elementary matrices.]

(4) Let us separate out the column operations so that we have:

F

_{1}F

_{2}*...*F

_{n}

[See Theorem 5, here for proof that applying elementary column operations is equivalent to a product of elementary matrices.]

(5) Thus, B = E

_{1}*...*E

_{n}AF

_{1}*...*F

_{n}

(6) Let P = E

_{1}*...*E

_{n}

(7) Let Q = F

_{1}*...*F

_{n}

(8) It is clear that P is invertible since it is the product (see Corollary 3.1, here) of E

_{i}which are each invertible (See Lemma 4, here).

(9) It is also clear that Q is invertible since it is the product (see Corollary 3.1, here) of F

_{i}which are each invertible (see Lemma 4, here).

QED

Corollary 1.1: If A and C are row-column equivalent, then there exists a matrix B such that A is row equivalent to B and B is column equivalent to C.

Proof:

(1) Let A be row-column equivalent to C.

(2) Then, there exists P,Q that are invertible such that C = PAQ where P is a product of elementary row operations and Q is a product of elementary column operations. [By Theorem 1 above]

(3) Let B = PA. It is clear that A is row equivalent to B since P is a product of elementary row operations. [See Theorem 5, here]

(4) It is also clear that C = (PA)Q = BQ. Since Q is a product of elementary column operations, so that C is column equivalent to B. [See Theorem 5, here]

QED

Definition 3: I

_{r}+0

This is an m x n matrix where r ≤ min(m,n) and:

a

_{i,j}= 1 if i ≤ r and i=j

otherwise, a

_{i,j}= 0

As should be clear, I

_{r}+0 is an m x n matrix that is in RC-Canonical Form.

Definition 4: rank

Since the row rank of a matrix A = the column rank of matrix A (see Theorem 2, here) we can refer to this value as the rank of A.

Lemma 2: For each r where 0 ≤ r ≤ min(m,n), there is precisely one m x n matrix in RC-canonical form of rank r that equals I

_{r}+0.

Proof:

(1) I

_{r}+0 is an m x n matrix in RC-canonical form of rank r so there is at least one.

(2) Let C be an m x n matrix in RC-canonical form of rank r.

(3) Let's assume that C corresponds to I

_{s}+0

(4) Then the row rank of C is s since the first s rows of I

_{s}+0 are linearly independent.

[See Definition 1, here for definition of linear independence; see Definition 2, here for definition of row rank]

(5) But then s=r since the row rank of C is r and we have shown that C=I

_{s}+0 = I

_{r}+0 since both I

_{s}+0 and I

_{r}+0 are n x m matrices.

QED

Lemma 3: If A is row column equivalent to B, then rank A = rank C.

Proof:

(1) If A is row column equivalent to B, then there exists a matrix C such that A is row equivalent to C and C is column equivalent to B. [See Corollary 1.1 above]

(2) Now, A has the same row rank as C.

[See Corollary 1.1, here for proof that they have the same row space; See Definition 2 here to see that same row space implies same row rank]

(3) C has the same column rank as B.

[See Corollary 1.1, here for proof that they have the same column space; See Definition 3 here to see that same column space implies same column rank]

(4) The column rank of C = the row rank of C [See Theorem 2, here]

(5) So we can conclude that that the row rank of A = column rank of C which is the same as saying that the rank of A = rank of C.

QED

Theorem 4: Existence of RC-canonical form equivalent

Every m x n matrix is row-column equivalent to a matrix in RC-canonical form

Proof:

(1) Every m x n matrix is row equivalent to a unique matrix in reduced echelon form. [See Corollary 3.1, here]

(2) We can assume that a matrix B is in reduced echelon form and that it has a nonzero element b

_{h,k}which is not a leading entry.

If a reduced echelon form does not have such an element, then it is already in RC-canonical form.

(3) Let l(i) be the column for each leading entry in row i.

(4) We can now apply the elementary column operation: A

_{(kCj + Ci)}with k = -b

_{h,k}and j = l(i) to column i.

(5) Since B is in reduced echelon form, we know that column l(i) is all zeros except for row i where it is 1.

(6) In this way we can see that the column operation A

_{((-bh,k)Cl(i) + Ci)}results in b

_{h,k}=0 but does not change any other value in the column.

(7) After repeatedly applying this column operation, we have removed all nonzero values which are not leading entries.

(8) Since these nonzero elements which were not leading entries could not occur in the same column as a leading entry (see Definition 2, here for definition of reduced echelon form), we now are left with columns that are entirely zero.

(9) We can use the column operation A

_{(Ci ↔ Cj)}to move all the zero columns after the nonzero columns.

(10) We now have r nonzero rows where 1 ≤ l(1) is less than l(2) is less than ... is less than l(r) ≤ r.

(11) It is clear that B is in RC-canonical form.

QED

Theorem 5: For every matrix, there is only one unique RC-canonical form that is row-column equivalent.

Proof:

(1) Let B,C be two matrixes in RC-canonical form that are row-column equivalent to a matrix A.

(2) Let r = the rank of A.

(3) By Lemma 3 above, it is clear that rank of B = rank of A and likewise rank of C = rank of A so therefore rank of B = rank of C = r.

(4) But then, by Lemma 2 above, B = C since there is only one m x n matrix in rc-canonical form with rank = r.

(5) This shows that there is only one matrix in RC-canonical form which is RC-equivalent to A.

QED

Theorem 6:

All matrices are the product of elementary matrices

Proof:

(1) Let A be an m x n matrix.

(2) Let A

_{rc}be the unique rc-canonical form of A. [See Theorem 4 above for proof of its existence and Theorem 5 above for proof of its uniqueness]

(3) If the rank of A

_{rc}= r, then A

_{rc}=I

_{r}+0.

(4) Now, I

_{r}+0 = I

_{m}*I

_{(0Rr+1)}*I

_{(0Rr+2)}*...*I

_{(0Rm)}.

(5) So, we can see that I

_{r}+0 is the product of elementary matrices.

(6) Since A is row column equivalent to A

_{rc}, it follows that there exists invertible matrices P,Q such that

**A = P(A**

_{rc})Q. [See Theorem 1 above]

(7) Since P,Q are invertible, then they are composed of elementary matrices. [See Theorem 5, here]

(8) Therefore, we can conclude A = P(A

_{rc})Q is composed of elementary matrices since P is composed of elementary matrices, A

_{rc}is composed of elementary matrices, and Q is composed of elementary matrices.

QED

References

- Charles G. Cullen, Matrices and Linear Transformations, Dover Publications, Inc., 1972.
- Hans Schneider, George Philip Barker, Matrices and Linear Algebra, 1989.

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