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Echelon matrices, Row canonical form, Row equivalence

Echelon matrices: A matrix A is called an echelon matrix, or is said to be in echelon form, if the following two conditions:

(1) All zero rows, if any, are at the bottom of the matrix.

(2) Each leading non-zero entry in a row is to the right of leading non-zero entry in the preceding row.

That is, A=[aij] is an echelon matrix if there exist nonzero entries

a1j1, a2j2, ———–,arjr where j1>j2>—–>jr

with the property that

aij=0     for           {i) i≤r,   j<ji   ii) i>r

The entries a1j1, a2j2, ———–,arjr which are the leading non-zero elements in their respective rows are called pivots of the echelon matrix.

Row Canonical form:

A matrix A is said to be in row canonical form if it is an echelon matrix, that is, if it satisfies the above properties (1) and (2), and if it satisfies the following additional two properties:

(3) Each pivot is equal to 1.

(4) Each pivot is the only nonzero entry in its column.

Example:

Screenshot_5

 

 

Row equivalence:

A matrix A is said to be row equivalent to a matrix B, written

A~B

if B can be obtained from A by a sequence of elementary row operations.

Rank of a matrix:

The rank of a matrix A, written rank (A), is equal to the number of pivots in an echelon form of A.

 

Theorem: Let A be a square matrix. Then the following are equivalent

(i) A is invertible(nonsingular)

(ii) A is row equivalent to the identity matrix I.

(iii) A is a product of elementary matrices.

 

Proof: Suppose A is invertible and suppose A is row equivalent to matrix B in row canonical form. Then there exist elementary matrices E1, E2 ,……, Es such that Es, …..,E2, E1A=B. Since A is invertible and each elementary matrix is invertible, B is also invertible. But if B≠I, then B has a zero row; whence B is not invertible. Thus B=I, (i) implies (ii).

If (ii) holds, then there exist elementary matrices E1, E2 ,……, Es such that Es, …..,E2, E1A=I. Hence A=( E1, E2 ,……, Es)-1= E-11, E-2 2 ,……, E-1 s. But the

E-1i are also elementary matrices. Thus (ii) implies (iii) .

If (iii) holds, then A= E1, E2 ,……, Es. The Ei are invertible matrices; hence their product A is also invertible. Thus (iii) implies (i).                                     (Proved )

 

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