Elementary row matrices

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Elementary row matrices

Recall the elementary row operations on matrices:

1 P_jk: Interchanging the jth and the kth rows. 2 E_jk(c): Adding c times the kth row to the jth row. 3 And M_j(λ): Multiplying the jth row by a scalar λ 6= 0.

Denote by Ejk the standard basic matrix whose (jk)th entry is 1 and the

rest 0’s. We need only m × m or square matrices in the following discussion. Let I = Im(=

m

X

j =1

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Elementary row matrices, contd.

For a scalar c and 1 ≤ j 6= k ≤ m let us define some m × m matrices as follows:

1 _P_jk_{: The matrix P}_jk _{is the matrix obtained by interchanging the j}th

and the kth rows of I.

2 _E_jk_{(c): The matrix E}_jk_{(c) = I + cE}_jk_{, j 6= k i.e. the matrix obtained}

from I by adding c times the kth row to the jth row.

3 _M_j_{(λ): The matrix M}_j_{(λ) is the matrix obtained from I by}

multiplying its jth row by λ(6= 0).

The above matrices are known as the elementary row matrices (of order m if you like.)

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Examples of ERM

0

s

Among 3 × 3 matrices we notice that

P12 =   0 1 0 1 0 0 0 0 1.   E32(c) =   1 0 0 0 1 0 0 c 1.   M2(λ) =   1 0 0 0 λ 0 0 0 1.  

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ERM

0

s contd.

Theorem 13

Let A be any m × n matrix and Pjk, Ejk(c) and Mj(λ) be the m × m

ERM’s (j 6= k, λ 6= 0).

1 The product P_jkA is the m × n matrix obtained by interchanging the

jth and the kth rows of A.

2 _{The product E}_jk_{(c)A is the m × n matrix obtained by adding c times}

the kth row of A to the jth row of A.

3 _{The product M}_j_{(λ)A is the m × n matrix obtained by multiplying the}

jth row of A by λ. Proof: Omitted.

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Reduced Row Echelon form

Theorem 14

Let A be an m × n matrix. There exist ERM’s E1, E2, ..., EN of order m

such that the product EN· · · E2E1A is a row echelon form of A.

Proof: Obvious.

Reduced Row Echelon form: Once an echelon form of A is obtained, we can by further row operations ensure that (i) each pivot becomes 1 and (ii) all the entries above each pivot become zeroes. This is called the Reduced Row Echelon form of A and it is unique.

Example: Suppose A =    1 2 3 4 0 5 6 0 0 0 0 7  

is in REF. Then by EROps we

can further reduce it to    1 2 3 4 0 1 1.2 0 0 0 0 1  

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Reduced Row Echelon form contd.

   1 0 0.6 0 0 1 1.2 0 0 0 0 1  

which is THE reduced REF.

Theorem 15 (REF of square matrices)

Let A be a square matrix, say n × n. There exist ERM’s E1, E2, ..., EN of

order n such that the product EN· · · E2E1A is either (i) the n × n identity

matrix I or (ii) its last row is 0. Proof:

Consider the reduced row echelon form of A. Let p(≤ n) be the number of pivots. (i) If p = n then perforce each row and column will have a pivot and the reduced REF must be I. (ii) If there are p < n pivots, then the last n − p ≥ 1 rows must vanish.

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Inverse of a square matrix

Definition 16

(Inverse of a square matrix) Let A be a square matrix. Its inverse is another square matrix B, if it exist, satisfying AB = BA = I. We say that A is invertible with an inverse B.

Properties:

1 _{If A has an inverse, then it is unique. It is denoted by A}−1_. 2 If A and B are invertible, then so is AB and (AB)−1 = B−1A−1. 3 Each ERM is invertible. In fact (i) P−1

jk = Pjk,

(ii) Ejk(c)−1= Ejk(−c) and (iii) Mj(λ)−1= Mj(1/λ). Observe that

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Gauss Jordan Method

Theorem 17

A square matrix A is invertible if and only if it is a product of ERM’s. Proof: If A is a product of ERM’s, then it is clearly invertible.

For the converse let A be invertible and consider its reduced REF, ˆA say. ˆ

A = EN· · · E2E1A =

( I or

the last row is 0.

In the first case A = E₁−1E₂−1· · · E_N−1 is a product of ERM’s as required while the second case can not occur for invertible A. WHY?

ˆ

A being a product of invertible matrices is invertible. ∃B such that ˆ

AB = I. then the last row of LHS is [0 0 ... 0] while that of RHS is [0 0 ... 1] a contradiction.

Exercise: If kth _{row of any matrix M is 0, then so is the case with MP for} any matrix P which can multiply M from the right.

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Gauss Jordan Method

Example: Write a b

c d

as a product of 4 elementary row matrices if D = ad − bc 6= 0. Case 1 (a 6= 0): a b c d E21(−c_a ) 7−→ a b 0 D/a M1(1_a),M2(_Da) 7−→ 1 b/a 0 1 E12(−b_a ) 7−→ I₂. Hence

A = E21(c/a)M1(a)M2(D/a)E12(b/a)

= 1 0 c/a 1 1/a 0 0 1 1 0 0 D/a 1 b/a 0 1 . Case 2: (a = 0) =⇒ D = −bc 6= 0 =⇒ c 6= 0: Then A = P12M2(b)M1(c)E12(d /c). (Exercise)

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Finding inverse by Gauss Jordan Method

Gauss-Jordan Method:

Let A be a square n × n matrix.

(1) Augment it on the right by I to obtain n × 2n matrix [A|I].

(2) Apply row operations to [A|I] till the left half becomes a REF of A. (3a) If the last row of the left half is 0, STOP as A is not invertible. (3b) Else continue till the left half becomes I.

(4) The right half is A−1.

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An example of Gauss Jordan Method

Example: Find the inverse of A =   −1 1 2 3 −1 1 −1 3 4  . Solution:      −1 1 2 1 0 0 3 −1 1 0 1 0 −1 3 4 0 0 1      E21(3),E31(−1) −→      −1 1 2 1 0 0 0 2 7 3 1 0 0 2 2 −1 0 1      E32(−1) −→      -1 1 2 1 0 0 0 2 7 3 1 0 0 0 -5 −4 −1 1     

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Example contd.

The REF of A obtained has the last row non zero, hence CONTINUE.

M1(−1),M2(1/2),M3(−1/5) −→      1 −1 −2 −1 0 0 0 1 7/2 3/2 1/2 0 0 0 1 4/5 1/5 −1/5      E13(2),E23(−7/2) −→      1 −1 0 3/5 2/5 −2/5 0 1 0 −13/10 −1/5 7/10 0 0 1 4/5 1/5 −1/5      E12(1) −→      1 0 0 −7/10 1/5 3/10 0 1 0 −13/10 −1/5 7/10 0 0 1 4/5 1/5 −1/5      January 4, 2017 41 / 42

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Example contd.

It follows that A−1= 1 10   −7 2 3 −13 −2 7 8 2 −2  .

Also putting inverses of all the row ops together in reverse order, A−1= E12(1)E13(2)E23(−7/2)

×M1(−1)M2(1/2)M3(−1/5)E32(−1)E21(3)E31(−1)

A =E31(1)E21(−3)E32(1)

×M3(−5)M2(2)M1(−1)E23(7/2)E13(−2)E12(−1)