Explanation Given: The given matrix is, A = [ 0 1 − 1 0 − 2 2 0 − 1 1 ] Approach: The null space of A is given by the solution space of A x = 0 . The null space of A T is given by the solution space of A T x = 0 . The column space of A is a subspace of R m consisting of all vectors of the form A x , x ∈ R n . To show that N ( A ) = R ( A T ) ⊥ , show that N ( A ) is orthogonal to all vectors of R ( A T ) . To show that N ( A T ) = R ( A ) ⊥ , show that N ( A T ) is orthogonal to all vectors of R ( A ) . Two vectors u and v are orthogonal if, 〈 u , v 〉 = 0 Calculation: By Gauss-Jordon elimination, the reduced row echelon form of A is, A = [ 0 1 − 1 0 0 0 0 0 0 ] Substitute this value in A x = 0 as follows, [ 0 1 − 1 0 0 0 0 0 0 ] [ x 1 x 2 x 3 ] = 0 The following system of equations is obtained from the above expression, x 2 − x 3 = 0 … ( 1 ) From equation ( 1 ) , x 2 = x 3 Let x 3 = t and x 1 = s . So the form the system of the solution is, [ x 1 x 2 x 3 ] = [ s t t ] = [ s 0 0 ] + [ 0 t t ] = s [ 1 0 0 ] + t [ 0 1 1 ] So the basis for null space of A is, { [ 1 0 0 ] , [ 0 1 1 ] } Therefore, the null space of A is N ( A ) = span { [ 1 0 0 ] , [ 0 1 1 ] } . The transpose matrix of A is, A T = [ 0 0 0 1 − 2 − 1 − 1 2 1 ] By Gauss-Jordon elimination, the reduced row echelon form of A T is, A T = [ 1 − 2 − 1 0 0 0 0 0 0 ] Substitute this value in A T x = 0 as follows, [ 1 − 2 − 1 0 0 0 0 0 0 ] [ x 1 x 2 x 3 ] = 0 The following system of equations is obtained from the above expression, x 1 − 2 x 2 − x 3 = 0 Let x 3 = t and x 2 = s . So the from the system of equations is, [ x 1 x 2 x 3 ] = [ 2 s + t s t ] = [ 2 s s 0 ] + [ t 0 t ] = s [ 2 1 0 ] + t [ 1 0 1 ] So the basis for null space of A T is, { [ 2 1 0 ] , [ 1 0 1 ] } Therefore, the null space of A T is N ( A T ) = span { [ 2 1 0 ] , [ 1 0 1 ] }

Bundle: Elementary Linear Algebra, Loose-leaf Version, 8th + MindTap Math, 1 term (6 months) Printed Access Card

8th Edition

ISBN: 9781337131216

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 5.3, Problem 68E

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The bases for null space of A and AT, column space of A and AT and show that N(A)=R(AT)⊥ and N(AT)=R(A)⊥ if A=[01−10−220−11] is given.

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Chapter 5.3, Problem 67E

Chapter 5.3, Problem 69E

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The bases for null space of A and A T , column space of A and A T and show that N ( A ) = R ( A T ) ⊥ and N ( A T ) = R ( A ) ⊥ if A = [ 0 1 − 1 0 − 2 2 0 − 1 1 ] is given.

Solution Summary: The author explains how the null space of A and AT are given by the solution space.

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To find:

The bases for null space of A and AT, column space of A and AT and show that N(A)=R(AT)⊥ and N(AT)=R(A)⊥ if A=[01−10−220−11] is given.

Want to see the full answer?

Chapter 5 Solutions

The bases for null space of A and A T , column space of A and A T and show that N ( A ) = R ( A T ) ⊥ and N ( A T ) = R ( A ) ⊥ if A = [ 0 1 − 1 0 − 2 2 0 − 1 1 ] is given.

Solution Summary: The author explains how the null space of A and AT are given by the solution space.

Videos

To find: The bases for null space of A and AT, column space of A and AT and show that N(A)=R(AT)⊥ and N(AT)=R(A)⊥ if A=[01−10−220−11] is given.

Want to see the full answer?

Chapter 5 Solutions

To find:

The bases for null space of A and AT, column space of A and AT and show that N(A)=R(AT)⊥ and N(AT)=R(A)⊥ if A=[01−10−220−11] is given.