Explanation Given: The matrix is A = [ 1 2 6 2 1 6 0 1 2 1 0 2 ] . Approach: Row space of a matrix A is the span of linearly independent rows of the reduced Echelon form of the matrix A . Column space of a matrix A is the span of linearly independent columns of the matrix A . Null space of a matrix A is the span of solution of the linear system A x = 0 . The normalized vector corresponding to a vector v is given by v = v ‖ v ‖ . 5) Let { u 1 , u 2 , ... , u n } be a set of linearly independent vectors in the vector space V . Then, the orthogonal basis for the set spanned by these vectors are given by v 1 = u 1 v 2 = u 2 − 〈 u 2 , v 1 〉 ‖ v 1 ‖ 2 v 1 v 3 = u 3 − 〈 u 3 , v 1 〉 ‖ v 1 ‖ 2 v 1 − 〈 u 3 , v 2 〉 ‖ v 2 ‖ 2 v 2 ⋮ ⋮ Calculation: Consider the matrix [ 1 2 6 2 1 6 0 1 2 1 0 2 ] Apply R 2 → 1 2 R 2 [ 1 2 6 1 1 2 3 0 1 2 1 0 2 ] Apply R 2 → R 2 − R 1 , R 4 → R 4 − R 1 [ 1 2 6 0 − 3 2 − 3 0 1 2 0 − 2 − 4 ] Apply R 2 → − 2 3 R 2 [ 1 2 6 0 1 2 0 1 2 0 − 2 − 4 ] Apply R 3 → R 3 − R 2 , R 4 → R 4 + 2 R 2 [ 1 2 6 0 1 2 0 0 0 0 0 0 ] Therefore, the row space is the span of the set { ( 1 , 2 , 6 ) , ( 0 , 1 , 2 ) } Let u 1 = ( 1 , 2 , 6 ) and u 2 = ( 0 , 1 , 2 ) . The orthogonal vectors are v 1 = ( 1 , 2 , 6 ) v 2 = ( 0 , 1 , 2 ) − 0 + 2 + 12 ( 1 + 4 + 36 ) 2 ( 1 , 2 , 6 ) = ( 0 , 1 , 2 ) − 14 41 ( 1 , 2 , 6 ) = ( − 14 41 , 13 41 , − 2 41 ) Normalize the above vectors v 1 = 1 41 ( 1 , 2 , 6 ) v 2 = 41 3 ( − 14 41 , 13 41 , − 2 41 ) = 1 41 ( − 14 3 , 13 3 , − 2 3 ) So, the orthonormal basis for the row space is { 1 41 ( 1 , 2 , 6 ) , 1 41 ( − 14 3 , 13 3 , − 2 3 ) } The column space is the span of the set { ( 1 , 2 , 0 , 1 ) , ( 2 , 1 , 1 , 0 ) } Let u 1 = ( 1 , 2 , 0 , 1 ) and u 2 = ( 2 , 1 , 1 , 0 )

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Author: ANNIN

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Chapter 5.5, Problem 4AP

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The orthonormal basis for the row space, column space and null space.

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The orthonormal basis for the row space, column space and null space.

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The orthonormal basis for the row space, column space and null space.