For the following matrices, first find a basis for the column space of the matrix. Then...

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For the following matrices, first find a basis for the columnspace of the matrix. Then use the Gram-Schmidt process to find anorthogonal basis for the column space. Finally, scale the vectorsof the orthogonal basis to find an orthonormal basis for the columnspace.
(a) [1 1 1, 1 0 2, 3 1 0, 0 0 4 ] b) [?1 6 6, 3 ?8 3, 1 ?2 6, 1?4 ?3 ]

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We presume that the triplets mentioned are the columns of the matrix.

(a). Let A =

1	1	3	0
1	0	1	0
1	2	0	4

The RREF of A is

1	0	0	4/5
0	1	0	8/5
0	0	1	-4/5

Thus, a basis for col(A) is { v₁,v₂,v₃} = { (1,1,1)^T,(1,0,2)^T,(3,1,0)^T} .

Let u₁=v₁=(1,1,1)^T,u₂=v₂–proj_u1(v₂)=v₂–[(v₂.u₁)/(u₁.u₁)]u₁=v₂–[(1+0+2)/(1+1+1)]u₁=(1,0,2)^T-(1,1,1)^T= (0,-1, 1)^Tand u₃=v₃–proj_u1(v₃)=v₃–[(v₃.u₁)/(u₁.u₁)]u₁–[(v₃.u₂)/(u₂.u₂)]u₂=v₃–[(3+1+0)/(1+1+1)]u₁-[(0-1+0)/ (0+1+1)]u₂ = (3,1,0)^T –(4/3)( 1,1,1)^T -(1/2)( 0,-1, 1)^T= (5/3,-5/6,-5/6)^T. Then { u₁,u₂,u₃}= {(1,1,1)^T, (0,-1, 1)^T, (5/3,-5/6,-5/6)^T } is an orthogonal basis for col(A).

Further, let e₁ = u₁/||u₁|| =(1/?3,1?3,1/?3)^T e₂ = u₂/||u₂|| =(0,-1?2, 1/?2)^Tand e₃ = u₃/||u₃|| = (5/3,-5/6,-5/6)^T= (?2/?3, 1/?6,-1/?6)^T. Then {e₁,e₂,e₃} = {(1/?3,1?3,1/?3)^T, (0,-1?2, 1/?2)^T, (?2/?3, 1/?6,-1/?6)^T} is an orthonormal basis for col(A).

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