a) Apply the Gram-Schmidt orthonormalization process to transform the set of vec

Question

a) Apply the Gram-Schmidt orthonormalization process to transform the set of vectors S into an orthonormal basis for R3.

b) Use the orthonormal basis above to find the QR factorization of S.

c) Multiply QR above to verify that you return to the vectors in S.

--------------------------------------

I keep getting stuck at part c (which may be due to me being wrong at a), as multiplying QR doesn't result in S again.

When I use WolframAlpha, I enter gram schmidt [(1,1,0,1),(1,0,1,1),(0,0,0,1)] which results in this matrix:

I get that same answer when I do it out by hand.

When I asked this question earlier, the person had the same first and second vector as I did. However, their third vector is much different than mine. Theirs is [-sqrt(2/5), -1/sqrt(10), -1/sqrt(10), -sqrt(2/5)] and I have no idea how they got that.

I had the same equation set up as them, u3 = (0,0,0,1)-1/3(1,1,0,1)-1/15(1,-2,3,1) but when I do that out by hand or with WolframAlpha I get (-2/5, -1/5, -1/5, 3/5) but they got (-2/5, -1/5, -1/5, -2/5).

They then multiplied that by sqrt(5/2) which gave them a completely different third vector than me. https://www.chegg.com/homework-help/questions-and-answers/use-set-vectors-apply-gram-schmidt-orthonormalization-process-transform-set-vectors-s-orth-q28158859

The problem is that with my answer, when I continue with it through the QR factorization/decomposition and multiple them together, they don't result in the original vector set S. However, the other person's answer does work out, but I have no idea how they got their third vector.

Can someone please walk me through this? Either explain how they got that answer, or provide a different answer. I've been trying to figure this out all day, and I don't know what I'm doing wrong. I've followed each step in our book and it still isn't working.

Explanation / Answer

Let v1 = (1,1,0,1)T, v2 = (1,0,1,1)T and v3 = (0,0,0,1)T. Also, let u1 = v1 = (1,1,0,1)T,   u2 = v2-proju1(v2)                                              = v2-[(v2.u1)/(u1.u1)]u1= v2-[(1+0+0+1)/(1+1+0+1)]u1 =(1,0,1,1)T- (2/3)(1,1,0,1)T =(1/3,-2/3,1,1/3)T and u3 = v3-proju1(v3)- proju2(v3)= v3-[(v3.u1)/(u1.u1)]u1-[(v3.u2)/(u2.u2)]u2=v3-[(0+0+0+1)/(1+1+0+1)]u1-[(0+0+0+1/3) /(1/9+4/9+1+1/9)]u2=(0,0,0,1)T-(1/3)(1,1,0,1)T-(1/5)(1/3,-2/3,1,1/3)T= (-2/5,-1/5,-1/5,3/5)T.

Then {u1,u2,u3} =   { (1,1,0,1)T, (1/3,-2/3,1,1/3)T, (-2/5,-1/5,-1/5,3/5)T}is an orthogonal basis for R3.

Now, let e1=u1/||u1||=(1/?3,1/?3,0,1/?3)T,e2=u2/||u2||=(1/3,-2/3,1,1/3)T/[?(5/3)]=(1/?15,-2/?15, ?(3/5), 1/?15)T and e3 = u3/||u3||=(-2/5,-1/5,-1/5,3/5)T/[?(3/5)]=(-2/?15,-1/?15,-1/?15,?(3/5) )T.

Thus, your calculations are correct. However, there is another method to compute a vector u3 orthogonal to both u1 and u2. Let u3=(x,y,z,w)T.Thenu3.u1=0or,x+y+w=0andu3.u2=0or, x/3-2y/3+z+w/3=0. The coefficient matrix of this linear system is A =

1

1

0

1

1/3

-2/3

1

1/3

The RREF of A is

1

0

1

1

0

1

-1

0

so that y –z = 0 or, y = z and x +y+w = 0 or, x+z+w = 0 so that x = -z-w. Then X = (-z-w, z, z,w)T = z(-1,1,1,0) +w(-1,0,0,1)T. Now, by assigning arbitrary value (like 0,1,1,0 or, 1,1) to z and w, we get u3. Computing u3 by the Gram-Schmidt requires a lot of calculations so some persons may adopt an alternative method. It is very useful when, as per the Gram-Schmidt process, u3 is a zero vector. Since a zero vector cannot be normalized, we need an alternate process.

Incidentally, the vector (-2/5, -1/5, -1/5, -2/5) cannot be u3 as it is not orthogonal to u1. It must be a computational error.

1

1

0

1

1/3

-2/3

1

1/3

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