LA : Just a quick question. Easy answer.

  • Thread starter JonathanT
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Ok so I just got my test back for Linear Algebra and I was told to find a basis for the ker(A) where A = \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 \\ -1 & 1 & -1 & -2 \end{array}

now I computed Ax = 0 and found x = [-t, 0 ,-t, t]^T. where x4 = t.

Then I wrote the basis as [1,0,1,-1]^T because I was under the impression that the basis is unique to any scalar multiple of the basis. So I simply took the scalar -1 to make as much of the basis positive as possible as I understand the convention is.

Am I wrong is thinking that the basis for the Ker(A) is [1,0,1,-1]^T as well as [-1,0,-1,1]^T?

I was marked of 6 points for this problem and I just wanted to make sure my logic was correct before I took a trip to the TA and looked like an idiot.

Thanks for your time.
 
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  • #2
Ray Vickson
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Ok so I just got my test back for Linear Algebra and I was told to find a basis for the ker(A) where A = \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 \\ -1 & 1 & -1 & -2 \end{array}

now I computed Ax = 0 and found x = [-t, 0 ,-t, t]^T. where x4 = t.

Then I wrote the basis as [1,0,1,-1]^T because I was under the impression that the basis is unique to any scalar multiple of the basis. So I simply took the scalar -1 to make as much of the basis positive as possible as I understand the convention is.

Am I wrong is thinking that the basis for the Ker(A) is [1,0,1,-1]^T as well as [-1,0,-1,1]^T?

I was marked of 6 points for this problem and I just wanted to make sure my logic was correct before I took a trip to the TA and looked like an idiot.

Thanks for your time.

I have never heard of any "convention" about making something as "positive as possible". Either [1,0 1 -1]^T or [-1,0,-1,1]^T or [1/4,0,1/4,-1/4]^T or [-1/√3, 0,-1/√3, 1/√3]^T, [150,0,150,-150]^T etc., etc. are all perfectly good answers to the question, but it would be most "reasonable" to go with the first, second or fourth possibility above.

Is 6 not a good mark on the question?

RGV
 
  • #3
jbunniii
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Ok so I just got my test back for Linear Algebra and I was told to find a basis for the ker(A) where A = \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 \\ -1 & 1 & -1 & -2 \end{array}

now I computed Ax = 0 and found x = [-t, 0 ,-t, t]^T. where x4 = t.

Then I wrote the basis as [1,0,1,-1]^T because I was under the impression that the basis is unique to any scalar multiple of the basis. So I simply took the scalar -1 to make as much of the basis positive as possible as I understand the convention is.

Am I wrong is thinking that the basis for the Ker(A) is [1,0,1,-1]^T as well as [-1,0,-1,1]^T?

I was marked of 6 points for this problem and I just wanted to make sure my logic was correct before I took a trip to the TA and looked like an idiot.

Thanks for your time.

If the problem asked for an orthonormal basis, then you would have to divide your vector by its norm. If not, then your answer is correct as stated. Any nonzero scalar multiple of your vector would also be correct.
 
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I have never heard of any "convention" about making something as "positive as possible". Either [1,0 1 -1]^T or [-1,0,-1,1]^T or [1/4,0,1/4,-1/4]^T or [-1/√3, 0,-1/√3, 1/√3]^T, [150,0,150,-150]^T etc., etc. are all perfectly good answers to the question, but it would be most "reasonable" to go with the first, second or fourth possibility above.

Is 6 not a good mark on the question?

RGV

Thanks both of you. I'm glad I'm not crazy. Sorry I guess I didn't mean convention. I guess it would have been better to say that its the way its done by people like me who are "anal." I'll go talk to my TA about this and get my 6 points back. He only gave me 4/10 for the question.

And it was just a regular basis not an orthonormal one. Thanks for the help. Any theorem or proof I could show for this?

EDIT: Nvm, clearly Ax = 0 is a good enough proof.
 
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