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Is this a subspace.

  1. Oct 27, 2012 #1
    Which of the following is a subspace of M2x2 (the vector space of 2x2 matrices. and explain why or why not:
    1) Set of 2x2 matrices A such that det(A)=1
    2) set of 2x2 matrices B such that B[1 -1]^t=0 vector



    To check if something is a subspace I must satisfy 3 conditions (applied for matrix A):
    1) 0 matrix is A
    2) If U and V are in A then U+V is in A
    3) if V is in A then cV is in A for some scalar c.
    The above is analogous for matrix B.

    For 1) Set of 2x2 matrices A such that det(A)=1
    The 0 matrix is not in this set because the determinant is 0 which ≠1, thus the set of 2x2 matrices A is not a subspace.

    Is this correct?



    For 2) set of 2x2 matrices B such that B[1 -1]^t=0 vector
    The 0 matrix is in this set because the matrix 2x2 consisting of all 0s multiplied by [1 -1]^t is =0.

    Now I want to make sure I'm correctly applying the latter 2 conditions.
    If U and V are in this set, then the following is true.
    If U*[1 -1]^t=0 and V*[1 -1]^t=0
    U+V=(U+V)[1 -1]^t. Since U and V are 0, U+V=0. thus U+V=(0+0)[1 -1]^t.=0, thus U+V is in B?

    For condition 3: if V1*[1 -1]^t=0 is in the set, then cV must be in the set for it to be a subspace.
    cV1*[1 -1]^t=c*0*[1 -1]^t=0, thus cV is in the set?

    Thus the set of 2x2 matrices B such that B[1 -1]^t=0 vector is a subspace.
     
    Last edited: Oct 28, 2012
  2. jcsd
  3. Oct 28, 2012 #2

    haruspex

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    First, B is not the set of interest. It is used as an example of one of the matrices. No letter has been assigned to the set, so let's call it S.
    If U is in S it means U*[1 -1]^t=0. U is not of the form U1*[1 -1]^t, and it is not usually 0.
     
  4. Oct 28, 2012 #3
    Thanks.

    I edited my original post. Does it look better?
     
  5. Oct 28, 2012 #4

    micromass

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    Please do not edit your original post and just reply.
     
  6. Oct 28, 2012 #5

    haruspex

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    Yes.
    No it doesn't
    No they aren't
    No it isn't.
    Stop and think... what do you have to prove about U+V that would imply it is in S?
     
  7. Oct 28, 2012 #6
    I have to prove that the addition of two things in S must be in S in order for it to be a subspace.

    Instead of U+V=(U+V)[1 -1]^t. , I should've written:
    U*[1 -1]^t +V*[1 -1]^t=0=(U+V)*[1 -1]^t=0
    (0+0)*[1 -1]^t=0
    Thus U*[1 -1]^t +V*[1 -1]^t is in S.

    Is this the right idea?
     
  8. Oct 28, 2012 #7

    haruspex

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    Yes, and if you just rearrange that sequence you will have a logical proof. But what has the next line to do with anything?
    No, you're not trying to prove U*[1 -1]^t +V*[1 -1]^t is in S. You want to show U+V is in S.
     
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