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Linear Algebra, Strang 2.7

  1. Jun 10, 2008 #1
    what???

    2.7

    7. True or False

    a. The block matrix [0 A / A 0 ] is automatically symmetric.

    If it's symmetric along it's diagonal, A = A^T. So why isn't this true? It also holds true for a 3 x 3.

    b. If A and B are symmetric then their product AB is symmetric.

    Seems like it to me.

    c. If A is not symmetric then A^-1 is not symmetric.

    Yay, got 1 right :p

    d. When A, B, C are symmetric, the transpose of ABC is CBA.

    Hmm ... I did all that work and wrong? I want to sleep!

    Ok, I got all these wrong except c. I kinda figured when I got True for all. LOL, someone plz help me :)

    [​IMG]
     
    Last edited: Jun 10, 2008
  2. jcsd
  3. Jun 10, 2008 #2
    I wrote b down incorrect in the 2nd step, but I still got a symmetric matrix after fixing it.
     
  4. Jun 10, 2008 #3

    Dick

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    You can't prove things by picking specific examples, right? I mean for b) A=[[1,0],[0,0]], B=[[0,1],[1,0]]. They are both symmetric. AB is NOT symmetric. You are picking very specific examples which are largely multiples of the identity matrix. Many things are true for multiples of the identity which aren't true of matrices in general. For example a multiple of the identity commutes with EVERYTHING.
     
    Last edited: Jun 10, 2008
  5. Jun 10, 2008 #4
    Ok, I see your point with b and I can also apply it to d. However, with A, I don't see why it's not automatically symmetric. The variable A symmetric along the diagonal, and transposing it, I get A = A^T.

    My choices for b & d were too specific as you said, but I was told what to use for a. argh ...
     
  6. Jun 10, 2008 #5

    Dick

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    It's a block matrix. The transpose of [[0,A],[A,0]] is [[0,A^T],[A^T,0]]. Pick an A that's not symmetric. I don't think you were told A is symmetric.
     
    Last edited: Jun 10, 2008
  7. Jun 10, 2008 #6

    Dick

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    And you didn't get c) correct for the right reason. 2x2 matrices don't necessarily prove anything either. Try just using (A^T)^(-1)=(A^(-1))^T.
     
  8. Jun 10, 2008 #7
    Ok, since A wasn't symmetric, I found that (A^T)^(-1) did not equal (A^(-1))^T.
     
  9. Jun 10, 2008 #8
    Block matrix ... hmm ... block matrix ... must look this up.
     
  10. Jun 10, 2008 #9

    Dick

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    Good idea! :)
     
  11. Jun 10, 2008 #10
    Hey Dick, how did you come up your examples? Did you already know your matrices were not going to be symmetric as a product? Simply through trial and error?

    I've had similar problems like that in previous sections, and I did them correctly but this time I completely failed :(
     
  12. Jun 10, 2008 #11

    Dick

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    Pick matrices that have no more properties than they actually NEED to have. Avoid examples that are diagonal. That's what went wrong in b). One of the matrices is diagonal and automatically commutes. Pick one that isn't diagonal. The odds are really high you'll find the result isn't symmetric.
     
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