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Homework Help: Orthogonal vector proof

  1. Dec 12, 2012 #1
    Let A be an n × n invertible matrix. Show that if
    i ≠ j, then row vector i of A and column vector
    j of A-1 are orthogonal.



    I'm lost in regards to where to lost.

    I want to show that a vector from row vector i from A is orthogonal to a column vector j from A.
    Orthogonal means the dot product is 0 or the angle between the 2 vectors is 90 degrees.

    After stating the obvious I'm stuck. I think I need to start with figuring out what the relationship between a row vector from A and a column vector from A^-1 is, but how do I do that?
     
  2. jcsd
  3. Dec 12, 2012 #2
    How would you show that the dot product of (row vector i of A) and (column vector j of A-1) is equal to 0? And in what kind of operation would you perform several such calculations between row vectors of one matrix and column vectors of another?
     
  4. Dec 12, 2012 #3
    If v represents the row vector and w represents the column vector.
    I must show that v*wt=0. * means dot product.

    The dot product would be then computed by multiplying each element in v by each corresponding element in w and taking the sum of all those products. But I'm not sure how this will show what the problem wants.
     
  5. Dec 12, 2012 #4
    Apart from calculating dot products, when does one multiply the elements of a row with the elements of a column and take the sum of the results?

    If that's no help, try this: how would you prove that A-1 really is an inverse of A? What kind of calculation could you perform to show this?
     
  6. Dec 12, 2012 #5
    a-1*a=i?
     
  7. Dec 12, 2012 #6
    How does one multiply one matrix by another? Say you wanted to multiply A by A-1. Can you explain how you'd obtain the element in row i and column j of the resulting matrix?
     
  8. Dec 12, 2012 #7
    OH! I see what you're saying. You'd multiply the row by the column and sum up the products.
    This would explain how to multiply the 2 vectors together.
     
  9. Dec 12, 2012 #8
    All you need now is some criterion to decide which elements in AA-1 can be zero and which can't.
     
  10. Dec 12, 2012 #9
    Okay Thanks for the help.

    I am still having trouble. Or maybe I understand it but just not realize it.

    In AA-1=I, everything off the diagonal is 0. Everything on the diagonal is nonzero, but that's when i=j. Everything off the diagonal is 0 b/c i≠j.

    Is that the idea?
     
  11. Dec 13, 2012 #10
    That's exactly the right idea. You can safely ignore the diagonal, since your initial question was only about i≠j.
     
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