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A matrix satisfies A^2 - 4A + 5I = 0, then n is even.

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data

    1) Let A be an n x n matrix with A^2 -4A +5I = 0. Show that n must be even.

    2) Let A be an m x n matrix where m<n. Show that det(AT x A) = 0

    3. The attempt at a solution

    1) (A-2I)^2 +I=0

    Not sure what to do after this though


    Thanks in advance
     
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2

    AKG

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    An n x n matrix with real entries? If it can have complex entries, this isn't true. Anyways, what do you know about minimal polynomials and characteristic polynomials?
    What do you know about the relationships between rank, invertibility, and determinants.
    Okay, that's not bad. So (A-2I)2 = -I. Compute the determinant of both sides.
     
  4. Apr 21, 2007 #3
    Cheers I've got the first question now. Was easier than i thought.

    I still can't do 2) though.

    Could you explain further please?
     
  5. Apr 21, 2007 #4

    AKG

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    What are the dimensions of the matrix ATA? What can you say about the rank of ATA?
     
  6. Apr 22, 2007 #5
    ATA is n x n
    We haven't covered ranks yet

    I know that ATA can reduced so that it has one row of zero's hence det=0. But I don't know how to show it in general.
     
  7. Apr 22, 2007 #6
    It may help to think of matrix multiplication with a vector as a linear combination of the columns of the matrix

    i.e. For [tex]A\vec{c} = \vec{b}\\[/tex] b is a linear combination of the columns of A

    And hence a matrix multiplication with a vector will produce a matrix whose columns are a linear combination of the columns of the first matrix.

    i.e. For [tex]AB = C\\[/tex] C's columns are linear combinations of the columns of A

    Sorry if the Latex is less than desirable, as you can see, I'm new here.
     
    Last edited: Apr 22, 2007
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