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Two related questions about matrix algebra

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data

    a.) If A is an 'n x n' matrix and X is an 'n x 1' nonzero column matrix with

    AX = 0

    show, by assuming the contrary, that det(A) = 0

    b.) Using the answer in 'a' show that the scalar equation which gives the values of λ that satisfy the matrix equation AX = λIX is:

    det(A - λI) = 0

    2. The attempt at a solution

    a.) If det(A) ≠ 0 then A^-1 exists.

    X = A^-1 x (AX) = A^-1 x (0) = 0

    This is a contradiction because x nonzero so det(A) = 0.... this bit I understand however the next part

    b.) AX = λIX -> X(A - λI) = 0

    X ≠ 0 so for the equation to be true A - λI = 0

    I'm not sure how to apply the first result to the second question? It's from Bimore & Davies Calculus: Concepts and Methods book.

    Any help would be appreciated!
     
  2. jcsd
  3. Feb 17, 2013 #2
    X ≠ 0, does not implies that A - λI = 0
    For example
    [itex]\begin{pmatrix}
    2 &0 \\
    0 &1
    \end{pmatrix}[/itex]
    It's clear that it's eigenvalues are 1 and 2, but...
    A-2I=
    [itex]\begin{pmatrix}
    0 &0 \\
    0 &-1
    \end{pmatrix}[/itex]
    And if we take X=
    \begin{pmatrix}
    1\\
    0
    \end{pmatrix}
    Then A.X=0
    I think, your mistake it's that that it's some X (more precisely, the eigenvectors that the matrix has), but not to all X, otherwhise the matrix has to be λI and thus, A-λI=0 is true, but it isn't necessary.



    I think you're pretty close to the answer, you have to think what happens when there's some X such as:
    AX = λIX
    Then, as you wrote:
    (A-λI)X=0
     
    Last edited: Feb 17, 2013
  4. Feb 17, 2013 #3
    How about this:

    1.) AX = 0 .... Det(A)≠ 0 -> A^(-1) exists Edit: I had X not equal to 0

    X = A^(-1) x AX = A^(-1) x 0 = 0 so det(A) = 0

    2.) AX = λIX -> X(A - λI) = 0 .... Let B = A - λI .... If det(B) not equal to 0 then:

    X = B^(-1) x BX = B^(-1) x 0 = 0

    so det(B) = 0 -> det(A - λI) = 0????
     
    Last edited: Feb 17, 2013
  5. Feb 17, 2013 #4
    I recommend you to use dots instead x to the matrix product.
    And, while I understand what you tried to say here:
    AX = λIX -> X(A - λI)
    X is a column vector, so you can only use the product of the square matrix with the X at the right (You can lose points for doing this).
     
    Last edited: Feb 17, 2013
  6. Feb 17, 2013 #5
    Thanks for the input! Not sure I completely follow what you say here.

    Is this incorrect?

    AX = λIX -> AX - λIX = 0 -> X(A - λI) = 0?


    Edit: oh great thanks!
     
  7. Feb 17, 2013 #6
    The proof of b) is right, but you don't need to repeat the proof, you can just say that:
    (A - λI)X=0 them by a), det(A - λI)=0

    AX = λIX -> AX - λIX = 0 -> X(A - λI) = 0
    Again:
    AX = λIX -> AX - λIX = 0 -> (A - λI)X = 0
    :P
    And yes, it's correct because the matrix product to right (to left) is distributive with the matrix addition.
     
    Last edited: Feb 17, 2013
  8. Feb 17, 2013 #7
    No, this is not correct. AX = λIX -> (A - λI)X = 0. Which is not the same as X(A - λI) = 0.

    Now, let (A - λI) = A', so we have A'X = 0. Can you apply part (a) to this?
     
  9. Feb 17, 2013 #8
    Oh right you guys are saying:

    X(A-λI) ≠ (A-λI)X

    Yeah I just have a habit to put the X in front of brackets all the time. You think I would lose marks for that?

    I guess technically I should.
     
  10. Feb 17, 2013 #9
    It is not just that X(A-λI) ≠ (A-λI)X, it is much worse. When X is a column, and A is a matrix, XA simply does not make sense, it is not defined. It is a pretty severe mistake to make, but it is also very easily avoidable: do not change the order of entities in a matrix product, matrix multiplication is not commutative, except in very special cases.
     
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