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Repeating eigenvalues and diagonalizing

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A=[0 -9; 1 -6]
    Can this matrix be diagonalized?

    2. Relevant equations
    det(A-[itex]\lambda[/itex]I)=0

    3. The attempt at a solution
    det(A-[itex]\lambda[/itex]I)=0 gives the eigenvalues of the matrix and yields two eigenvalues that are equal, [itex]\lambda[/itex]= -3

    A matrix with repeating eigenvalues are defective and can therefore NOT be diagonalized.
    I would further say that rank(A)=1 and nullity(A)=n-rank(A)=2-1=1.... Matlab does not agree with me.... what is wrong with my reasoning here?
     
  2. jcsd
  3. Sep 5, 2012 #2

    micromass

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    This is not true. A matrix with repeating eigenvalues may still be diagonalizable (or it may be that it can not be diagonalized). What you need to do is find the eigenspace belonging to the eigenvalue of -2. If this eigenspace has dimension 2 (that is: if there exist two linearly independent eigenvectors), then the matrix can be diagonalized.
     
  4. Sep 5, 2012 #3

    Mark44

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    Don't you mean the eigenvalue λ = -3?

    If so, the eigenspace of this eigenvalue is one-dimensional and consists of multiples of <3, 1>.
     
  5. Sep 5, 2012 #4

    micromass

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    Yes, I meant -3, thank you!!

    All I wanted to make clear is that being diagonalizable does not depend on there being repeated eigenvalues, but that we need to find the eigenspaces.

    Anyway, let's continue with the rank and nullity. What is the determinant of the matrix? What does this imply about the rank?
     
  6. Sep 5, 2012 #5
    Thanks, I think I went into a trap of reasoning here... If A then B, is not the same as B then A.

    Can one still conclude that a matrix is diagonalizable if it has distinct eigenvalues, since distinct eigenvalues ensures linearly independence?
    ... full rank when det(A)!=0, forgot about that one.
     
  7. Sep 5, 2012 #6

    HallsofIvy

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    Caution! This is NOT true! For an obvious example, the matrix
    [tex]\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}[/tex]
    has repeating eigenvalues but is already diagonalized.

    What is true is that an n by n matrix with fewer than n independent eigenvectors cannot be diagonalized. If an n by n matrix has n distinct eigenvalues, the eigenvectors corresponding to each are independent so the matrix is diagonalizable. If there are repeating eigenvalues, you don't' know if there are n independent eigenvectors until you check the eigenvectors themselves.

    In this particular case, saying that -3 is an eigenvalue means that
    [tex]\begin{bmatrix}0 & -9 \\ 1 & -6\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-9y \\ x- 6y\end{bmatrix}= \begin{bmatrix}-3x \\ -3y\end{bmatrix}[/tex]
    which is equivalent to -9y= -3x and x- 6y= -3y which are both equivalent to x= 3y. That is, every eigenvector corresponding to eigenvalue -3 is a multiple of <3, 1>. So, although your reasoning is wrong, your conlusion is true: this matrix has only one independent eigenvector and so is not diagonalizable.

    But being diagonalizable has NOTHING to do with "rank". As long as a matrix is invertible, not diagonalizable, it has full rank. Because this matrix does not have 0 as an eigenvalue, there is NO vector, v, such that Av= 0, it is invertible and has rank 2.
     
    Last edited: Sep 5, 2012
  8. Sep 5, 2012 #7

    Ray Vickson

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    The determinant of A is not zero, so A is invertible, and hence has rank 2. However, you need to look instead at the matrix B = A - λI = A + 3*I (where I = 2x2 identity matrix) and determine its rank, etc.

    RGV
     
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