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Math Help - Diagonalizing matrix problem with parameters

  1. #1
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    Diagonalizing matrix problem with parameters

    Hi,
    I have this problem I'm stuck on:
    A is a 3x3 matrix
    A= Diagonalizing matrix problem with parameters-gif.gif
    a,b,c are real numbers
    1) determine for what values of a,b,c the matrix A is diagnoizable. (advice: distinguish between the cases a=c and a=\=c)

    My attempt at a solution has been finding det(tI-A)=0 Diagonalizing matrix problem with parameters-codecogseqn.gif, the characteristic polynomial is (t-c)(t-a)^2 and therefore the eigenvalues are a and c.

    for t=c we get the matrix Diagonalizing matrix problem with parameters-codecogseqn-2-.gif let it be C. for Cx=0 we get the system of linear equations:
    (c-a)x1-x3=0
    (c-a)x2 -bx3=0
    likewise, for t=a we get
    -1x3=0
    -bx3=0
    (a-c)x3=0

    Now, when I tried to find the eigenspaces that correspond to a and c to find out their dimensions I got pretty lost.
    I have no idea what to do. help would be very welcomed!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Diagonalizing matrix problem with parameters

    Case 1. a\neq c. Then, \lambda=a (double), \lambda=c (simple). By a well known theorem, \dim V_c=1=\mbox{ multiplic }(c). On the other hand:

    \dim V_a=3-\mbox{rank}\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{b}  \\{0}&{0}&{c-a}\end{bmatrix}=3-1=2=\mbox{ multiplic }(a). That is, A is diagonalizable.

    Case 2. a=c. Then, \lambda=a (triple). So,

    \dim V_a=3-\mbox{rank}\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{b}  \\{0}&{0}&{0}\end{bmatrix}=3-1=2\neq \mbox{ multiplic }(a). In this case, A is not diagonalizable.
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  3. #3
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    Re: Diagonalizing matrix problem with parameters

    Got it. thank you!
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  4. #4
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    Re: Diagonalizing matrix problem with parameters

    we can try to proceed with your original approach, and "see what happens". i often do this "just out of curiosity".

    it's clear that you found the characteristic equation properly, it's (t - c)(t - a)2.

    now let's find some eigenvectors. first let's tackle c:

    (A - cI)x = 0 \iff \begin{bmatrix}a-c&0&1\\0&a-c&b\\0&0&0 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}

    it's clear that x3 is "free", and provided a ≠ c:

    x_2 = \frac{b}{c-a}x_3

    x_1 = \frac{1}{c-a}x_3

    which means that any eigenvector (belonging to c) is a scalar multiple of (1,b,c-a) (we have a one-dimensional eigenspace).

    (note that if we DID have a = c, we we get instead is: x3 = 0. more on this later).

    next we look at a:

    \begin{bmatrix}0&0&1\\0&0&b\\0&0&c-a \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}

    gives:

    x3 = 0
    bx3 = 0 (and thus x3 = 0, this just duplicates the first equation)
    (c-a)x3 = 0 (another duplication).

    thus we are free to choose x1 and x2 independently, so we can pick (1,0,0) and (0,1,0) as linearly independent eigenvectors.

    so if {(1,0,0),(0,1,0),(1,b,c-a)} are linearly independent, A is diagonalizable (we have an eigenbasis). it should be clear that this will be true if and only if c -a ≠ 0, that is: if a ≠ c.

    on the other hand, if a = c, we only get an eigenspace of dimension 2: namely {(x,y,0) in R3}, so there is no eigenbasis, and A is not diagonalizable.
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