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Linear Algebra - diagonalizable matrix

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    For the following matrix, find the value of t, if any, so that the following matrix is diagonalizable

    [tex]A=
    \begin{pmatrix}
    5 & -2 & 4\\
    0 & 3 & t\\
    0 & 0 & 5
    \end{pmatrix}
    [/tex]


    3. The attempt at a solution
    In order for A to be diagonalizable, we need 3 linearly independent eigenvectors, that is, 3 linearly independent eigenvalues

    [tex]det(A-xI)=det
    \begin{pmatrix}
    5-x & -2 & 4\\
    0 & 3-x & t\\
    0 & 0 & 5-x
    \end{pmatrix}
    [/tex]
    [tex]=(5-x)det
    \begin{pmatrix}
    3-x & t\\
    0 & 5-x
    \end{pmatrix}
    [/tex]
    [tex]= (5-x)((3-x)(5-x)-0t)[/tex]

    The eigenvalues are 3 and 5.
    Obviously, it doesn't matter what t is, we will not be able to get the matrix A to be diagonalizable.

    My professor said that he thought there was one correct value for t (but he wasn't sure). Is what I've done correct?
     
  2. jcsd
  3. Nov 24, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    It is not a matter of whether or not you have a repeated eigenvalue; what is important is whether the geometric and algebraic multiplicities of an eigenvalue are the same. In other words, the dimensionality of the eigenspace for eigenvalue λ=5 needs to be determined. If λ=5 has two linearly-independent eigenvectors, then the matrix will be diagonalizable, so you need to find eigenvectors.

    RGV
     
  4. Nov 25, 2011 #3
    There is a value of t for which the matrix is diagonalizable.
     
  5. Nov 25, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If [itex]\begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex] is an eigenvector for A, with eigenvalue 5, then
    [tex]\begin{pmatrix}5 & -2 & 4 \\ 0 & 3 & t \\ 0 & 0 & 5\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= 5\begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex]
    [tex]\begin{pmatrix}5x- 2y+ 4z \\ 3y+ tz \\ 5z\end{pmatrix}= \begin{pmatrix}5x \\ 5y \\ 5z\end{pmatrix}[/tex]

    and so must satify the equations 5x- 2y+ z= 5x, 3y+ tz= 5y, 5z= 5z.

    Note that the "5x" terms cancel in the first equation and then there is NO "x" in the equations. One eigenvector with eigenvalue is <1, 0, 0>. There is a single value of t that makes the first two equations identical, giving more than one dimension for the eigenspace.
     
  6. Nov 26, 2011 #5
    So for eigenvalue 3, if [tex]\begin{pmatrix}x' \\ y' \\ z'\end{pmatrix}[/tex] is also an eigenvector for A

    [tex]\begin{pmatrix} 5&-2&4\\0&3&t\\0&0&5\end{pmatrix}\begin{pmatrix}x'\\y'\\z'\end{pmatrix} = 3\begin{pmatrix}x'\\y'\\z'\end{pmatrix}[/tex]

    Where there are no free variables, no matter what t is, and [tex]5x'-2y'+4z'=3x'[/tex] and [tex]3y'+tz'=3z'[/tex] and [tex]5z'=3z'[/tex] means that [tex] x'=y'=z'=0[/tex] correct?

    So if t=4, the matrix A will always be diagonalizable?

    We can say that the 3 linearly independent eigenvectors are (if for the second eigenvector, we take z=1 and x=1 and for the third eigenvector we take z=2 and x=2 because both are free):
    [tex]\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 2 \\ 1 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 4 \\ 2 \end{pmatrix}[/tex]

    But there are infinitely many eigenvectors, I just chose these 3. Is this correct?
     
  7. Nov 26, 2011 #6
    But i just realized that the eigenvector [tex]\begin{pmatrix} 0\\0\\0 \end{pmatrix}[/tex]
    isn't a proper eigenvector because it is all 0. So I could take the eigenvectors (if for the first eigenvector, we take z=3 and x=1):
    [tex]\begin{pmatrix} 1 \\ 6 \\ 3 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 2 \\ 1 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 4 \\ 2 \end{pmatrix}[/tex]

    Is this now correct?
     
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