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Diagonalizable matrix

  1. Jan 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the LTI (A,B,C,D) system
    $$
    \dot{x}=
    \begin{pmatrix}
    0.5&0&0&0\\
    0&-2&0&0\\
    1&0&0.5&0\\
    0&0&0&-1
    \end{pmatrix}
    x+
    \begin{pmatrix}
    1\\
    1\\
    0\\
    0
    \end{pmatrix}
    u
    $$
    $$
    y=
    \begin{pmatrix}
    0&1&0&1
    \end{pmatrix}
    x
    $$
    Determine if A is diagonalizable
    2. Relevant equations


    3. The attempt at a solution
    The characteristic equation is given by
    $$
    \lambda I-A=
    \begin{pmatrix}
    \lambda-0.5&0&0&0\\
    0&\lambda+2&0&0\\
    1&0&\lambda-0.5&0\\
    0&0&0&\lambda+1
    \end{pmatrix}
    $$
    The determinant is given by
    $$
    det(\lambda I-A)=(\lambda-0.5)(\lambda+2)(\lambda-0.5)(\lambda+2)
    $$
    What do I have to do next to determine if A is diagonalizable? I am waiting hopefully for an answer!
     
  2. jcsd
  3. Jan 15, 2016 #2

    Mark44

    Staff: Mentor

    You have a mistake in your work. The last line above should be ##| \lambda I - A| = (\lambda-0.5)(\lambda+2)(\lambda-0.5)(\lambda + 1)##.
    Set the determinant to 0 to find the three eigenvalues (one is repeated).
    The matrix is diagonalizable if these eigenvalues yield four linearly independent eigenvectors.

    BTW, what does LTI (A,B,C,D) mean?
     
  4. Jan 15, 2016 #3

    Mark44

    Staff: Mentor

    I should also mention that since your matrix ##\lambda I - A## is lower triangular (all entries above the main diagonal are zero), its determinant is the product of the entries on the diagonal.
     
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