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Matrix diagonalisation with complex eigenvalues

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Is there a basis of R4 consisting of eigenvectors for A matrix?
    If so, is the matrix A diagonalisable? Diagonalise A, if this is possible. If A is not diagonalisable because some eigenvalues are complex, then find a 'block' diagonalisation
    of A, involving a 2 × 2 block corresponding to a pair of complex-conjugate eigenvalues.

    A=
    0 1 0 0
    0 0 1 0
    0 0 0 1
    1 0 0 0

    2. Relevant equations



    3. The attempt at a solution

    I worked out the eigenvalues 1,-1,i,-i. I also worked out real eigenvectors: (-1,1,-1,1)^T and (1,1,1,1)^T. Whenever I try to work out complex eigenvectors I get no free variables..? How come A is "not diagonalisable because some eigenvalues are complex?" What does it mean by find block disgonalisation?
     
  2. jcsd
  3. May 7, 2009 #2
    [tex] A [/tex] is diagonalizable, just not over [tex] \mathbb{R} [/tex]. As for the block-diagonalization, what they mean is that given a real [tex] 2 \times 2 [/tex] matrix [tex] B [/tex] with eigenvalues [tex] \alpha \pm i \beta [/tex], there exists a real, nonsingular matrix [tex] S [/tex] such that
    [tex]
    S^{-1} B S = \begin{pmatrix}
    \alpha & \beta\\
    -\beta & \alpha
    \end{pmatrix} \textrm{.}
    [/tex]
    This is sometimes called the real canonical form of [tex] B [/tex]. The way you find the matrix [tex] S [/tex] is by first finding some complex eigenvector of [tex] B [/tex] corresponding to the eigenvalue [tex] \alpha + i \beta [/tex], say [tex] \mathbf{w} = \mathbf{u} + i\mathbf{v} [/tex], where [tex] \mathbf{u} [/tex] and [tex] \mathbf{v} [/tex] are real vectors. Then [tex] S [/tex] is the matrix with columns [tex] \mathbf{u} [/tex] and [tex] \mathbf{v} [/tex].

    In your case, the real canonical form of the matrix will turn out to be
    [tex]
    \begin{pmatrix}
    1 &0&0&0\\
    0& -1 &0&0\\
    0&0&0& 1\\
    0&0& -1 &0\\
    \end{pmatrix} \textrm{.}
    [/tex]
    This is what is meant by block-diagonalization. A basis that puts your original matrix into this form will be [tex] \{ \mathbf{w}_{+1}, \mathbf{w}_{-1}, \mathbf{u}, \mathbf{v} \} [/tex], where [tex] \mathbf{w}_{\lambda} [/tex] indicates an eigenvector corresponding to the eigenvalue [tex] \lambda [/tex] and [tex] \mathbf{u} + i \mathbf{v} [/tex] is an eigenvector with eigenvalue [tex] + i [/tex].
     
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