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Matrix diagonalisation with complex eigenvalues

  • Thread starter misterau
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  • #1
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Homework Statement


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

Homework Equations





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?
 

Answers and Replies

  • #2
139
12
[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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