# Matrix diagonalisation with complex eigenvalues

1. May 7, 2009

### misterau

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. May 7, 2009

### VKint

$$A$$ is diagonalizable, just not over $$\mathbb{R}$$. As for the block-diagonalization, what they mean is that given a real $$2 \times 2$$ matrix $$B$$ with eigenvalues $$\alpha \pm i \beta$$, there exists a real, nonsingular matrix $$S$$ such that
$$S^{-1} B S = \begin{pmatrix} \alpha & \beta\\ -\beta & \alpha \end{pmatrix} \textrm{.}$$
This is sometimes called the real canonical form of $$B$$. The way you find the matrix $$S$$ is by first finding some complex eigenvector of $$B$$ corresponding to the eigenvalue $$\alpha + i \beta$$, say $$\mathbf{w} = \mathbf{u} + i\mathbf{v}$$, where $$\mathbf{u}$$ and $$\mathbf{v}$$ are real vectors. Then $$S$$ is the matrix with columns $$\mathbf{u}$$ and $$\mathbf{v}$$.

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