1
$\begingroup$

The complex matrix

$\left[\begin{array}{rrrr} i&1&0&0\\ 0&i&0&0\\ 0&0&-i&0\\ 0&0&0&-i\end{array}\right]$

has repeated eigenvalues $\lambda_1=i$ and $\lambda_2=-i$ each of multiplicity two. The first eigenvalue is defective, having an eigenspace

$\mathbb{E}_{\lambda=i} = \mbox{span}\left\{\left[\begin{array}[c]\\1\\0\\0\\0 \end{array}\right]\right\}$ whose dimension is less than the multiplicity of the eigenvalue,

while the second eigenvalue is complete having an eigenspace

$\mathbb{E}_{\lambda=-i} = \mbox{span}\left\{\left[\begin{array}[c]\\0\\0\\1\\0 \end{array}\right], \left[\begin{array}[c]\\0\\0\\0\\1 \end{array} \right]\right\}$ whose dimension equals the multiplicity of the eigenvalue.

I'm trying to find an example of a real-valued matrix that has repeated complex eigenvalues, at least one of which is complete. Does such a matrix exist?

If such a matrix exists, it would have to be a $4\times4$ matrix or larger.

I think perhaps it doesn't exist and the reason is related to the fact that complex eigenvalues of real matrices come in conjugate pairs, each of which have an eigenvector that is the other's conjugate. Something about that relationship may be preventing a non-defective cases from arising. But, if that's the case, I'm not sure why.

If helpful, here's a real-valued matrix I found in a similar question that has repeated complex eigenvalues, but both of these eigenvalues are defective.

$\left[\begin{array}{rrrr} 0&0&0&-1\\ 1&0&0&0\\ 0&1&0&-2\\ 0&0&1&0 \end{array}\right]$

Finally, I think another way I could pose the question is: are there real-valued diagonalizable matrices with repeated complex eigenvalues.

$\endgroup$

1 Answer 1

2
$\begingroup$

You can take$$\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix},$$which is similar to$$\begin{bmatrix}i&0&0&0\\0&i&0&0\\0&0&-i&0\\0&0&0&-i\end{bmatrix}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .