Is a matrix with real and/or complex conjugate eigenvalues necessarily real?

Question

A real matrix has only real and/or complex conjugate eigenvalues. Is the inverse true? In other words, is a matrix with real and/or complex conjugate eigenvalues necessarily real?

Answer 1

I found a few cases where this is not necessarily true, so I thought of answering my own question. Consider the case of diagonizable matrices. For these matrices we can write $A=V^{-1}\Lambda V$, where $A$ is the diagonizable matrix, $V$ is the matrix of eigenvectors and $\Lambda=diag(\lambda_1, ...,\lambda_N)$ with $\lambda_1, ...,\lambda_N$ the $N$ eigenvalues of $A$. By reshuffling the matrices as $V A V^{-1}=\Lambda$, it is clear that if the eigenvalues are complex (right hand side of the equal), the eigenvectors (on the left hand side of the equal) need to be complex too if $A$ is real. Therefore, it is not necessarily true that matrices with real and/or complex conjugate eigenvalues are real.

Answer 2

@ Ozz,

in fact your question is poorly asked: indeed the answer is obviously NO because the non-real matrix $\begin{pmatrix}0&i\\0&0\end{pmatrix}$ has real eigenvalues.

A more interesting question is as follows: let $A$ be a complex matrix s.t. $A$ has real eigenvalues and, eventually, non-real eigenvalues $(\lambda_i,\overline{\lambda_i})$ s.t. for every $i$, $\lambda_i$ and $\overline{\lambda_i}$ have same multiplicity ; is $A$ similar to a real matrix ?

The answer is NO ; yet the answer is YES if we add the following condition: for every $i$, the number of Jordan blocks corresponding to $\overline{\lambda_i}$ and their orders are identical to the number and the orders of Jordan blocks corresponding to $\lambda_i$. The last condition is equivalent to: for every $k$, $\dim(\ker(A-\lambda_iI)^k)=\dim(\ker(A-\overline{\lambda_i}I)^k)$.

In particular, my above example is similar to the real matrix $\begin{pmatrix}0&1\\0&0\end{pmatrix}$