Let $A$ be a real square matrix of order $n$ and if $A$ has complex(non-zero imaginary part) eigenvalues then their corresponding eigenvectors must be complex.
I found one obvious example:
\begin{equation}A=
\begin{pmatrix}
0 & -1 \\
1 & 0\\
\end{pmatrix}
\end{equation}
which has two distinct eigenvalues that are $i $ and $-i $
and corresponding eigenvectors are also complex.
I believe that it is true for any matrix but am unable to prove it.
Am I correct? If yes then can someone help to prove it? And if it is not true then give one counterexample.
Thanks in advance.
Best Answer
First note that since all real numbers are also complex numbers, so any $n\times n$ matrix has exactly $n$ complex eigenvalues (counted with multiplicity). So what you're asking literally is trivially true.
What I think you mean by 'complex' here is 'complex with nonzero imaginary part' or 'complex and not real'.
In that case, then
if the entries/elements of $A$ are all real ($A\in \mathbb R^{n\times n}$), then the answer is yes: if all eigenvectors are complex, then all eigenvalues are complex. We can prove this by contradiction: Let $x$ be an eigenvector with non-real eigenvalue $\lambda$. Assume $x$ is real. Then since $Ax=\lambda x$, the rhs is a non-real vector, and the lhs is a real vector. Thus we have a contradiction, so $x$ must be non-real.
if we allow the entries of $A$ to be complex on ($A\in \mathbb C^{n\times n}$), in general, then the answer is no