Can a matrix have no eigenvectors?

Question

Is it possible for a matrix to have no eigenvectors? One way this could happen: say $A$ is an invertible matrix. And the characteristic polynomial, $|A - \lambda I| = 0$ has only one solution, $\lambda = 0$. Then, when you try to find an eigenvector via $(A-\lambda I)v=0$, you get $Av=0$ for which $v=0$ is the only solution. If this is impossible, are there any other ways there could be no eigenvector at all?

Answer 1

If $A$ is a square matrix of order $n$ its characteristic polynomial has the form

$$ \det(A-tI_n)=(-1)^nt^n+\text{terms of degree $<n$}.$$

Such a polynomial has always a root $\tau$ over the algebraic closure of the field of coefficients of $A$. Since $\det(A-\tau I_n)=0$ the matrix $A-\tau I_n$ has rank $<n$ and a non-zero vector $v$ in its nullspace (in fact the kernel of a linear transformation) is an eigenvector for $A$.

Of course it is possibile that the characteristic polynomial has no root in the field of coefficients of the matrix. For instance a $2\times2$ matrix representing a rotation of the real plane different from an integer multiple of $\pi$ has no real eigenvectors.

Answer 2

Consider the real matrix $A=\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$. There do not exist real eigenvalues of $A$ hence it doesn't have eigenvectors in the field of real numbers. However, in the field of complex numbers, it will definitely possess eigenvectors.