eigenvectors of invertible matrices over the complex numbers

Question

Suppose we have a matrix $A\in GL_n(\mathbb{C})$. Does $A$ always have at least one eigenvector? Specifically for 2x2 matrices, The rotation matrix has no real eigenvectors but it has complex eigenvectors. The matrix $\begin{pmatrix} 1&&0\\1&&1\\ \end{pmatrix}$ only has $\begin{pmatrix} 0\\1\\ \end{pmatrix}$ as an eigenvector, so clearly $A$ does not need to have $n$ eigenvectors.

I get that any invertible matrix will have a nonzero determinant, so if you write out the characteristic equation you will get at least one nonzero eigenvalue, but does this eigenvalue have to correspond to an eigenvector?

Answer 1

The characteristic polynomial of $A\in GL_n(\mathbb C)$ is of degree $n$ and therefore has at least one root (if of course $n > 0$). Suppose $\lambda_0$ is this root. Then $0 = \det (A - \lambda_0I_n)$ and consequently the matrix $A - \lambda_0I_n$ is singular, which means that the dimension of the column space of $A - \lambda_0I_n$ is greater than $0$, so it has at least one nonzero vector. Such a vector is exactly an eigenvector of $A$, because $$ (A - \lambda_0I_n)x = 0 \Leftrightarrow Ax = \lambda_0x. $$

Moreover, if $A\in GL_{2n+1}(\mathbb R)$, the theorem remains true! Every polynomial with real coefficients of odd degree has a real root.

Answer 2

A matrix

$A \in GL(n, \Bbb C) \tag 1$

always has at least one eigenvector, seen as follows: the linear "eigen-equation" is

$A \vec v = \lambda \vec v, \; \lambda \in \Bbb C,\; 0 \ne \vec v \in \Bbb C^n; \tag 2$

we write this as

$(A - \lambda I)\vec v = 0, \tag 3$

which has a non-zero solution $\vec v$ precisely when

$\chi_A(\lambda) = \det(A - \lambda I) = 0; \tag 4$

so for any $\lambda$ satisfying (4), of which there are at most $n$, we obtain at least one eigenvector $\vec v \in \Bbb C^n$. It is well-known that eigenvectors associated with distinct eigenvalues are linearly independent; thus, there are at least as many independent eigenvectors as there are distinct eigenvalues; if (4) has $n$ distinct zeroes, then $A$ has $n$ linearly independent eigenvectors.

The real rotation matrices such as

$R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \tag 5$

generally have complex eigenvalues, for

$\det(R(\theta) - \lambda I) = \det \left ( \begin{bmatrix} \cos \theta - \lambda & \sin \theta \\ -\sin \theta & \cos \theta - \lambda \end{bmatrix} \right )$

$= (\cos \theta - \lambda)^2 + \sin^2 \theta = \lambda^2 - (2\cos \theta) \lambda + 1 = 0 \tag 6$

typically has complex roots, given as they are by the quadratic formula

$\lambda =\dfrac{2\cos \theta \pm \sqrt{4\cos^2 \theta - 4}}{2} = \dfrac{2\cos \theta \pm 2\sqrt{\cos^2 \theta - 1}}{2} $

$=\cos \theta \pm \sqrt{-\sin^2 \theta} = \cos \theta \pm i\sin \theta = e^{\pm i\theta}; \tag 7$

since the eigenvalues are in general complex, so are the eigenvectors, the exceptions being where $\sin \theta = 0$, that is, when $\theta = n \pi$, $n \in \Bbb Z,$ when they take the form $(1, 0)^T$, $(0, 1)^T$ as $R(n\pi) = \pm \, I$.

It doesn't matter whether an eigenvalue is $0$ or not; it will always have at least one eigenvector provided we are working over an algebraically closed field such as $\Bbb C$. Indeed, the $0$-eigenspace of a matrix $A$ is simply $\ker A$.