# eigenvectors of invertible matrices over the complex numbers

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?

• Any square matrix has at least one eigenvector because it has at least one eigenvalue. (Note that if $v$ is an eigenvector then so is $t v$ for $t \neq 0$, so any matrix has lots of eigenvectors. I presume you are considering the entire ray to be the same eigenvector as such.) – copper.hat Feb 9 at 23:13
• Are you asking if any matrix has a real eigenvector? This is not true. Take $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. – copper.hat Feb 9 at 23:14
• If $\lambda$ is an eigenvalue then $A - \lambda I$ is not invertible and so has a kernel of dimension at least one. However, this null space need not have a direction with all real (as opposed to complex) components. – copper.hat Feb 9 at 23:16
• It depends on the field: In $\mathbb{C}$ any polyomial (hence also any characteristic function) has a root, and thus an eigenvector, but in $\mathbb{R}$ there are polynomials which do not have a root - thus also some characteristic function will not have an eigenvalue, and consequently no eigenvector – Maksim Feb 9 at 23:18
• "but does this eigenvalue have to correspond to an eigenvector?" Eigenvalues always correspond to eigenvectors. $\lambda$ is an eigenvalue for $A$ if and only if $Av = \lambda v$ has a non-zero solution for $v$. Similarly, a vector $v$ is an eigenvector if and only if it solves the above for some $\lambda$. The definition mentions nothing about characteristic polynomials; that's a theorem! An eigenvalue without an eigenvector is not an eigenvalue at all. – Theo Bendit Feb 9 at 23:51

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 $$\ker A$$.

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.