# Do all square matrices have eigenvectors?

I came across a video lecture in which the professor stated that there may or may not be any eigenvectors for a given linear transformation. And, so far I thought every matrix has eigenvectors. Please clarify.

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Think of rotation in $\Bbb R^2$. –  David Mitra Jan 28 at 11:51
Your question says linear transformation, while your title says square matrices. These are not the same things. –  Marc van Leeuwen Jan 28 at 17:04

## 8 Answers

It depends over what field we're working. For example, the real matrix

$$A=\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}$$

has no eigenvalues at all (i.e., over $\;\Bbb R\;$ ), yet the very same matrix defined over the complex field $\;\Bbb C\;$ has two eignevalues: $\;\pm i\;$

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So when we talk about eigenvectors we must specify the domain? –  Saurabh Shringarpure Jan 28 at 11:58
Always, @SaurabhShringarpure . –  DonAntonio Jan 28 at 12:05
And of course over the rationals $\mathbb{Q}$ even a matrix like $$\begin{pmatrix} 0 & 2 \\ 1 & 0 \\ \end{pmatrix}$$ lacks eigenvalues (and eigenvectors). –  Jeppe Stig Nielsen Jan 28 at 17:36

Over an algebraically closed field, every matrix has an eigenvalue. For instance, every complex matrix has an eigenvalue. Every real matrix has an eigenvalue, but it may be complex.

In fact, a field $K$ is algebraically closed iff every matrix with entries in $K$ has an eigenvalue. You can use the companion matrix to prove one direction. In particular, the existence of eigenvalues for complex matrices is equivalent to the fundamental theorem of algebra.

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Every matrix? (Certainly only square ones). But the question says linear transformations. –  Marc van Leeuwen Jan 28 at 16:58

Take a look at the matrix $$A=\begin{bmatrix}0 &1\\-1 & 0\end{bmatrix}.$$

This matrix has characteristic polynimial of $p(\lambda) = \lambda^2 + 1$, meaning that for real $\lambda$, every matrix $A-\lambda I$ has rank $2$, therefore the matrix has no eigenvectors.

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So when we talk about eigenvectors we must specify the domain? –  Saurabh Shringarpure Jan 28 at 11:55
Of course. An eigenvector of matrix $A$ which maps a vector space $V$ onto itself iz by definition such a vector $x$ in $V$ that $Ax=\lambda x$. Mind you, for complex matrices, you always have at least one eigenvector. –  5xum Jan 28 at 11:57

No, but you can build some.

A matrix in a given field (or even commutative ring) may or may not have eigenvectors. It has eigenvectors if and only if it has eigenvalues, by definition. The Cayley-Hamilton theorem provides an easy characterization of whether a matrix has eigenvalues: the eigenvalues are exactly the roots of the characteristic polynomial. Thus a matrix has eigenvectors if and only if the characteristic polynomial has at least one root. For example, the following matrix over $\mathbb{R}$ has no eigenvectors, because its characteristic polynomial $X^2+1$ has no real root: $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$$ This is a rotation matrix: it represents a planar transformation that transforms any vector into a vector that makes a specific angle with the original (a right angle, in this case), and in particular the result cannot possibly be parallel with the original.

Thus it is certainly possible for a matrix not to have any eigenvectors. However, given a matrix over a field, it is possible to construct a larger field in which the matrix has eigenvectors. Any extension field in which the characteristic polynomial has at least one root will do. In particular, in an algebraically closed field such as $\mathbb{C}$, every matrix has at least one eigenvalue and therefore has eigenvectors. For example, the matrix above, when taken as a matrix over $\mathbb{C}$, has the eigenvalues $i$ and $-i$ and eigenvectors of the form $\{(\pm i z,z) \mid z\in\mathbb{C}\}$.

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+1 for the only complete explanation :) –  nbubis Jan 29 at 9:02

Here are some cases of endomorphisms of a vector space that don't have any eigenvectors. (Note that a linear transformation has to be from a space to itself, i.e. a vector space endomorphism, for the notion of eigenvalue to even be defined. Note also that "linear transformation" implies working over a given field$~K$, which one cannot extend at will, unlike the case of matrices that can be interpreted over any ring that contains their entries.)

• The left-shift operator $\def\N{{\mathbf N}}(a_i)_{i\in\N}\mapsto(a_{i+1})_{i\in\N}$ in the $2$-dimensional $\def\Q{{\mathbf Q}}~\Q$ vector space of sequences in $\Q^\N$ that satisfy the Fibonacci recurrence $a_{i+2}=a_i+a_{i+1}$. (Here eigenvectors do exist for the corresponding endomorphism of the corresponding space of sequences of real numbers.)

• Rotation about an angle $\def\Z{{\mathbf Z}}\alpha\notin\pi\Z$ in a $2$-dimensional Euclidean space. (Here the characteristic polynomial only has complex roots, which do not give eigenvectors in this $\def\R{{\mathbf R}}\R$ vector space.)

• Multiplication by some fixed non-constant polynomial in $K[X]$, or by some fixed non-constant series in $K[[X]]$, for any field $K$. (Here the infinite dimension is the essential aspect.)

• The unique linear endomorphism of any $0$-dimensional $K$-vector space.

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You can make any polynomial (especially those with no real root) be the characteristic polynomial of a matrix.

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The question is equivalent to asking if all polynomials have roots.

Every square matrix has a characteristic polynomial. In the domain of real numbers, not every polynomial has real roots and so not every matrix has an eigenvalue, eigenvector pair. In the domain of complex numbers, every polynomial has at least one root (indeed, that's why complex numbers where originally used), so every matrix has at least one eigen-pair.

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even further, it can be proved that for this complex eigenvalue its conjugated is also an eigenvalue as well the conjugation of the eigenvector is also a eigenvector for the other eigenvalue. –  janmarqz Jan 29 at 14:31

Suppose we consider the field of complex numbers (which is the richest). The Fundamental Theorem of Algebra states that any polynomial with complex coefficients of degree $n$ has exactly $n$ complex roots (taking into account root multiplicities which in our context are also called "algebraic multiplicities"). Therefore any $n$ by $n$ generaly complex matrix has $n$ eigenvalues counting multiplicities.

The question about eigenvectors is not so simple and is equivalent to weather or not the matrix is diagonalizable. The theorem goes: An $n$ by $n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors (that therefore span $\mathbf{C}^n$). If all eigenvalues are distinct this is guaranteed.

Suppose now that not all eigenavues are distinct. Consider an eigenvalue $p$ and define the eigenspace of $p$ as

$$E(p)=\{{x:Ax=px},x\ne0\}$$

The dimension of $E(p)$ is also called the geometric multiplicity of $p$. It can be shown that if $p$ has algebraic multiplicity $1$, $E(p)$ is just a line. If that is the case for all eigenvalues then our matrix has $n$ linearly independent eigenvectors and hence is diagonalizable. Suppose now that $p$ has algebraic multiplicity $k\ge2$. Then the dimension of $E(p)$ (the geometric multiplicity) can be anything in $\{{1,2,\dots,k}\}$ ie a line, a plane and so on. In this case the theorem must be modified: A matrix is diagonalizable if and only if the algebraic multiplicity of each eigenvalue matches the geometric multiplicity.

A good example (the Jordan canonical form matrices) is

$$A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}.$$

The characteristing polynomial is $-p^3$ which has only one root ($0$) but with multiplicity $3$. And it has only $1$ eigenvector

$$e=\begin{pmatrix}1\\0\\0\end{pmatrix}.$$

One eigenvector is not enough to span $\mathbf{C}^3$ and therefore the matrix is not diagonalizable. I think I made this reply longer than it is but that's how you count complex eigenvectors.

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