Are there operators on arbitrarily large vector spaces with no eigenvalues?

Question

So, we know that operators (linear transformations, square matrices, whatever you wanna call 'em) on complex vector spaces always have an eigenvalue. Moreover, operators on real vector spaces of odd dimension always have an eigenvalue (for a similar reason). We also know that some operators in R^2 for example have no eigenvalues (real eigenvalues, that is), for example, rotations (except the 180 degree rotation). My question therefore is, Are there operators on arbitrarily large real vector spaces (of even dimension) with no eigenvalues whatsoever? Can we construct them?

Answer 1

Of course. Consider the transformation of $\mathbb R^{2n}$ represented by the following $2n\times 2n$ matrix: $$M_\theta:=\left[\begin{array}{cccccc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\\&&\cos\theta&-\sin\theta\\&&\sin\theta&\cos\theta\\&&&&\ddots\\&&&&&\cos\theta&-\sin\theta\\&&&&&\sin\theta&\cos\theta\end{array}\right].$$ And pick, for example, $\theta=\pi/2$. Now, it is clear that $M_\theta$ is a block diagonal matrix with diagonal elements being $2\times 2$ rotation matrices, which as you observed have no (real) eigenvalues. It follows that the same is true for $M_\theta$.

Answer 2

Take a $2 \times 2$ rotation matrix whose characteristic polynomial doesn't split (for example, rotation by $\pi/4$ radians); call this matrix $R$. Now, for any positive integer $k$, consider the $2k \times 2k$ block diagonal matrix with only $R$ on its diagonal. Its characteristic polynomial will be $(\chi_{R}(t))^k$, and hence will not split over $\Bbb{R}$; i.e it has no real eigenvalues.

Answer 3

If $f(x) = x^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0$ is a polynomial, then the $n\times n$ matrix $$ A = \left(\begin{array}{ccccr} 0 & 0 & \cdots & 0 & -a_0\\ 1 & 0 & \cdots & 0 & -a_1\\ 0 & 1 & \cdots & 0 & -a_2\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{n-1} \end{array}\right)$$ is called the companion matrix of $f$. It is not hard to show that the characteristic polynomial of $A$ is $(-1)^nf(x)$. (If you know about the minimal polynomial, the minimal polynomial of $A$ is $f(x)$ as well).

That means that the matrix $A$ has an eigenvalue if and only if $f$ has a root.

So for any value of $n$ for which you can find a monic polynomial of degree $n$ with no roots, you can find an $n\times n$ matrix that has no eigenvalues.

Of course, if every polynomial of degree $n$ has a root in your field, then every $n\times n$ matrix over that field will have an eigenvalue.

This gives you, for instance:

• Every square matrix over $\mathbb{C}$ has at least one eigenvalue (in $\mathbb{C}$).
• There are real matrices of any even order that have no eigenvalues (in $\mathbb{R}$).
• Every real matrix of odd order has at least one real eigenvalue.
• There are $n\times n$ matrices with rational coefficients that have no rational eigenvalues for any $n\gt 1$.
• There are $n\times n$ matrices with coefficients in the field $\mathbb{F}_q$ of $q$ elements that have no eigenvalues in $\mathbb{F}_q$, for any $n\gt 1$.
• Every matrix with coefficients in $K$ (a field) has at least one eigenvalue in $K$ if and only if $K$ is algebraically closed.