# Left and right eigenvalues

A quaternionic matrix $A$ gives rise to a function $\mathbb{H}^n \to \mathbb{H}^n$ given by $x \mapsto A \cdot x$. This is real linear, but not complex- or quaternionic-linear (in general) if we consider $\mathbb{H}^n$ as a left $\mathbb{C}$ or $\mathbb{H}$ module, but is pretty good if we use right actions.

A right eigvenvalue of $A$ is a quaternion $q$ such that $A\cdot x = x \cdot q$ for some $x\in \mathbb{H}^n$; a left eigenvalue is quaterion $q$ such that $A \cdot x = q\cdot x$ for some $x\in \mathbb{H}^n$.

The algebra of right eigenvalues is pretty good, but the algebra of left eigenvalues is quite interesting. For example, it is not hard to see that there are matrices $A$ with infinitely many left eigenvalues, even for $2$-by-$2$ matrices.

Now let's assume that $A\in Sp(n)$, so that the left eigenvalues are all contained in $S^3\subseteq \mathbb{H}$. What sort of geometric properties must the set $L(A)$ of left eigenvalues have?

EDIT: An example is $$\left$\begin{array}{cc} 0 & 1 \cr -1 & 0 \end{array} \right$ \cdot \left$\begin{array}{c} 1 \cr q \end{array} \right$ = q \cdot \left$\begin{array}{c} 1 \cr q \end{array} \right$$$ for any $q\in S^3 \subseteq \mathbb{H}$ with zero real part, since then $q^2 = -1$.

EDIT 2: Examples like this show that for some symplectic matrices, the set of left eigenvalues is a union of copies of $S^2$.

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What's an example of a two-by-two quaterninonic matrix with infinitely-many left eigenvalues? –  José Figueroa-O'Farrill Jul 17 '10 at 17:48
It's misleading to call the left gadgets "eigenvalues". –  Victor Protsak Jul 17 '10 at 18:54
I tried for clarity by always using "left eigenvalues" and never "eigenvalues." –  Jeff Strom Jul 17 '10 at 20:33

In fact, except if $A\in Sp(n)$ is an involution, there are also infinitely many right eigenvalues (all in $\mathbb{S}^3$). I realize that the question is about left eigenvalues, but as the following is too long for a comment, I post it as answer. Apologies.

If $\lambda$ is such an eigenvalue, $Ax=x\lambda$, $x\in\mathbb{S}^{4n-1}$, then for any $\mu\in\mathbb{S}^3$, $Ax\mu=x\mu\cdot\overline{\mu}\lambda\mu$, so all the conjugates $\overline{\mu}\lambda\mu$ are also eigenvalues.

To see if there are eigenvalues at all, one can consider critical values of the function $x\mapsto \mathrm{Re}(x^*Ax)$ from $\mathbb{S}^{4n-1}$ to $\mathbb{R}$. Let $t\in[-1,1]$ be such a value, taken at the vector $x$. Then if $t=\pm 1$, $Ax=\pm x$, and if $|t|<1$, $x$ and $Ax$ are independent over $\mathbb{R}$, and it is easy to see that the real plane they generate is invariant under $A$ (in fact, $(A+A^{-1})x=2tx$). Letting $t=\cos(\theta)$, you can then find orthogonal $v_1,v_2$ of norm one in this plane such that $A$ has the standard rotation matrix $R_\theta$ in this basis. Then $v=v_1-v_2i$ is a right eigenvector with eigenvalue $e^{i\theta}$, and one can replace $i$ with any norm one imaginary quaternion.

As in the unitary case, the (quaternionic) orthogonal hyperplane to $v$ is invariant, and you can induct on dimension to find a complete basis.

In short : conjugacy classes in $Sp(n)$ are parameterized by $n$ real numbers in [-1,1].

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