# Does every eigenspace of the exterior power $\bigwedge^k A$ corresponds to an invariant subspace?

Let $$V$$ be an $$n$$-dimensional real vector space, and let $$1 be fixed. Given an automorphism $$A \in \text{GL}(V)$$, consider its $$k$$-th exterior power $$\bigwedge^k A \in \text{GL}(V)$$.

Suppose $$\bigwedge^k A$$ admits an eigenvector. (equivalently, $$\bigwedge^k A$$ admits a non-zero eigenvalue).

Does $$A$$ has a $$k$$-dimensional invariant subspace?

If the eigenvector $$v$$ of $$\bigwedge^k A$$ is decomposable, then the answer is positive:

Write $$v=v_1 \wedge v_2 \dots \wedge v_k$$; Then $$\bigwedge^k A(v_1 \wedge v_2 \dots \wedge v_k)=\lambda v_1 \wedge v_2 \dots \wedge v_k \Rightarrow \text{span}(Av_1,\dots,Av_k)=\text{span}(v_1,\dots,v_k),$$

so $$\text{span}(v_1,\dots,v_k)$$ is $$A$$-invariant.

However, I am not sure that every eigenspace $$\bigwedge^k A$$ of should be decomposable.

This question looks somewhat related to this nice question, which is still not fully answered.

Nice question. First, note that over $$\mathbb{C}$$, any operator can be represented with respect to an appropriate basis by an upper triangular matrix. This implies that any operator $$A$$ has invariant subspaces of all possible dimensions so the question is not interesting over $$\mathbb{C}$$.

To construct a counterexample over $$\mathbb{R}$$, I will use the following observations:

1. If $$n$$ is even and the characteristic polynomial of $$A$$ has no real roots, then $$A$$ has no odd-dimensional $$A$$-invariant subspaces. The reason is that if you restrict $$A$$ to an odd-dimensional $$A$$-invariant subspace, you get an operator which must have an eigenvector (with a real eigenvalue), contradicting the fact that all the roots of the characteristic polynomial of $$A$$ aren't real.
2. If the (possibly complex) roots of the characteristic polynomial of $$A$$ are $$(\lambda_i)_{i=1}^n$$ (with multiplicity) then the roots of the characteristic polynomial of $$\Lambda^k(A)$$ are $$(\lambda_{\alpha})$$ where $$\alpha = (i_1 < \dots < i_k)$$ runs over all possible multi-indices and $$\lambda_{\alpha} := \lambda_{i_1} \dots \lambda_{i_k}$$. To see this, assume first that $$A$$ is a complex operator and choose an ordered basis $$(e_i)_{i=1}^n$$ with respect to which $$A$$ is represented by an upper triangular matrix with $$Ae_i = \lambda_i e_i \mod \operatorname{span} \{ e_j \}_{j < i}.$$ Then $$\Lambda^k(A)$$ is represented with respect to the induced ordered basis $$(e_{\alpha})$$ (where the order on the multi-indices is the lexicographical one) by an upper triangular matrix with $$\Lambda^k(A)(e_\alpha) = \lambda_{\alpha} e_{\alpha} \mod \operatorname{span} \{ e_{\beta} \}_{\beta < \alpha}.$$ The result for real operators follows by complexification using the fact that exterior power and complexification commute.

Now, let $$\theta = \frac{2\pi}{3}$$ and set $$\alpha = e^{i\theta}$$. Consider the operator $$A \colon \mathbb{R}^6 \rightarrow \mathbb{R}^6$$ which is represented with respect to the standard basis by the block diagonal matrix $$\begin{pmatrix} \cos \theta & -\sin \theta & 0 & 0 & 0 & 0 \\ \sin \theta & \cos \theta & 0 & 0 & 0 & 0 \\ 0 & 0 & \cos \theta & -\sin \theta & 0 & 0 \\ 0 & 0 & \sin \theta & \cos \theta & 0 & 0 \\ 0 & 0 & 0 & 0 & \cos \theta & -\sin \theta \\ 0 & 0 & 0 & 0 & \sin \theta & \cos \theta \end{pmatrix}.$$

The characteristic polynomial of $$A$$ is $$(z - \alpha)^3(z - \overline{\alpha})^3 = (z^2 - (2 \Re{\alpha})z + |\alpha|^2)^3 = (z^2 + z + 1)^3$$ with roots $$\alpha, \overline{\alpha}, \alpha, \overline{\alpha}, \alpha, \overline{\alpha}.$$ The roots aren't real so $$A$$ doesn't have a three-dimensional invariant subspace. However $$\alpha^3 = 1$$ is a real root of the characteristic polynomial of $$\Lambda^3(A)$$ (of multiplicity two) so $$\Lambda^3(A)$$ has two linearly independent eigenvectors which are necessarily indecomposable.

Remark: One can show using primary decomposition that if a real operator has a real eigenvalue then it has invariant subspaces of all possible dimensions. Hence, counterexamples are possible only in even dimensions. It is a nice exercise to see why you can't have a counterexample in dimension four so this is a minimal counterexample in terms of dimension.