# How do I find the common invariant subspaces of a span of matrices?

Let $G_1, \ldots, G_n$ be a set of $m\times m$ linearly-independent complex matrices.

Let $\mathcal{G} = \operatorname{span}\left\{ G_1, \ldots , G_n\right\}$ be the vector space that spans the set of $G$ matrices, and let $\mathcal{S}$ be any linear subspace of $\mathcal{G}$.

Finally, let $\mathcal{I}$ be a linear subspace of $\mathbb{C}^m$.

Is there a way to determine all pairs $\left( \mathcal{S}, \mathcal{I}\right)$ such that for all $S \in \mathcal{S}$, the subspace $\mathcal{I}$ is an invariant subspace of $S$? In other words, is there a way to determine the linear subspaces of $\mathcal{G}$ whose elements share a (non-trivial) invariant subspace?

This problem is vaguely related to this one posted in 2013.

I know that if we arbitrarily pick an $\mathcal{I}$, we can determine the largest subspace $\mathcal{S}$ for which $\mathcal{I}$ is invariant. The problem is that there are infinitely many possible $\mathcal{I}$ to choose from. Likewise, if we arbitrarily choose an $\mathcal{S}$, we can deduce $\mathcal{I}$, but there are infinitely many $\mathcal{S}$ to choose from.

I haven't made any progress trying to determine both subspaces simultaneously. Any assistance would be greatly appreciated. Even some suggestions on where I might look to find a solution would be much appreciated.

• – David E Speyer Feb 9 '15 at 15:07
• – David E Speyer Feb 9 '15 at 15:10
• @DavidSpeyer - Many thanks for the heads up. I'll look into it. – COTO Feb 11 '15 at 23:56
• Does this help? – Fabby Feb 12 '15 at 19:48
• How about computing products $G_1G_2\cdots G_k$, $\forall k \in \{1,N\}$ and multiply a random vector with that. Then put all vectors with some large enough norm into a new matrix and then iterate by the new matrix with the sequence of matrices. Then iterate the procedure. This (I hope) will give something similar to the "power method" for finding largest eigenvalue, eigenvector pair. Oh yes, probably some re-normalization should be done also. – mathreadler Feb 13 '15 at 16:48