So I'm trying to figure out given a linear transformation $T$ how to find a plane that $T$ maps onto itself. One way to state this problem is solving for vectors $u$ and $v$ such that $T(uv)=(uv)\lambda$ where $\lambda$ is two by two and $(uv)$ is the $n\times2$ matrix whose columns are $u$ and $v$. This is clearly an analogue to eigenvalues and eigenvectors, so I tried to come up with a polynomial that will give me $\lambda$ so we can then solve for $u$ and $v$, as we do with eigenvectors. Of course, we will get a different $\lambda$ if we choose a different basis of the same plane, so really we want to solve for $\lambda$ up to similarity.
What I came up with is $$T(uv)=(uv)\lambda\implies \det|T^2-tr(\lambda)T+\det(\lambda)I|=0$$ where $tr(\lambda)$ is the trace. This is a single polynomial in 4 variables, which is a little concerning. I wonder whether some additional clever algebra could give another equation. Alternatively, maybe this one constraint is enough to give $\lambda$ up to similarity.
At this point I tried to find an answer online. What I found is that there is a theory of generalized eigenspaces, but they take a totally different approach. I found a couple textbooks that talk about this, but their approach is completely different. Can anyone explain how the approach I took here fits into the bigger picture?
Edit: I want to point out that typically a fixed plane is not spanned by eigenvectors. For example, $T$ might rotate a plane. In that case no single vector is fixed, but the entire plane is.
Your linear transformation is represented by a matrix $A$ such that $T(V) = AV$
If $U$ and $V$ are two linearly independent eighenvectors of $A$ with eigenvalues $\lambda _1$ and $\lambda _2$ respectively, then $$T(\alpha U + \beta V) = A( \alpha U + \beta V) = \alpha AU + \beta AV = (\alpha \lambda_1)U + ( \beta \lambda_2)V$$
That is the plane spanned by $U$ and $V$ maps to itself under the linear transformation.