# Counting invariant subspaces of a Vector space

Well, I was reading about invariant subspaces and related things and this question came to my mind: If I choose a vector space and fix a linear transformation on itself, then how many invariant subspaces will there be? Is there any formula or materials to read?

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It depends. Have you tried any examples? –  Qiaochu Yuan May 31 '12 at 18:24
No I did not tried, but did some problems on invariant spaces, but did not find anything like " If $dim V=n$ and $T$ is a linear operator on $V$ then these are the or this number of invariant subspaces will be there". For example if $T$ is a linear operator on $\mathbb{R}^3$ then t fixes a line i mean it has one dimensional invariant subspace. –  La Belle Noiseuse May 31 '12 at 18:30
Try some examples. –  Qiaochu Yuan May 31 '12 at 18:30

Consider a linear transformation $T$ on a finite-dimensional vector space over the complex numbers (or any algebraically closed field). If $T$ has an eigenvalue $\lambda$ with two linearly independent eigenvectors $u$ and $v$, then the span of $u + c v$ is invariant for any scalar $c$, so there are infinitely many invariant subspaces. So now assume each eigenvalue has only a one-dimensional eigenspace. Let the characteristic polynomial of $T$ be $\prod_{j=1}^k (\lambda - \lambda_j)^{n_j}$ where $\lambda_1,\ldots,\lambda_k$ are the distinct eigenvalues. Then there are $(n_1+1)(n_2+1)\ldots(n_k+1)$ invariant subspaces. Namely, an invariant subspace is the direct sum of spaces $V_j$, $j=1 \ldots, k$, where $V_j = \{x: (T - \lambda_j I)^i x = 0\}$ for some $i \in \{0, 1, \ldots, n_k\}$.
It all depends on the vector space (=v.s.) and on the linear transformation (=transf.) $\,$itself: for example, over any v.s., all the subspaces are invariant under the zero transf., but over $\,\mathbb{R}^2\,$ the map $\,\,(x,y)\to (y,-x)\,\,$ has no non-trivial invariant subspaces, as you can check.