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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. –  Une Femme Douce May 31 '12 at 18:30
Try some examples. –  Qiaochu Yuan May 31 '12 at 18:30

3 Answers 3

up vote 2 down vote accepted

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 $i \in \{0, 1, \ldots, n_k\}$.

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It depends on your background in this area, and on what you mean by counting. Some of the relevant theorems about those kinds of questions will be the matrix diagonalization, primary decomposition theorem and the Jordan theorem.

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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.

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Hi Don: That's a lot of abbreviations! I had to pause at each one to figure it out... could you expand them for the benefit of lower level readers? Thanks! –  rschwieb May 31 '12 at 18:46

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