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The first question is:

Let $f$ be an endomorfism of an $n$-dimensional vector space $V$ with $n$ different eigenvalues. How does one show that $V$ has precisely $2^n$ subspaces which are $f$-invariant?

The second questions is:

Let $f$ be a nilpotent endomorfism of a 3-dimensional $\mathbb{R}$-vectorspace $V$. How does one show that $f$ has infinitely many $f$-invariant subspaces iff $f^2 = 0$.

I tried answering these questions by playing around with the using the definitions of the relevant concepts (endomorfism, nilpotence, et cetera) but I couldn't figure them out.

Max Lonysa Muller
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  • For question 1, see the final comments by Arturo Magadin in his answer here: http://math.stackexchange.com/questions/31158/finding-a-invariant-subspaces-for-a-specific-matrix – user43208 Oct 08 '13 at 13:44
  • Perhaps even more pertinently, see Magadin's explanation here, which is very closely related to your question 1: http://math.stackexchange.com/questions/62338/diagonalizable-transformation-restricted-to-an-invariant-subspace-is-diagonaliza – user43208 Oct 08 '13 at 13:48

1 Answers1

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The first question:

A subspace $V_K$ spanned on a set $K$ of eigenvectors has the dimension $k=|K|$ and has not any eigenvector outside $|K|$. So all such $2^n$ subspaces are different. From the answer of Zorn in Diagonalizable transformation restricted to an invariant subspace is diagonalizable it follows that there are not other invariant subspaces.

The second question:

If $f^2=0$ then $\dim {\rm im}f+\dim \ker f=3$ and ${\rm im} f\subseteq \ker f$, so $\dim \ker f\ge 2$ and all its subspaces are invariant. Conversely, if $f$ has infinitely many invariant subspaces then it has infinitely many $1$-dimensional invariant subspaces (since every $2$-dimensional invariant subspace contains infinitely many $1$-dimensional invariant ones) and all of them must be in $\ker f$. Hence $\dim \ker f=2$ and $\dim {\rm im}f=1$. If ${\rm im}f\not\subset\ker f$ then ${\rm im}f$ is invariant and $f$ is not nilpotent. Hence ${\rm im}f\subset\ker f$ and $f^2=0$.

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Boris Novikov
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