Let $f$ be a nilpotent endomorphism of a 3 dimensional $\mathbb{R}$-vector space $V$. My book says that $f$ has infinitely many $f$-invariant subspaces if and only if $f^2=0$, but it does not show me why.
Can somebody prove this for me, since it doesn't seem that trivial to me and I have no clue as to why this would hold?
Edit: Can you prove this without using Jordan forms?
Suppose that $f$ is a nilpotent endomorphism of a $3$-dimensional vector space. Then its only eigenvalue is $0$, and the three possible Jordan forms of $f$ are $$ \pmatrix{0&1\\&0&1\\&&0}, \quad \pmatrix{0\\&0&1\\&&0},\quad \pmatrix{0\\&0\\&&0} $$ The form satisfying $J^2 \neq 0$ does not have infinitely many invariant subspaces, since it only has a one-dimensional eigen-space. The other two forms have an eigenspace of dimension $2$, and therefore have infinitely may $1$-dimensional eigenspaces.
Without Jordan canonical form:
Suppose that $f$ is a nilpotent endomorphism of a $3$-dimensional vector space. Then, its only eigenvalue is $0$.
Suppose that there is a $v$ such that $f^2(v) \neq 0$. Show, then, that $f$ has the form $$ \pmatrix{0&1\\&0&1\\&&0} $$ With respect to the basis $\{f^2(v),f(v),v\}$. If $f$ has no such vector, show that it must map a two-dimensional eigenspace to $0$.