nilpotent linear transformation and invariant subspaces

Question

I'm trying to proof a biconditional statement about a nilpotent linear transformation, and I think I already proved it one way,but I'm stuck on the other way.

The statement is as follows:

Let $\phi: \mathbb{R}^3 \rightarrow \mathbb{R}^3$, be a nilpotent linear transformation, then $\phi^2 = 0$ if and only if $\phi$ has an infinite amount of $\phi$-invariant subspaces

I already proved that if $\phi^2 = 0$, then there exist an infinite amount of invariant subspaces, my proof is as follows:

We know that $\phi^2 = 0$, so $\phi$ is nilpotent and therefore there exists a basis $B = \{ v_1, v_2,v_3\}$ such that for all $v_i \in B$ we have $\phi(v_i) = v_{i+1}$ or $\phi(v_i)=0$. One can easily find that because $\phi^2 = 0$, we have that $\phi(v_1) = v_2$ and that $\phi(v_2)=\phi(v_3)=0$ . Consider now all hyperplanes such that they are generated by $v_2$ and a vector $b$ such that $b = \lambda \cdot v_1 + \mu \cdot v_3$ with $\lambda \neq 0$. All these hyperplanes consist the line $L_{v_2} = \{ \lambda \cdot v_2 \shortmid \lambda \in \mathbb{R}\}$. By a straightforward calculation we also find that all those hyperplanes are projected onto this line, so all these hyperplanes are $\phi$ invariant and there are an infinite amount of these hyperplanes.

I'm not quite sure whether this proof is correct, and I'm stuck at the reverse: I don't even know where to start with that statement . Also, I'd love to have some good tips on how to proof the converse statement without using jordan canonical form considering that hasn't come up yet in the textbook that I'm using.

Any help would be greatly appreciated.

Answer 1

Your first proof is correct. In other words, if $\phi^2 = 0$, then the dimension of the kernel is at least $2$.

On the other hand, if $\phi^2$ fails to be zero. Then, every invariant subspace of $\phi$ contains an eigenspace, but the only eigenvalue $\phi$ can have is $0$. So, every invariant subspace of $\phi$ contains $\ker(\phi)$, a $1$-dimensional subspace. So, there is exactly one $1$-dimensional invariant subspace.

Now, consider any vector for which $\phi^2(v) \neq 0$. We could use such a vector to build a basis $(v,\phi(v),\phi^2(v))$. So, the only invariant subspace containing $v$ is $\Bbb R^3$.

So, any two-dimensional subspace is a subset of $\ker \phi^2$, which is a $2$-dimensional subspace. So, there is exactly one $2$-dimensional subspace.

Answer 2

If $\phi^2 = 0$ and $\phi \ne 0$, the Jordan canonical form of $\phi$ must be $$ \pmatrix{0 & 1 & 0\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$$ which has two-dimensional null space; all the infinitely many one-dimensional subspaces of this are invariant under $\phi$. Explicitly, these are $$ \text{span}\left(\pmatrix{a \cr 0\cr b\cr}\right)$$ for all $a, b$ not both $0$.

Also, $A^T$ has two-dimensional null space, and the two-dimensional hyperplanes orthogonal to each of these are invariant under $\phi$. Explicitly, these are the spaces of the form $$ \left\{ \pmatrix{x\cr y\cr z\cr}\; : \; a y + b z = 0\right\}$$ for $a, b$ not both $0$.

EDIT: For the converse (without using Jordan form):

Suppose $\phi: \mathbb R^3 \to \mathbb R^3$ is nilpotent but $\phi^2 \ne 0$. Since $\phi$ is nilpotent, $\phi^m = 0$ for some $m$.
If $C(x)$ is the characteristic polynomial of $\phi$, we also have $C(\phi) = 0$, and $C(x)$ has degree $3$. The greatest common divisor $g(x)$ of $x^m$ and $C(x)$ must also have $g(\phi) = 0$; the only monic nonconstant polynomials of degree $\le 3$ dividing $x^m$ are $x$, $x^2$ and $x^3$. But since we're assuming $\phi^2 \ne 0$, the only possibility is $\phi^3 = 0$.

Now there must be a vector $u$ such that $\phi^2 (u) \ne 0$. Let $v = \phi(u)$: this is a vector such that $\phi(v) \ne 0$ but $\phi^2(v) = \phi^3(u) = 0$. Let $w = \phi(v)$, a nonzero vector such that $\phi(w) = 0$. It's not hard to show that $u, v, w$ are linearly independent, so they are a basis of $\mathbb R^3$.

A one-dimensional subspace of $\mathbb R^3$ invariant under $\phi$ is spanned by an eigenvector of $\phi$. Since $\phi$ is nilpotent, the corresponding eigenvalue must be $0$, so the eigenvector is in the null space of $\phi$. But since $\phi(a u + b v + c w) = a v + b w$, the null space of $\phi$ is one dimensional, the span of $w$. And thus there is only one one-dimensional subspace invariant under $\phi$.

For a two-dimensional subspace of $\mathbb R^3$ invariant under $\phi$, the orthogonal complement is a one-dimensional subspace invariant under $\phi^T$, i.e. the span of a member of the null space of $\phi^T$. But by the same argument used for $\phi$, the null space of $\phi^T$ is one-dimensional, and so there is only one two-dimensional subspace invariant under $\phi$.