Solution space for quadratic equations with nilpotent matrices

Question

Let ${\bf w}\in\mathbb{R}^3$ and ${\bf N}\in\mathbb{R}^{3\times 3}$ be a nilpotent matrix with degree 3. Consider the following system of quadratic equations,

$$ \begin{align} {\bf w}^\top{\bf w} &= 1, \\ {\bf w}^\top{\bf N}{\bf w} &= 0, \\ {\bf w}^\top{\bf N}{\bf N}{\bf w} &= 0, \end{align} $$

Hypothesis: Every solution ${\bf w}$ can be generated from a single basis vector ${\bf w}_0$ by means of applying ${\bf N}$ (so three in this case, ${\bf w}_0, {\bf Nw}_0, {\bf NNw}_0$).

This is straight-forward to see if $\bf N$ is the shift operator, i.e. if

$${\bf N} = \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$

Question: Is this true for general nilpotent ${\bf N}$ and in more dimensions? And if so, how would you go about a proof? If nilpotent is not enough, can one proof the hypothesis for shift-operators?

Answer 1

$N$ satisfies $N^3 = 0$ but $N^2 \ne 0$. The null space ${\mathscr N}(N)$ must be one-dimensional and ${\mathscr N}(N^2)$ two-dimensional.

There are three cases:

1. If $N w = 0$, your second and third equations are obviously true. Now $w$ is a unit vector in the $1$-dimensional space ${\mathscr N}(N)$, and there are two of them.
2. If $N^2 w = 0$ but $N w \ne 0$, your third equation is true. $w$ and $N w$ are both in the two-dimensional space ${\mathscr N}(N^2)$, with $N w$ a nonzero member of its one-dimensional subspace ${\mathscr N}(N)$. The second equation says $w$ is orthogonal to $Nw$ and thus to ${\mathscr N}(N)$. The orthogonal complement of ${\mathscr N}(N)$ in ${\mathscr N}(N^2)$ is a one-dimensional space. Again we have two unit vectors there.
3. If $N^2 w \ne 0$, then $N w$ and $N^2 w$ are linearly independent vectors in the two-dimensional space ${\mathscr N}(N^2)$. The second and third equations say $w$ is orthogonal to those two vectors, and therefore to all of ${\mathscr N}(N^2)$. The orthogonal complement of ${\mathscr N}(N^2)$ is one-dimensional, and again has two unit vectors.

Putting these together, we have $6$ solutions.

However, the solutions are not necessarily of the form you gave. For a random example, try $$N = \pmatrix{4 & 8 & -4\cr 1 & 2 & -16\cr 1 & 2 & -6\cr}$$ The six solutions are $$ \pm \pmatrix{1/\sqrt{41}\cr 2/\sqrt{41}\cr -6/\sqrt{41}}, \ \pm \pmatrix{6/\sqrt{205}\cr 12/\sqrt{205}\cr 5/\sqrt{205}}, \ \pm \pmatrix{2/\sqrt{5} \cr 1/\sqrt{5}\cr 0} $$ But $$ N \pmatrix{ 1/\sqrt{41}\cr 2/\sqrt{41}\cr -6/\sqrt{41}} = \pmatrix{44/\sqrt{41}\cr 101/\sqrt{41}\cr \sqrt{41}\cr}$$ which is not a solution.

More generally, consider an $n \times n$ matrix $N$ over $\mathbb R$ that is nilpotent of degree $n$, i.e. $N^n = 0$ but $N^{n-1} \ne 0$. Then $\mathscr N(N^j)$ is $j$-dimensional for $1 \le j \le n-1$. Suppose $w$ is a unit vector with $w^T N^j w = 0$ for $j = 1 \ldots n-1$.
There are $n$ cases, corresponding to $k = 1 \ldots n$. I claim that for each such $k$, there are two solutions where $N^k w = 0$ but $N^{k-1} w \ne 0$. Thus there are $2n$ solutions in all.

Proof of claim: If $N^k w = 0$ but $N^{k-1} w \ne 0$, we certainly have $w^T N^j w = 0$ for $j \ge k$. The $k-1$ vectors $ Nw,\; \ldots N^{k-1} w$ are linearly independent members of ${\mathscr N}(N^{k-1})$, so they form a basis of that $k-1$-dimensional space. The equations $w^T N^j w$ then say that $w$ is orthogonal to ${\mathscr N}(N^{k-1})$. Thus $w$ is a unit vector in a one-dimensional space, the orthogonal complement of ${\mathscr N}(N^{k-1})$ in ${\mathscr N}(N^{k})$, and there are exactly two of these.