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I'm considering a $d \times d$ square matrix $A$ over $\mathbb{R}$, and a given vector $v \in \mathbb{R}^d$, and I'm wondering when the matrix $$\Sigma = [A^0v \quad A^1 v \quad A^2v\quad\ldots\quad A^{d-1}v]$$ has full rank (with $A^0$ being the identity matrix $I_d$).

The vector $v$ is considered given, so (I think) this is not just a question of whether a such $v$ exists, which I've seen discussed a few places. (e.g. here).

As this answer explains, if $A$ is diagonalizable, this is equivalent to $A$ having $d$ distinct eigenvalues and that $v$ cannot be written as a linear combination of $m < d$ of the eigenvalues of $A$.

But in the case that $A$ is not diagonalizable, I cannot come up with a similar criterion. Inspired by the diagonalizable case, one could think that again distinct eigenvalues would be the key. But if $$ A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \qquad v = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, $$ $A$ has eigenvalues $1, 1$ and $2$, but $[v \quad Av\quad A^2v] = \begin{pmatrix}1 & 2 & 4 \\ 1 & 2 & 3 \\ 1 & 1 & 1\end{pmatrix}$ has full rank still.

(Observe that if one removes the off-diagonal in $A$, it becomes diagonalizable and so by the non-unique eigenvalues, it doesn't have full rank).

Can anyone phrase conditions on $A$ and $v$ to determine whether the cyclic subspace has full rank?

Benjamin Wang
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Nikolaj Theodor Thams
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1 Answers1

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$A$ admits a cyclic vector if and only if its Jordan form consists of only one Jordan block for each of its distinct eigenvalues.

Given that $A$ admits a cyclic vector, let $J$ denote the Jordan form of $A$, and $S$ be a similarity transformation for which $S^{-1}AS = J$. Suppose that $$ J = J_{m_1}(\lambda_1) \oplus \cdots \oplus J_{m_k}(\lambda_k), $$ where $J_m(\lambda)$ denotes the size-$m$ Jordan block associated with $\lambda$, and $\oplus$ denotes a direct sum.

Claim: Given that $A$ satisfies the above, we can state the following:

  • $v$ is a cyclic vector of $J$ if and only if the $j$th component of $v$ is non-zero for each $j$ of the form $j = \sum_{p=1}^q m_p$ for $q=1,\dots,k$. In other words, $v$ is cyclic iff the components corresponding to the final column of each Jordan block are non-zero.
  • $w$ is a cyclic vector of $A$ if and only if $S^{-1}w$ is a cyclic vector of $J$.

It is easy to show that the first statement implies the second, but proving the first is perhaps a bit trickier. In this statement, the difficult direction of proof is showing that every such $v$ is indeed cyclic. One proof strategy for this implication be divided into the following steps:

  1. Let $K = J_m(\lambda)$. Show that $v$ is a cyclic vector of $K$ iff its final component is non-zero.
  2. Show that if $p(\lambda) \neq 0$, then $p(K) v$ is also a cyclic vector of $K$.
  3. Let $p_1(x) = (x - \lambda_2)^{m_2} \cdots (x - \lambda_k)^{m_k}$. Show that if $v$ satisfies the hypotheses of the statement, then $p_1(J)v$ has the form $(w, 0)$, where $w$ is a cyclic vector of $J_{m_1}(\lambda_1)$.
  4. Argue inductively: with the hypothesis that the vector $(v_{m_1 + 1}, \dots v_{d})$ is a cyclic vector of $J_{m_2}(\lambda_2) \oplus \cdots \oplus J_{m_k}(\lambda_k)$, deduce that $v$ is a cyclic vector of $J$.
Ben Grossmann
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