When $v, Av, A^2v$ are linearly independent Suppose $A \in \mathbb{R}^{3 \times 3}$ and $v \in \mathbb{R}^3.$ I'm looking for some general conditions on when the set $\{ v, Av, A^2v\}$ are linearly independent. In particular, a solution even for the case when $A$ is diagonalizable with 3 distinct eigenvalues would be a good start, but I'm not quite sure where to begin. Thanks!
 A: If the minimal polynomial of $A$ has degree $\le 2$, then it can't happen.  Conversely, if the minimal polynomial has degree $3$, almost every $v$ will work.
In the case where $A$ has $3$ distinct eigenvalues $\lambda_j$ with corresponding eigenvectors $v_j$, and thus minimal polynomial $(z - \lambda_1)(z-\lambda_2)(z-\lambda_3)$, any vector $v = c_1 v_1 + c_2 v_2 + c_3 v_3$ with $c_1, c_2, c_3$ all nonzero will work.  This is because
for any polynomial $p(z)$, $p(A) v_j = p(\lambda_j) v_j$.  If $p(z) = a_0 + a_1 z + a_2 z^2$ has degree $\le 2$, at least one $\lambda_j$ is not a root of $p(z)$, and the corresponding $p(A) v_j \ne 0$.  Since $v_1$, $v_2$ and $v_3$ are linearly independent, $a_0 v + a_1 A v + a_2 A^2 v = c_1 p(\lambda_1) v_1 + c_2 p(\lambda_2) v_2 + c_3 
p(\lambda_3) v_3 \ne 0$.
A: By appealing to the scalar triple product, we see that the triple $({\bf v}, A{\bf v}, A^2 {\bf v})$ is linearly independent iff $$p(A, {\bf v}) := \operatorname{det}\pmatrix{{\bf v}&A{\bf v}&A^2 {\bf v}} \neq 0 .$$ We analyze this condition for each possible Jordan normal form and then interpret the result geometrically to derive a condition that applies to all matrices with that normal form.
Case I: $A$ is diagonalizable
Subcase I: All eigenvalues of $A$ are real.
In any basis $\mathcal{B} = ({\bf e}_i)$ of eigenvectors of $A$, $[A]_{\mathcal B}$ is some diagonal matrix $\Lambda$ whose diagonal entries are the eigenvalues $\lambda_k$. If we decompose ${\bf v} = \sum_{i = 1}^3 v^i {\bf e}_i$, then
$$p(A, {\bf v}) = v^1 v^2 v^3 (\lambda_2 - \lambda_3) (\lambda_3 - \lambda_1) (\lambda_1 - \lambda_2).$$
Interpreting geometrically the nonvanishing conditions $(\lambda_2 - \lambda_3) (\lambda_3 - \lambda_1) (\lambda_1 - \lambda_2) \neq 0$ and $v_1 v_2 v_3 \neq 0$, it is sufficient and necessary for $({\bf v}, A{\bf v}, A^2 {\bf v})$ to be linearly independent that both:

*

*All $3$ eigenvalues of $A$ are distinct.

*The vector ${\bf v}$ is not contained in any $2$-dimensional sum of eigenspaces of $A$. Put another way,
${\bf v} \not\in \operatorname{span}\{{\bf e}_2, {\bf e}_3\} \cup \operatorname{span}\{{\bf e}_3, {\bf e}_1\} \cup \operatorname{span}\{{\bf e}_1, {\bf e}_2\}$.

Subcase II: $A$ has a pair of nonreal, complex conjugate eigenvalues.
Suppose the eigenvalues are $\alpha \pm \beta i$, $\lambda$, with $\alpha, \beta, \lambda \in \Bbb R$ (in particular, $\beta \neq 0$). Then, we can choose a (real) basis $\mathcal B = ({\bf e}_i)$ such that $$A_{\mathcal B} = \pmatrix{\alpha&-\beta\\\beta&\alpha\\&&\lambda} .$$
Computing gives that $$p(A, {\bf v}) = ((v^1)^2 + (v^2)^2) v^3 \beta [(\alpha - \lambda)^2 + \beta^2] .$$ So, the triple is linearly independent if $\bf v$ is contained neither in the unique $2$-dimensional invariant subspace of $A$ (namely, $\operatorname{span}\{{\bf e}_1, {\bf e_2}\} = \{v^3 = 0\}$ nor the $\lambda$-eigenspace (that is, $\operatorname{span}\{{\bf e_3}\} = \{v^1 = v^2 = 0\}$).
The nondiagonalizable cases are similar, and in those cases all eigenvalues must be real.
Case II: The largest Jordan block of $A$ has dimension $2$.
Choosing a suitable basis ${\mathcal B}$ of generalized eigenvectors gives
$$A_{\mathcal B} = \pmatrix{\lambda&1\\&\lambda\\&&\mu},$$ where $\lambda$ and $\mu$ are the eigenvalues of algebraic multiplicity $1, 2$, resp. Decomposing $\bf v$ as before, we find that
$$p(A, {\bf v}) = -(v^2)^2 v^3 (\lambda - \mu)^2,$$
so in this case a sufficient and necessary condition for the nonvanishing of $p(A, {\bf v})$ is that both:

*

*The $2$ eigenvalues of $A$ are distinct.

*The vector ${\bf v}$ is contained in neither the direct sum of the eigenspaces of $\lambda$ and $\mu$ (namely, $\operatorname{span}\{{\bf e}_1, {\bf e}_3\} = \{v^2 = 0\}$) nor the generalized eigenspace of $\lambda$ (namely, $\operatorname{span}\{{\bf e}_1, {\bf e}_2\} = \{v^3 = 0\}$).

Case III: The largest Jordan block of $A$ has dimension $3$
In the remaining case, that the Jordan block of $A$ has dimension $3$, say, with eigenvalue $\lambda$, so $\Bbb R^3$ is itself the generalized eigenspace of $\lambda$, in a suitable basis we have $$A_{\mathcal B} = \pmatrix{\lambda&1\\&\lambda&1\\&&\lambda} ,$$ and we have
$$p(A, {\bf v}) = -(v^3)^3,$$ so in this case the condition is simply that ${\bf v}$ is not in the image of $A - \lambda I$ (namely, $\operatorname{span}\{{\bf e}^1, {\bf e}^2\} = \{v^3 = 0\}$), equivalently, the preimage of the $1$-dimensional eigenspace $\{{\bf e}^1\}$ under $A$.
Remark The quantity $p(A, {\bf v})$ is a nontrivial homogeneous polynomial condition of total degree $6$ in the entries $a_{ij}, v_k$ of $A, {\bf v}$, respectively. So, for almost every pair $(A, {\bf v}) \in \Bbb R^{3 \times 3} \times \Bbb R^3$ the triple is linearly independent, in the sense that the set of such pairs is nonempty and Zariski-open, hence dense, in $\Bbb R^{3 \times 3} \times \Bbb R^3 \cong \Bbb R^{12}$.
