This questsion is an extension of a previously asked question that I think needs a more rigorous proof (please see the link and the answer by user Aaron).

The question is, given a little bit of different notation (in accordance with control theory), for $b\in\mathbb{R}^{n}$ and $A\in\mathbb{R}^{n\times n}$, what is a sufficient condition for

\begin{align} b,Ab,A^{2}b,\dots,A^{n-1}b \end{align}

all being linearly independent? Furthermore, if we assume that $A$ has $n$ distinct eigenvalues $\lambda_{i}$, $i=1,\dots,n$, is this condition sufficient for the above to hold and if so how do we prove it?

In the above linked question Aaron's answer was:

If the characteristic polynomial of $A$ has no repeated roots, then we have a basis $\{v_{i}\;\vert\;1\leq i\leq n\}$ of eigenvectors for $\mathbb{C}^{n}$, and we can express $x=\sum \alpha_{i}v_{i}$ in terms of the basis. If all of the $\alpha_{i}$ are non-zero, then all of your vectors will be linearly independent. In fact, with this condition on $A$, this is a necessary and sufficient condition on $x$.

If $A$ has an eigenvalue with geometric multiplicity greater than 1, then no $x$ will work.

If $A$ has repeated eigenvalues but all eigenvalues have geometric multiplicity of 1, then it is still possible to find such an $x$, but things are a bit more complicated.

I would be very happy if someone could fill in the blanks for his statements, it is not immediately evident that a $b$ with nonzero entries leads to all vectors $b,Ab,\dots,A^{n-1}b$ being linearly independent.

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    $\begingroup$ Aaron doesn't say $b$ has nonzero entries. Aaron says if the expression for $b$ as a linear combination of the eigenvectors has all nonzero coefficients, etc., etc. $\endgroup$ Commented Nov 17, 2020 at 23:50
  • $\begingroup$ Yes, you are correct, my fault. However if you make change of basis then those $\alpha_{i}$ will become coordinates for $b$, thus entries in the transformed case. In this case, it would be correct what I said first. Also, $A$ would take on a diagonal form. $\endgroup$ Commented Nov 17, 2020 at 23:56

1 Answer 1


No assumption is made on the field $k$ and $A\in M_n(k)$.

For $b\in k^n$, the $A^0 b,\ldots, A^{n-1}b$ are linearly dependent iff there is some non-zero polynomial $h\in k[x]$ of degree $<n$ such that $h(A)b=0$.

With $p\in k[x]$ the minimal polynomial of $A$ and $g=\gcd(p,h)$ then $h(A)b=0,p(A)b=0$ implies $g(A)b=0$.

Thus the necessary and sufficient condition for the $A^0 b,\ldots, A^{n-1}b$ to be linearly independent is that $\deg(p)=n$ and $f(A)b\ne 0$ for each $f\in k[x]$ strictly dividing $p$.

When the $\deg(p)=n$ condition is satisfied, factorize in irreducibles $p=\prod_j q_j^{e_j}\in k[x]$. Taking $b_j \in \ker(q_j(A)^{e_j}) ,\not \in \ker(q_j(A)^{e_j-1})$ then $b = \sum_j b_j$ works.

(take $f$ such that $p = q_j f$, then $0=f(A)b = f(A)b_j=\gcd(f,q_j^{e_j})(A)b_j=q_j(A)^{e_j-1}b_j$ is a contradiction)

  • $\begingroup$ Thank you. This is more complex than I thought it to be. $\endgroup$ Commented Nov 18, 2020 at 8:31
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    $\begingroup$ @SimpleProgrammer It can't be trivial because if some $b$ works then on the $A^0b,\ldots,A^{n-1}b$ basis $A$ acts as the companion matrix, so this is the main way to find if two matrices $A,A'$ are similar and if so to obtain a $P$ such that $A=PA' P^{-1}$. That said which step is complex ? $\endgroup$
    – reuns
    Commented Nov 18, 2020 at 14:11
  • $\begingroup$ Yes, I see. Well, linear algrbra is not my strong suit and these arguments all look very abstract to me (since I do not know the subject to well). For example, why I need this result is for purposes of control theory, but I did not expect control theory to dwell this deep into linear algebra (control theory can be very hands on / practical is what I mean, so this came as a surprise). $\endgroup$ Commented Nov 19, 2020 at 11:20

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