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Problem.
Let $A$ be a complex $n \times n$ matrix and $b$ be a complex $n \times 1$ vector. Prove Rank$\begin{bmatrix} b & Ab & A^2b & \cdots & A^{n-1} b \end{bmatrix} = n$ if and only if Rank $\begin{bmatrix} A - \lambda I & b \end{bmatrix} = n$ for every eigenvalue $\lambda$ of $A$.

My Question.
What is $\begin{bmatrix} A - \lambda I & b \end{bmatrix}$?? I feel like I should recognize this object, or it should have some significance, or it should make me think, "Aha! This is related to XYZ concept." But instead, nothing. I know that $\begin{bmatrix} A - \lambda I & b \end{bmatrix}$ is the augmented matrix of the system $(A - \lambda I)x = b$. But so what? Does that carry some special significance?

Marc van Leeuwen
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1Teaches2Learn
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    The rank of $A-\lambda I$ is at most $n-1,$ if $\lambda $ is an eigenvalue. – Ryszard Szwarc Mar 16 '22 at 22:46
  • I would interpret it as the matrix $A-\lambda I$ with the vector $b$ added as a column. – CyclotomicField Mar 17 '22 at 01:33
  • Your question starts with you saying "What is $[A-\lambda I b]$?", and ends with "I know that it is the augmented matrix of the system $(A-\lambda I)x = b$"... so... what exactly is your question? – 5xum Mar 17 '22 at 12:52
  • @5xum No my question ends, verbatim, "Does that carry some special significance?" That's my question. – 1Teaches2Learn Mar 18 '22 at 19:23
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    It gives a nice characterization of the existence of cyclic vectors. The eigenvalues should be single and the vector $b$ should satisfy appropriate assumptions. – Ryszard Szwarc Mar 19 '22 at 18:59

1 Answers1

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First observe that $${\rm Rank}\ [ b,\, Ab,\, \ldots, \, A^{n-1}b] ={\rm Rank}\ [b ,\, (A-\lambda I)b,\, \ldots ,\, (A-\lambda I)^{n-1} b] ,\qquad \qquad (1) $$ for any number $\lambda.$

Assume $${\rm Rank}\,[A-\lambda I,\, b]\le n-1$$ for some $\lambda.$ Therefore ${\rm Rank}\, [A-\lambda I]\le n-1,$ i.e. $\lambda $ is an eigenvalue of $A.$ One of two cases occur:

  1. $b\in {\rm Im}\, (A-\lambda I)$, i.e. $b$ is a linear combination of the columns of the matrix $A-\lambda I.$
  2. $b\notin {\rm Im}\, (A-\lambda I)$ and ${\rm Rank}\,[A-\lambda I]\le n-2.$

In the first case
$${\rm Rank}\ [b ,\, (A-\lambda I)b,\, \ldots ,\, (A-\lambda I)^{n-1} b]\le {\rm Rank} [A-\lambda I]\le n-1$$ In the second case we get $${\rm Rank}\ [b ,\, (A-\lambda I)b,\, \ldots ,\, (A-\lambda I)^{n-1} b]\le 1+{\rm Rank}\,[A-\lambda I] \le n-1$$ In view of $(1)$ we obtain $${\rm Rank}\ [ b,\, Ab,\, \ldots, \, A^{n-1}b]\le n-1.$$

For the converse implication assume ${\rm Rank }\,[A-\lambda I,\, b]=n$ for every eigenvalue $\lambda.$ Hence $b\notin {\rm Im}(A-\lambda I)$ and ${\rm Rank }\,[A-\lambda I]=n-1.$ Therefore $\dim \ker(A-\lambda I)=1$ for every eigenvalue $\lambda.$ From now on the Jordan normal form of $A$ can be applied in order to finish the proof.

Ryszard Szwarc
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  • I'm afraid I still don't understand. Looking at the "converse implication": I get that the Jordan form has only one Jordan block per eigenvalue. I get that the characteristic polynomial must equal the minimal polynomial. I get that this means that there exists a cyclic vector. But how do we know that $b$ specifically is a cyclic vector? I think that's the piece I've been missing all along, and unfortunately it is the one piece not supplied by your otherwise excellent proof! – 1Teaches2Learn Mar 26 '22 at 21:21
  • You are right. Still it takes some effort to complete the converse implication. The Jordan decomposition leads to a decomposition of the entire space into subspaces corresponding to the eigenvalues. In this way $b$ is decomposed into say $b_1+b_2+\ldots +b_k.$ The assumption about rank imply that $b_j\neq 0$ for every $j.$ This and the fact of one block for every eigenvalue are crucial to show that $b$ is a cyclic vector. Still proving it requires some nontrivial reasoning. I think the best would be by contradiction. – Ryszard Szwarc Mar 26 '22 at 22:57
  • I forgot to add that the assumptions concerning the rank imply that $ b_j$ is cyclic vector the submatrix $A_j ,$ the $j$th block of $A.$ – Ryszard Szwarc Mar 26 '22 at 23:15
  • Shall I extend the answer to include the full proof of the converse implication ? It requires typing additional 10 lines or so. – Ryszard Szwarc Mar 27 '22 at 22:56
  • Thank you so much for the offer! I was able to eventually figure this out. I truly appreciate your help with this. – 1Teaches2Learn Mar 29 '22 at 16:29