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?
Best Answer
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:
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.