The problem I am working on asks: Let $A$ be a complex $n \times n$ matrix and suppose $y$ is a nonzero complex $1 \times n$ vector. Show that there is a nonzero polynomial $q(x)$ and an eigenvalue $\lambda$ of $A$ such that $y \ q(A)$ is nonzero, and $(y \ q(A)) A = \lambda ( y \ q(A))$.
My strategy: Take the set $(y, yA, yA^2, \dots, yA^n)$. These $n + 1$ vectors thus being linearly dependent, there exist complex coefficients $a_0, \dots, a_n$, not all zero, such that $0 = \sum_{i=0}^n a_iyA^i = cy(A - \lambda_0I)\dots(A - \lambda_mI)$, where $c$ is a nonzero complex number introduced when factoring the polynomial. My question is: Can I reorder these factors?
Thanks in advance for any help!
(By way of additional explanation, the reason I'd want to reorder these factors comes from the rest of my proof: The expression $0 = cy(A - \lambda_0I)\dots(A - \lambda_mI)$ implies that the vector $y$ will be transformed by this series of operators and, somewhere along the way, one of them, say $(A - \lambda_kI)$, must map the input vector to zero. So I let $q(A)$ be the polynomial formed by the factors preceding this factor, i.e. $q(A) = c(A - \lambda_0I)\dots(A - \lambda_{k-1}I)$, which can easily be converted into a polynomial of complex numbers, i.e. the required $q(x)$. Then $y \ q(A) \neq 0$ while $y \ q(A) (A - \lambda_k I) = 0$, which I rearrange to $ (y \ q(A)) A = \lambda_k (y \ q(A))$, as desired. But, what if the first factor $(A - \lambda_0I)$ is the one that maps the vector to zero? Then I have no polynomial constructed from the preceding factors, so I'd like to rearrange the factors so that the one that sends the vector to zero comes later. I also note that I think there are other problems to address with this proof strategy, like what if ALL of the factors map $y$ to zero, but first things first!)