A question related to period of a $2 \times 2$ matrix

Question

Let $b,c \in \mathbb{R}^2$ and $A \in \mathbb{R}^{2 \times 2}$. I want to find if there exists a triplet $(c,A,b)$ such that: \begin{align*} c^Tb=1&, c^T A^3 b=1, &\ldots \\ c^TAb=0&, c^T A^4 b=0,& \ldots\\ c^T A^2 b=0&, c^T A^5 b=0, & \ldots \\ \end{align*} In otherwords, $c^TA^{n}b=1$ whenever $n(\mod 3)=0$ and $0$ otherwise. I strongly feel that there doesn't exist any such triplet but unable to provide a proof. The reason I feel this way is because I think somehow $A$ should be related to permutation matrices and a $2 \times 2$ permutation matrix can only have period $2$.

Can anyone suggest me a way to approach this problem? For the sake of simplicity I assumed that both $c=b=[1 \ 0]^T$. Then the question is equiavalent to asking does there exist a matrix $A$ such that $(A^n)_{11}=1$ when $n(\mod{3})=0$. But this doesn't seem to help in solving the problem.

Thanks for any help!

Answer 1

If $$Ab, A^2b \in C=\{ b \in \mathbb R^2 \mid c^Tb = 0 \}$$ since $C$ is 1-dimensional vector subspace of $\mathbb R$, vector $A^2b = \lambda Ab$. But then, $$A^3b = A(A^2b)=\lambda A^2b \in C$$, which means that $c^t A^3b=0$.

Answer 2

The answer is no: suppose that $c^TAb = c^TA^2b = 0$. Then $Ab$ is perpendicular to $c$, and so is $A^2b$. In two dimensions, this means that $A^2b$ is a multiple of $Ab$, which is to say that $A(Ab) = \lambda Ab$.

It would follow that for $k \geq 2$, we have $A^kb = A^{k-1}(Ab) = \lambda^{k-1}b$, which implies that $c^TA^kb = 0$ for $k \geq 2$.

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As for the generalization: Suppose $A$ is $n \times n$, and $0 = c^TAb = c^TA^2b = \cdots = c^TA^{n}b$. Let $S$ denote the span of $\{Ab,A^2b,\dots,A^{n}b\} \subset c^\perp$. By the Cayley-Hamitlon theorem, we have $$ A^{n+1}b = A^{n}Ab = (d_0 I + d_1A + \cdots + d_{n-1}A^{n-1})Ab \in S $$ $A^{n+1}b \in S \subset \{c\}^\perp$, so $c^TA^{n+1}b = 0$.