Existence of a non-zero sequence For integers $i \geq 1$, let fixed real constants $a_i, b_i, c_i, d_i$ be given, such that

*

*$(a_i,b_i)\neq(0,0)$ and $(c_i,d_i)\neq(0,0),$

*for every $m\in\mathbb{N}$ there exist finite sequences $s_1,\dots,s_m$ and $t_1,\dots,t_m$ of real numbers, such that $s_i=a_is_{i+1}+b_it_{i+1},$ $t_i=c_is_{i+1}+b_it_{i+1}$ and $(s_i,t_i)\neq(0,0)$  for every $i$.
Note that with every choice of the integer m, the finite sequences $s_1,\dots,s_m$ and $t_1,\dots,t_m$ might be redefined. The following question asks whether we can avoid redefining these numbers.
Question: Do there necessarily exist (infinite) sequences $(s_i)_{i\in\mathbb{N}}$ and $(t_i)_{i\in\mathbb{N}}$ of real numbers,   such that $s_i=a_is_{i+1}+b_it_{i+1}$, $t_i=c_is_{i+1}+b_it_{i+1}$ and $(s_i,t_i)\neq(0,0)$  for every ${i\in\mathbb{N}}$?
At first, I thought the answer is yes, however when I tried to justify the answer I could not do it. The difficulty of this is the infinite recurrent definition of $s_1$ and $t_1$.
I thought of using Zorn's Lemma and the Axiom of choice however I failed to set up the problem in those settings.

A simpler version of this question was answered here.
A more general setting for this question has been posted here.
 A: If we set  $A_i\in\mathbb{R}^{2\times 2}$ and $x_i\in\mathbb{R}^{2}$ and
$$
A_i=\begin{pmatrix}
a_i & b_i \\ c_i & d_i
\end{pmatrix} \;\;\text{and}\;\;
x_i=\begin{pmatrix}
s_i \\ t_i
\end{pmatrix}
$$
then $s_i =a_is_{i+1}+b_it_{i+1},\;t_i = c_is_{i+1}+d_it_{i+1}$ can be rewritten as $x_i=A_ix_{i+1}.$ This means
$$
x_1= \left(\prod_{i=1}^{m} A_i\right)\,x_{m+1}
$$
Therefore, restriction 2. can be written as
$$
\prod_{i=1}^{m} A_i \neq 0\;\;\forall\; m\in\mathbb{N}
$$
If $\prod_{i=1}^{m} A_i $ was $0$, we would always get $x_1=0.$
If $\prod_{i=1}^{m} A_i \neq 0,$ we can choose any vector $y$ with $\left(\prod_{i=1}^{m} A_i\right) y \neq 0$ and set $x_{m+1}=y,$ which will result in $x_i\neq 0\;\forall i\in\{1,\ldots,m+1\}.$
Note that this also means that all matrices $A_i$ have at least rank $1,$ there are no null matrices among the $A_i.$
If there are no singular matrices among the $A_i$, we simply set $x_1 = (1\;\; 0)^T$ and calculate all the other $x_i$ according to $x_{i+1}=A_i^{-1}x_i.$
If there is only a finite number of singular matrices among the $A_i$, then we do the following: Let $m$ be the largest index such that $A_m$ is singular. As described above, we can choose any vector $y$ with $\left(\prod_{i=1}^{m} A_i\right) y \neq 0$ and set $x_{m+1}=y,$ which will result in $x_i\neq 0\;\forall i\in\{1,\ldots,m+1\}.$ We calculate all the other $x_i$ according to $x_{i+1}=A_i^{-1}x_i$ for $i\geq m+1.$
The interesting case is the one in which there is an infinite number of singular (rank $1$) matrices among the $A_i.$ In the following, we assume that we have an infinite number of singular matrices among the $A_i.$
Now we chop the sequence $A_i$ into portions
$(A_{i_0+1},\ldots A_{i_1}),$
$(A_{i_1+1},\ldots A_{i_2}),$
$(A_{i_2+1},\ldots A_{i_3})$ etc.
such that each of those groups contains at least one singular matrix $A_i.$ Note that we have set $i_0=0$ for convenience.
We calculate the product of the elements within those groups:
$$
B_k = \prod_{i=i_{k-1}+1}^{i_k} A_i
$$
Each $B_k$ is a rank-$1$ matrix and as such can be written as $B_k=u_kv_k^T$ with
$u,v\in\mathbb{R}^2.$ We have
$$
0\neq \prod_{i=1}^{i_m} A_{i} =
 \prod_{k=1}^{m} B_k =
 \prod_{k=1}^{m} u_kv_k^T  \\
= u_1 \left(\prod_{k=1}^{m-1} v_k^Tu_{k+1}\right) v_m^T
$$
which means that $v_k^Tu_{k+1}\neq 0\;\forall k\in\mathbb{N}$
Now we set
$$
x_{i_m+1} = \frac{u_{m+1}}{\prod_{k=1}^{m} v_k^Tu_{k+1}}\;\; \text{for}\;\;m\in\{0\}\cup\mathbb{N}
$$
For a given $m$, we calculate $x_{i_m},x_{i_m-1},x_{i_m-2},\ldots, x_{i_{m-1}+1}$ using $x_i = A_ix_{i+1}$. As the last step, we only have to confirm that this last value $x_{i_{m-1}+1}$ is consistent with the choice we made above. This can be shown as follows:
$$
x_{i_{m-1}+1} = B_m x_{i_m+1} 
= \frac{ (u_mv_m^T) u_{m+1}}{\prod_{k=1}^{m} v_k^Tu_{k+1}}
= \frac{u_m (v_m^T u_{m+1})}{\prod_{k=1}^{m} v_k^Tu_{k+1}}
= \frac{u_m}{\prod_{k=1}^{m-1} v_k^Tu_{k+1}}
$$
