This problem comes from condensed matter physics. Matrices are acting on $L^2$ space, since wavefunctions are normalizable. Right, left eigenvectors are a pair of Majorana representation of Fermions, singular values are the energy. Our interest is to look at zero energy modes.
Statement 1.a
Let's define a matrix with diagonal always equal to one.
$$ M(a,n) \equiv \begin{pmatrix}
1 & -a & & & \\
& 1 & -a & & \\
& & 1 & -a & \\
& & & 1 & \dots \\
& & & & \dots
\end{pmatrix}_{n\times n}$$
For any finite dimension $n$, matrix $M$ is full rank, nullity is zero.
Interesting case happens if we consider infinite dimension ($n\to \infty $):
when $|a|<1$ and , nullity is still zero.
when $|a|>1$ and , nullity is one!
Statement 1.b
Now let $a_i$ becomes random. ( only consider $a>0$ case ).
Statement 1.a can be modified as:
when $ \prod_i a_i <1$ , nullity is zero.
when $\prod_i a_i>1$ , nullity is one.
$$ M(a_i,n) \equiv \begin{pmatrix} 1 & -a_1 & & & \\ & 1 & -a_2 & & \\ & & 1 & -a_3 & \\ & & & 1 & \dots \\ & & & & \dots \end{pmatrix}_{n\times n}$$
Statement 2.a
Adding $b$ terms to statement 1.a
$$ M(a,b,n) \equiv \begin{pmatrix}
1 & -a & -b & & \\
& 1 & -a & -b & \\
& & 1 & -a & \dots \\
& & & 1 & \dots \\
& & & & \dots
\end{pmatrix}_{n\times n}$$
Let's only consider $a>0$ and $b>0$ case. The statement generalizes to:
when $b<1-a$ , nullity is still zero.
when $ 1 -a < b <1 + a$ , nullity is one.
when $ 1 + a < b $ , nullity is two.
My question is, what will be the Statement 2.b when $a_i, b_i$ become random?
Statement 1.a, 1.b, 2.a are conclusions from physics. The zero singular values are in fact Majorana zero modes.
I hope to get some insights from mathematicians.
What math tools and concepts should I use?
Null space, the vector has to be normalizable. i.e. $L^2$.
Inspired by the discussion with @ancientmathematician $$\sum_{i=1}^{+\infty} M_{ij} r_i = 0$$ The iteration relation $r_{i+2} = -\frac{a_i}{b_i} r_{i+1} + \frac{h_i}{b_i} r_{i} $ with intial condition $r_0,r_1$ ,
diagonal $h_i \equiv 1$ now take any value $h$
The solution takes this form: $$r_{i+2} = \big(\prod_{k=0}^i A_k \big) c_0 + \big(\prod_{k=0}^i B_k \big) c_1 $$ $$ A_i,B_i = \frac{-a_i \pm \sqrt{a^2_i + 4 h_i b_i} }{2 b_i} $$
If the solution $\{r_i\} \in L^2 $ is not exploded. There is nullity is adding one.
Therefore the condition for
nullity=2 is : $$|\prod_{k=0}^\infty A_k |< 1 \quad , \quad |\prod_{k=0}^\infty B_k |< 1 $$
nullity=1 is : $$|\prod_{k=0}^\infty A_k |< 1 < |\prod_{k=0}^\infty B_k | $$
nullity=0 is : $$ 1 < |\prod_{k=0}^\infty A_k | \quad , \quad 1 < |\prod_{k=0}^\infty B_k | $$
Taking the log of the product chain, we got average: $$ \overline{ \log | \frac{ -a \pm \sqrt{a^2 + 4 h b} }{2 b} | } = 0 $$ This structure is Inverse hyperbolic functions.
For continues distribution $p(a), p(b), p(h)$ $$ \Lambda_{\pm} \equiv \int da \ db \ dh \log \frac{ \sqrt{a^2 + 4 h b} \pm a }{2 b} p(a) p(b) p(h) = 0 $$ gives the critical point.
counting the valid numbers of $\Lambda_{\pm} < 0$ will give the nullity
Update:
My derivation for the non-constant recurrance relation is not exact.
I found (1.6) to (1.8) from this paper:
Recurrence Relations and Benford’s Law
The auxiliary functions shift one site!
$$ \mu_i + \lambda_i = A_i + B_i = \frac{a_i}{b_i} $$
$$ \mu_i \lambda_{i-1} = A_i B_i = \frac{h_i}{b_i} $$
This is of course not the intial random field.
However if considering $\lambda_i$ being a constant, it is a beautiful and a strong result!