Bound for eigenvalues of some special matrix Let $Tridiagonal(a, c, b)= \begin{vmatrix}
c & b & 0 & \ldots & 0 \\
a &  c & b & \ldots & 0 \\
0 &  a & c & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots \\
0 &  0 &  0 &  a &  c
\end{vmatrix}$
and consider $A=Tridiagonal(1, -2, 1), B=Tridiagonal(-1, 0, 1), M=A^{-1}B$
Is there known any bound for the maximum of the eigenvalues of $M$?
The bound $\rho(M)\leq || A^{-1} || \cdot || B || $ is too crude.
 A: A slightly longer comment:
By examining the structure of a few $M:=A^{-1}B$ for various $n$, it seems to me that the rows of $M$ satisfy
$$\tag{1}
m_{ii}+m_{ij}=\begin{cases}-1&\text{for $j>i$,}\\+1&\text{for $j<i$,}\end{cases}
\quad
|m_{ii}|\leq 1, \quad i=1,\ldots,n,
$$
that is, the diagonal entries are between $-1$ and $1$, the off-diagonal entries in the same row are constant to left and to right from the diagonal entry and have a special relation w.r.t. the diagonal entry.
Assuming this can be somehow shown (probably by premultiplying this assumed form of $M$ by $A$ and comparing it to $B$) then
$$
\rho(M)\leq\|M\|_{\infty}\leq 1 + 2(n-1).
$$
For example, for $n=1000$, this gives
$$
\rho(M)\approx 318.6,
\quad
\|M\|_\infty=999,
\quad
1+2(n-1)=1999,
\quad
\|A^{-1}\|_2\|B\|_2\approx 2.0\cdot 10^5,
$$
so the estimate does not look so bad (overestimates the real spectral radius by less than an order of magnitude) and is not far from the real $\infty$-norm (by about a factor of 2). Of course, one would need to show (1).
