If we have the tridiagonal matrix
$$A = \begin{bmatrix} 4 & 1 & & & \\ 1 & 4 & 1 & & \\ & \ddots & \ddots & \ddots & \\ & & 1 & 4 & 1\end{bmatrix}$$
then no matter the size of $A$, its condition number, using the infinity norm, is $\kappa (A) = 3$. Is this correct? If so, why does this happen? Why doesn't the condition number change?
As Evinda stated, $$ \|A\|_\infty := \sup\{\|Ax\|_\infty : \|x\|_\infty \le 1 \} = \max_{1\le i\le n} \sum_{j=1}^n |A_{ij}| .$$
I think I can show that $K_\infty(A) \le 3$. Clearly $\|A\|_\infty = 6$. Now write $A = 4 (I + B)$, where $B$ has diagonal entries equal to $0$ and the upper and lower diagonal entries equal to $\frac14$. So $\|B\|_\infty = \frac12 < 1$. Hence we can compute $A^{-1}$ using the geometric series $$ A^{-1} = \frac14 \sum_{k=0}^\infty (-1)^k B^k $$ and hence $$ \|A^{-1}\|_\infty \le \frac14 \sum_{k=0}^\infty \|B\|_\infty^k = \frac14 \cdot \frac1{1-\frac12} = \frac12 .$$ Therefore the condition number of $A$ is less than or equal to $6 \times \frac12 = 3$.
Next, I did some numerical experiments using Mathematica and 50 digit arithmetic. The condition number at $n=100$ is not exactly $3$, rather it is roughly $2.9999999999999999999999999999038729245836257932320$. My guess is that as a function of $n$ it converges monotonically to $3$. Matlab would get $3$ as its answer because it uses IEEE double precision which is only about 17 digits of accuracy.
