# Description

Suppose we have the following non singular tridiagonal matrix $$B = \begin{bmatrix} 1 & a_1 \\ b_2 & 1 & a_2 \\ & \ddots & \ddots & \ddots \\ && b_{n-1} & 1 & a_{n-1} \\ &&&b_{n} & 1 \end{bmatrix}$$ We know that $$a_{i-1}b_{i} \leq \frac{1}{4}$$, which is the reason I know my matrix is nonsingular. In fact we can allow $$a_{i-1}b_{i} < \epsilon_i$$ for any $$\epsilon_i > 0$$ we choose. It is also true that all values $$a_i$$ and $$b_i$$ are strictly positive.

Let $$\Vert B \Vert$$ be the matrix 1-norm defined as $$\Vert B \Vert = \max(|col_i|, i = 1,..., n)$$, where $$col_i$$ is the $$ith$$ column of $$B$$, and $$|col_i|$$ is defined as $$\sum_{j=1}^{n}|B_{j,i}|$$.

We can then write $$B$$ as $$B = I + A$$, where $$I$$ is the identity matrix and $$A$$ is still a tridiagonal matrix.

I want to show that $$\Vert B^{-1} \Vert \leq \frac{1}{1- \Vert A \Vert}$$

# My attempt

The first result I noticed is that $$\Vert B \Vert = 1 + \Vert A \Vert$$ which is true since the way $$I$$ alters the 1-norm of $$A$$ is that it adds one to every $$|col_{i,A}|$$, and since $$\Vert A \Vert$$ is the maximum of all these values, said maximum is only incremented by one.

Then we of course have that,

$$1 = \Vert I \Vert = \Vert BB^{-1} \Vert \leq \Vert B^{-1} \Vert \Vert B \Vert$$

We can also write $$\Vert B \Vert - 2\Vert A \Vert = 1 - \Vert A \Vert$$

I have tried to combine the above three results in many ways, however I cannot get the result I am looking for.

Many thanks is advance for any hints/help!

• The inequality doesn't hold. Consider $A=\pmatrix{0&1&0\\ \frac14&0&\frac14\\ 0&1&0}$ and $B=I+A$. We have $\|A\|=2$ and hence $\|B^{-1}\|>0>-1=\frac{1}{1-\|A\|}$. Mar 20, 2022 at 11:10

As noted in the comments by user1551, this inequality can't hold if $$\| A \| \ge 1$$, which your assumptions don't seem to rule out.
Provided we actually have $$\| A \| < 1$$, you can use the Neumann series $$B^{-1} = \sum_{k=0}^{\infty} (I-B)^k=\sum_{k=0}^{\infty} (-A)^k,$$ the triangle inequality and the formula for the geometric series.
Note that this also holds if $$A$$ and $$B$$ are linear bounded operators between the same Banach space with $$B = I + A$$ and $$\| I - B\| < 1$$.