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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!

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    $\begingroup$ 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 $\|B^{-1}\|>0>-1=\frac{1}{1-\|A\|}$. $\endgroup$
    – user1551
    Mar 20, 2022 at 11:10

1 Answer 1

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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 $\| I - B\| < 1$.

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