# Inverse of a special matrix

1. Sep 26, 2014

### srz

Hi...can anyone please suggest whether the following inverse has a power series expansion
$$(I+\delta A)^{-1}$$
where $$\delta$$ is a constant and $$A =$$
$$\begin{pmatrix} T & T-1 & T-2 &... & 3 & 2 & 1\\ T-1 & T-1 & T-2 & ... & 3 & 2 & 1 \\ .. \\2 & 2 & 2 &... & 2 & 2 & 1 \\ 1 & 1 & 1 & ... & 1 & 1 & 1 \end{pmatrix}$$
Thanks!

2. Sep 26, 2014

### mathman

If delta is small enough.

3. Sep 27, 2014

### srz

Thanks. Is it true that $$det|I+A|=1+trace(A)+det|A|$$? If not then is there any general expression for $$det|I+A|$$

4. Sep 27, 2014

### lpetrich

srz, try A = -I in your first formula.

There does exist a general formula for det(I+A) in terms of traces of powers of A, but it's rather complicated. Determinant - Wikipedia has it.

5. Sep 27, 2014

### srz

Thanks lpetrich.

6. Oct 10, 2014

### platetheduke

For the inverse, you've got $(I+\delta A)^{-1}=I-\delta A+(\delta A)^2-(\delta A)^3+\ldots$. It should converge if $\vert\delta A\vert<1$.

Polynomials and rational functions of a single matrix behave very similarly to the single real variable case.