Formal expression for $A^{-1}$, given $A=1+B$

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I have a 2x2 real matrix $A$, with $\det A\not=0$ and I define it as $$A=1+B$$ where 1 is the identity and B another regular and real matrix. I need to express the inverse $A^{-1}$ in terms of B: I have used $A^{-1}A=1$ in order to derive a recurrence relation that yelds $$A^{-1}=1-B+B^{2}-B^{3}+B^{4}-...$$ which formally is a geometric series in B.

My questions are:

  • "When this series is properly defined?"
  • "Does exist a constraint on $B$ in order to deal with a convergent series?"
  • "If for some $B$ the series diverge, what is the expression for $A^{-1}$ in terms of $B$?"
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1
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Let $$B=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ Then $$A^{-1}\ =\ \begin{pmatrix}1+a & b \\ c & 1+d\end{pmatrix}^{-1}$$ $$=\ \frac1{(1+a)(1+d)-bc} \begin{pmatrix}1+d & -b \\ -c & 1+a\end{pmatrix}$$ $$=\ \frac1{1+(a+d)+(ad-bc)}\left[\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\,+\,\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}\right]$$ $$=\ \frac1{1+\mathrm{trace}(B)+\det(B)} \left[I+\det(B)\cdot B^{-1}\right]$$

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If you introduce a norm $\| \cdot \|$ on the space of matrices such that the addition and multiplication of matrices become continuous in that norm, then a sufficient condition for that series to be convergent is $\| B \| < 1$.

One such norm is the operator norm: if $A : \Bbb R^n \to \Bbb R^n$ is a matrix, then one defines

$$\| A \| = \sup _{\| v \| = 1} \| Av \|$$

which is perfect for your needs.

2
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You have conditions on the spectral radius $\rho$ :

  • if $\rho(B) < 1$ then your serie converge
  • if $\rho(B) > 1$ then your serie diverge

And if $\rho(B) = 1$ it can converge or diverge

The convergence is proved by knowing that $\rho(B) = \inf_N N(B)$ over all the operator norms

The divergence is simple : just take an engeinvector with eigenvalue $\lambda > \rho(B) > 1$ and apply this to the partial sum