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

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 $B$?"

Let $$B=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ Then $$A^{-1}\ =\ \begin{pmatrix}1+a & b \\ c & 1+d\end{pmatrix}^{-1}$$ $$=\ \frac1{(1+a)(1+d)-bc} \begin{pmatrix}1+d & -b \\ -c & 1+a\end{pmatrix}$$ $$=\ \frac1{1+(a+d)+(ad-bc)}\left[\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\,+\,\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}\right]$$ $$=\ \frac1{1+\mathrm{trace}(B)+\det(B)} \left[I+\det(B)\cdot B^{-1}\right]$$

• I am trying to generalize your result to a nxn matrix which is real and symmetric: using $AA^{-1}=1$, I get $$B^{2}-BTr(b)+det(B)1=0$$ which is verified only for n=2. Do you think it's possible to extend your result? – Yildiz Jan 8 '17 at 22:53

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

• OP might want to know that spectral radius can be calculated by formula $\rho(B) =\lim_n\|B^n\|^{1/n}$ for any matrix norm. – Ennar Jan 8 '17 at 18:07
• nice ${}{}{}{}{}$ – Jorge Fernández Hidalgo Jan 8 '17 at 18:10

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.

• Can we say something strong about necessary conditions? – Jorge Fernández Hidalgo Jan 8 '17 at 17:46
• @Jorge Based on $(1+B)(1-B+B^2-B^3+B^4-\cdots +(-1)^kB^k)=1+(-1)^kB^{k+1}$, $\|B\|>1$ should imply the series diverges, assuming $1+B$ is invertible. I'm not sure what to expect if $1+B$ is singular, or if $\|B\|=1$. – Dustan Levenstein Jan 8 '17 at 17:58
• I wonder how to express $A^{-1}$ in terms of $B$ when $||B||>1$. Does exist any analytic continuation of the geometric series that I can use when the $||B||>1$? – Yildiz Jan 8 '17 at 18:37
• @Yildiz: Notice that, if $z \in \Bbb C$ and $|z|<1$ then $\sum _{n \ge 0} z^n = \frac 1 {1-z}$, so the natural analytic continuation to $\Bbb C \setminus \{1\}$ is $\frac 1 {1-z}$. By the identity theorem for analytic functions, this one is also unique. If you try to apply this to your problem, though, you will see that it doesn't help you a bit, because it keeps you moving in circles. – Alex M. Jan 8 '17 at 18:42