What is inverse of $I+A$?

Question

Assume $A$ is a square invertible matrix and we have $A^{-1}$. If we know that $I+A$ is also invertible, do we have a close form for $(I+A)^{-1}$ in terms of $A^{-1}$ and $A$?

Does it make it any easier if we know that sum of all rows are equal?

Answer 1

If the series $S = I - A + A^2 - A^3 + \dotsb$ converges, expand $S(I + A)$ to find that it is equal to the identity. See here for details on when $S$ converges.

$S$ clearly converges if $A^k = 0$ for some positive integer $k$ (nilpotency).

As for invertibility, if $A$ is diagonalizable, i.e. $A = PDP^{-1}$ for some diagonal matrix $D = \operatorname{diag}(e_1, e_2, \dotsc, e_n)$, then by Sylvester's Determinant Theorem, $$ \det(I + (PD)P^{-1}) = \det(P^{-1}(PD) + I) = \prod_{i = 1}^n(e_i + 1) $$

Hence, $I + A$ is invertible if no eigenvalue $e_i$ has a value of $-1$.

If $A$ is nilpotent, then its only eigenvalue is $0$, so $I + A$ is invertible.

Is there a closed-form solution?

I believe not. Basically, a closed-form expression of $(I + A)^{-1}$ using $A$ and $A^{-1}$ would amount to a closed-form expression of $(1 + x)^{-1}$ using $x$ and $x^{-1}$, where $x$ is real (or complex). A semi-rigorous articulation of this argument follows:

Proposition: There exists no family of matrices $\{X_{ij}\}_{m \times n}$, where every $X_{ij}$ is either equal to $A$, $A^{-1}$ or a constant dependent on the dimension of $A$, such that $(I + A)^{-1} = \sum_{i = 1}^m(\prod_{j = 1}^nX_{ij})$ for all values of $A$.

Proof:

Assume there exists such a family. Let $A$ be the $1 \times 1$ matrix $x$. Note that $\sum_{i = 1}^m(\prod_{j = 1}^nX_{ij})$ is a polynomial $P(x)$, which apparently equals $(1 + x)^{-1}$ for all values of $x$. Hence, $P(x)$ must be the taylor series $1 - x + x^2 - x^3 + \dotsb$, which contradicts the fact that $P(x)$ has a finite number of terms.

Answer 2

It should be added that the inverse of $I-A$ is $$I + A + A^2 + A^3 + \cdots$$ provided that for some matrix norm of $A$, $||A|| < 1$. This is because $$||(I-A)^{-1}x|| \leq (1+||A|| + ||A||^2 + ||A||^3 + \cdots)||x||$$ and hence $||(I-A)^{-1}x|| < M ||x||$ for some $M$ by the convergence of the geometric series. This of course implies that $(I-A)^{-1}$ is a well defined matrix and everything converges.

In fact, this can be used to show that the set of all invertible matrices is an open set under any norm! That is, given $B$ invertible, consider $B+ A=B(I+B^{-1}A)$. Now for $||A||<1/||B^{-1}||$ we have $||B^{-1}A|| < 1$, which implies $B+A$ is invertible. This show openess.

Answer 3

Check this question. The first answer presents a recursive formula to retrieve the inverse of a generic sum of matrices. So yours should be a special case.