Why is $(I+A)/(I-A)$ unambiguous?

Question

The following question is from Roger W. Brockett, Finite Dimensional Linear Systems (Image of source):

Ordinarily one does not use division notation when dealing with matrices because $A/B$ might be interpreted as $AB^{-1}$ or $B^{-1}A$ and these are not necessarily the same. Why is the notation $(I+A)/(I-A)$ unambiguous?

I do not know what (I+A)/(I-A) is supposed mean, let alone proving it is unambiguous.

So the question is what that notation mean, and preferably why is unambigous?

Answer 1

Matrices $A$ and $B$ need not commute. So, $AB^{-1} \neq B^{-1}A$ in general. However, your text is claiming that matrices $(I+A)$ and $(I-A)^{-1}$ do commute for all matrices $A$ where $(I-A)$ is invertible. This can be proven thusly;

$$(I+A)(I-A)^{-1}=(I-A)^{-1}+A(I-A)^{-1}$$

$$\therefore (I-A)(I+A)(I-A)^{-1}=I+(I-A)A(I-A)^{-1} $$

$$\therefore(I-A)(I+A)(I-A)^{-1}=I+A(I-A)(I-A)^{-1}$$

$$\therefore(I-A)(I+A)(I-A)^{-1}=I+A$$

This yields

$$(I+A)(I-A)^{-1}=(I-A)^{-1}(I+A)$$

So the notation $(I+A)/(I-A)$ is unambiguous for matrices of this specific form.$$$$(Note: the division symbol assumes and requires that $(I-A)^{-1}$ exist)

Answer 2

From $$ (I-A)(I+A+A^2+A^3+A^4+\ldots)=I $$ follows $$ (I-A)^{-1}=I+A+A^2+A^3+A^4+\ldots $$ and commutes with $I+A$.

More generally, $(I-A)^{-1}$ is a function of $A$ and so commutes with $A$ and then with $I+A$.

Answer 3

A generalization of the answer by @Volk. If $A$ is some square matrix over a field $k$ and $p \in k[X]$ is a polynomial such that $p(A)$ is invertible, then $A$ and $p(A)^{-1}$ commute:

$$ \begin{eqnarray} & p(A)^{-1} A &=& A \, p(A)^{-1} \\ \Leftrightarrow \quad & p(A) p(A)^{-1} A &=& p(A) A \, p(A)^{-1}\\ \Leftrightarrow \quad & A &=& A \, p(A) p(A)^{-1}\\ \Leftrightarrow \quad & A &=& A \end{eqnarray} $$

Here we used that $A$ and $p(A)$ commute (to see this, expand $A \, p(A)$ and $p(A) A$). It follows that also all non-negative integral powers $A^n$ of $A$ commute with $p(A)^{-1}$. Then for any polynomial $q \in k[X]$ also $q(A)$ and $p(A)^{-1}$ commute. In your example $p(X) = 1 - X$ and $q(X) = 1+A$.