Matrix commutation of $\boldsymbol{M}$ and $[\boldsymbol{I} - \boldsymbol{M}]^{-1}$

Question

We consider a matrix $\boldsymbol{M}$. We suppose it is diagonalizable, with eigenvalues $\lambda_{i}$. We always assume that $\forall i \, , \,\lambda_{i} \neq 1$. As a consequence, the matrix $[ \boldsymbol{I} - \boldsymbol{M} ]$ is invertible.

I would like to prove that $\boldsymbol{M}$ and $[\boldsymbol{I} - \boldsymbol{M}]^{-1}$ commute one with another. I believe that this property is true for all such set $\{ \lambda_{i} \}$ but I can only prove it for specific situations.

Indeed, if we suppose that $\forall i\,,\, |\lambda_{i}| < 1$, then I believe, we have the converging infinite serie

$$ [\boldsymbol{I} - \boldsymbol{M}]^{-1} = \sum_{k = 0}^{+ \infty} \boldsymbol{M}^{k} \, .$$

Then $[\boldsymbol{I} - \boldsymbol{M}]^{-1}$ can be seen as an infinite polynom in $\boldsymbol{M}$, which will naturally commute with $\boldsymbol{M}$.

My questions are as follows :

• Is my reasoning for the case $|\lambda_{i}| < 1$ correct ? How can it be made perfectly rigourous ?

• If true, how can we prove the commutativity of $\boldsymbol{M}$ and $[\boldsymbol{I} - \boldsymbol{M}]^{-1}$ for any general set of eigenvalues ?

Answer 1

If $A$ commute with $B$ and $B$ invertible then $A$ commute with $B^{-1}$. because $AB=BA$ implies that $B^{-1}A=AB^{-1}$. In your case you have $M$ commute with $I-M$ so $M$ commute with $(I-M)^{-1}$.

Answer 2

We have

$$(-M)(I-M)^{-1} = (I-M)(I-M)^{-1} - I (I-M)^{-1} = I - (I-M)^{-1}$$

Similarly

$$(I-M)^{-1}(-M) = (I-M)^{-1}(I - M) - I (I-M)^{-1} = I - (I-M)^{-1}$$

Thus $(I-M)^{-1}$ and $(-M)$ commute. It follows that also $(I-M)^{-1}$ and $M$ commute.

Answer 3

Your reasoning is needlessly complicated. Start from $$\begin{align} & (I-M)(I-M)^{-1} = I = (I-M)^{-1}(I-M) \\ \implies & (I-M)^{-1} - M(I-M)^{-1} = (I-M)^{-1} - (I-M)^{-1}M \\ \implies & M(I-M)^{-1} = (I-M)^{-1}M \end{align}$$ You don't even need $M$ to be diagonalizable for that, just for $I-M$ to be invertible. Your reasoning does work if $|\lambda_i| < 1$ for all $i$, but you need to justify a few things (why the infinite sum exists, why you can say that $M$ commutes with a series...)

Note that if you do assume that $M$ is diagonalizable, it's simpler. The result is clearly true for diagonal matrices. Suppose $M = PDP^{-1}$ with diagonal $D$. Then: $$\begin{align} & M(I-M)^{-1} = PDP^{-1} (PP^{-1} - PDP^{-1})^{-1} \\ = & PD(I-D)^{-1}P^{-1} \color{red}{=} P(I-D)^{-1}DP^{-1} \\ = & (I-M)^{-1}M \end{align}$$