Why is this matrix invertible?

Question

I'm following Intro to stochastic processes by Lawler, page 27.

It says if we have a matrix Q such that $Q^n \rightarrow0$, then the eigenvalues of Q have absolute value less than $1$. That part I understand.

Then it says: "Hence, $I-Q$ is invertible." How does that follow?

P.S. I understand the conditions for invertibility, like det can't be zero etc.

Answer 1

Following through on a hint given in another answer:

Assume that zero is an eigenvalue of $I-Q.$ Then $(I-Q)\mathbf{x}= \mathbf{0}$ for some $\mathbf{x}$. So $Q\mathbf{x}=\mathbf{x}$ and $\lambda=1$ is an eigenvalue of $Q$, contradicting $|\lambda|<1$.

We also know that the product of the eigenvalues of a matrix equals its determinant. So, since the eigenvalues of $I-Q$ are all non-zero, their product is non-zero, thus $\det(I-Q)\ne0$, and $I-Q$ is invertible.

Answer 2

HINT: $ \sum_{k=0}^{k=\infty} a^k= \frac{1}{1-a}$

if $0<a<1$.

Answer 3

Hint: If $Q$ has no eigenvalue of $1$ then $I - Q$ has no eigenvalue of $0$.