If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular.

Question

I am reading "Matrix Calculus"(in Japanese) by Masahiko Saito.

There is the following proposition in this book:

Let $I$ be an identity matrix.
Let $C$ be a square matrix.
Let $A=\begin{pmatrix} I&B \\ O&C \end{pmatrix}$.
If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular and $$\lim_{p\to\infty} A^p = \begin{pmatrix}I&B(I-C)^{-1}\\O&O\end{pmatrix}.$$

Proof:
By assumption, $I-C$ is non-singular.
By the previous proposition, $$A^p=\begin{pmatrix} I&B(I-C)^{-1}(I-C^p) \\ O&C^p\end{pmatrix}.$$
By theorem 3.3.8, $$\lim_{p\to\infty} C^p = O.$$
Therefore, $$\lim_{p\to\infty} A^p = \begin{pmatrix}I&B(I-C)^{-1}\\O&O\end{pmatrix}.$$

We must show $I-C$ is non-singular, but by mistake, the author wrote it is non-singular by assumption.

I wanna know a proof of the following proposition:

If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular.

My attempt:
$(I-C)(I+C+\dots+C^{p-1}) = I - C^p$.
I know $\lim_{p\to\infty} C^p = O$.
So, $\lim_{p\to\infty} (I - C^p) = I$.

Answer 1

If $I-C$ is singular it has a zero eigenvalue. But the eigenvalues of $I-C$ are $1-\lambda$ as $\lambda$ runs through the eigenvalues of $C$. So $C$ has an eigenvalue $1$ - a contradiction.

Answer 2

First, I will use that $\lim_{n\to \infty} \mathbf{C}^n = \mathbf{O}$, proven in If modulus of each one of eigenvalues of $B$ is less than $1$, then $B^k\rightarrow 0$

Now, we consider the homogeneous system \begin{equation*} (\mathbf{I}-\mathbf{C})\mathbf{x}=\mathbf{0}, \end{equation*} where $\mathbf{0}$ is a column of $0$'s with the same length as the order of $\mathbf{C}$. Then, we can rewrite it as \begin{equation*} \mathbf{C}\mathbf{x}=\mathbf{x}. \end{equation*} However, this results in the equation \begin{equation*} \mathbf{C}^n\mathbf{x}=\mathbf{x}. \end{equation*} that holds for every $n\in \mathbb{N}\cup \{0\}$. Provided that $\mathbf{C}^n\to \mathbf{0}$ as $n\to \infty$, taking limits at both side, we get \begin{equation*} \lim_{n\to \infty}\mathbf{x}=\lim_{n\to \infty}\mathbf{C}^n\mathbf{x}=\mathbf{0}. \end{equation*} Hence, the unique solution to this system is the trivial solution, so the rank of the square matrix $(\mathbf{I}-\mathbf{A})$ is maximum (by Rouché-Frobenius' theorem) and we can define its inverse.