Prove that if A is a square matrix of size $n \times n$, then $A_k \rightarrow 0$ as $k \rightarrow \infty$ if and only if $\mid \lambda

Question

Prove that if A is a square matrix of size $n \times n$ , then $A_k \rightarrow 0$ as $k \rightarrow \infty$ if and only if $\mid \lambda_i \mid < 1 \ \forall i$

Solution

Norm is a value that indicates the magnitude of a matrix. If the positive power of a matrix results in a smaller matrix, it means that the magnitude of the resultant matrix has decreased after the exponentiation. Thus,

(1) $\begin{equation*} \left\ A_k \rightarrow 0 \ as \ k \rightarrow \infty \Rightarrow \lVert A_k \rVert < \lVert A \rVert \end{equation*}$

Let $\lambda$ be the spectral radius $(\rho(A))$ of A. Then, spectral radius $\rho(A_k)$ will be $\lambda_k$ .

Let $\lVert . \rVert$ be a consistent norm (induced norms are consistent). Then according to spectral radius formula:

(2) $\begin{equation*} \left\ \rho(A) \le {\lVert A_k \rVert}^{1/k} \end{equation*}$

(3) $\begin{equation*} \left\ \Rightarrow \lVert \lambda_k \rVert \le \lVert A_k \rVert \end{equation*}$

Similarly,

(4) $\begin{equation*} \left\ \lVert \lambda \rVert \le \lVert A \rVert \end{equation*}$

Applying (1) in (3),

(5) $\begin{equation*} \left\ \lVert \lambda_k \rVert < \lVert A \rVert \end{equation*}$

Apply (4) in (5),

$\begin{align*} \left \lVert \lambda_k \rVert < \lVert \lambda \rVert \end{align*}$

$\begin{align*} \left \Rightarrow \lVert \lambda_k - 1 \rVert < \lVert 1 \rVert \end{align*}$

Since $\lambda$ is the spectral radius, all other eigenvalues of A can only have values less than $\lambda$ . Thus, $\mid \lambda_i \mid < 1 \ \forall i$ .

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