10. Let $A$ be an $n imes n$ matrix with positive real eigenvalues $\lambda_{1}\lambda_{2}\cdots\lambda_{n}$. Let $\mathbf{x}_{i}$ be an eigenvector belonging to $\lambda_{i}$ for each $i$, and let $\mathbf{x}=\alpha_{1} \mathbf{x}_{1}+\cdots+\alpha_{n} \mathbf{x}_{n}$.
(a) Show that $A^{m} \mathbf{x}=\sum_{i=1}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}$.
(b) Show that if $\lambda_{1}=1$, then $\lim _{m ightarrow m} A^{m} \mathbf{x}=\alpha_{1} \mathbf{x}_{1}$.

Question

10. Let $A$ be an $n 	imes n$ matrix with positive real eigenvalues $\lambda_{1}>\lambda_{2}>\cdots>\lambda_{n}$. Let $\mathbf{x}_{i}$ be an eigenvector belonging to $\lambda_{i}$ for each $i$, and let $\mathbf{x}=\alpha_{1} \mathbf{x}_{1}+\cdots+\alpha_{n} \mathbf{x}_{n}$.
(a) Show that $A^{m} \mathbf{x}=\sum_{i=1}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}$.
(b) Show that if $\lambda_{1}=1$, then $\lim _{m ightarrow m} A^{m} \mathbf{x}=\alpha_{1} \mathbf{x}_{1}$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the matrix $A$ with positive real eigenvalues $\lambda_{1} > \lambda_{2} > \cdots > \lambda_{n}$ and corresponding eigenvectors $\mathbf{x}_{i}$, we need to show that $A^{m} \mathbf{x} = \sum_{i=1}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}$. ‖ ‖ step 2 ⋮ Since $\mathbf{x}_{i}$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda_{i}$, we have $A \mathbf{x}_{i} = \lambda_{i} \mathbf{x}_{i}$. ‖ ‖ step 3 ⋮ Applying $A$ repeatedly, we get $A^{m} \mathbf{x}_{i} = \lambda_{i}^{m} \mathbf{x}_{i}$. ‖ ‖ step 4 ⋮ Given $\mathbf{x} = \alpha_{1} \mathbf{x}_{1} + \cdots + \alpha_{n} \mathbf{x}_{n}$, we apply $A^{m}$ to both sides: $A^{m} \mathbf{x} = A^{m} (\alpha_{1} \mathbf{x}_{1} + \cdots + \alpha_{n} \mathbf{x}_{n})$. ‖ ‖ step 5 ⋮ Using linearity of $A^{m}$, we get $A^{m} \mathbf{x} = \alpha_{1} A^{m} \mathbf{x}_{1} + \cdots + \alpha_{n} A^{m} \mathbf{x}_{n}$. ‖ ‖ step 6 ⋮ Substituting $A^{m} \mathbf{x}_{i} = \lambda_{i}^{m} \mathbf{x}_{i}$, we obtain $A^{m} \mathbf{x} = \alpha_{1} \lambda_{i}^{m} \mathbf{x}_{1} + \cdots + \alpha_{n} \lambda_{n}^{m} \mathbf{x}_{n}$. ‖ ‖ step 7 ⋮ Therefore, $A^{m} \mathbf{x} = \sum_{i=1}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}$. ‖ ‖ step 8 ⋮ For part (b), if $\lambda_{1} = 1$, we need to show that $\lim_{m \rightarrow \infty} A^{m} \mathbf{x} = \alpha_{1} \mathbf{x}_{1}$. ‖ ‖ step 9 ⋮ From part (a), $A^{m} \mathbf{x} = \sum_{i=1}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}$. ‖ ‖ step 10 ⋮ As $m \rightarrow \infty$, $\lambda_{i}^{m} \rightarrow 0$ for all $i \geq 2$ since $\lambda_{i} < 1$. ‖ ‖ step 11 ⋮ Therefore, $\lim_{m \rightarrow \infty} A^{m} \mathbf{x} = \alpha_{1} \lambda_{1}^{m} \mathbf{x}_{1} + \sum_{i=2}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}$. ‖ ‖ step 12 ⋮ Since $\lambda_{1} = 1$, $\lambda_{1}^{m} = 1$ for all $m$. ‖ ‖ step 13 ⋮ Thus, $\lim_{m \rightarrow \infty} A^{m} \mathbf{x} = \alpha_{1} \mathbf{x}_{1} + \sum_{i=2}^{n} \alpha_{i} \cdots 0 \mathbf{x}_{i} = \alpha_{1} \mathbf{x}_{1}$. ‖ ∻Answer∻ ⚹ $\alpha_{1} \mathbf{x}_{1}$ ⚹ ∻Key Concept∻ ⚹Eigenvalues and Eigenvectors⚹ ∻Explanation∻ ⚹The solution leverages the properties of eigenvalues and eigenvectors, particularly how powers of a matrix act on its eigenvectors, and the behavior of eigenvalues less than 1 as their powers increase.⚹◊What is the significance of having positive real eigenvalues in a matrix $A$ in linear algebra?⍭ Generate me a similar question◊

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