91Ó°ÊÓ

For an $\mathbf{n}\mathbf{×}\mathbf{n}$matrix A and scalar $\mathbf{λ}$,$\mathbf{λ}$is an eigenvalue of A, if there exist a non zero vector $\stackrel{\mathbf{→}}{\mathbf{v}}$in${\mathbf{R}}^{\mathbf{n}}$such that,

$\mathbf{A}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{=}\mathbf{λ}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{orA}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{-}\mathbf{λ}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{0}}\mathbf{orA}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{-}\mathbf{λ}\stackrel{\mathbf{→}}{{\mathbf{I}}_{\mathbf{n}}\mathbf{v}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{0}}\mathbf{or}\mathbf{(}\mathbf{A}\mathbf{-}{\mathbf{λ\pm õ}}_{\mathbf{n}}\mathbf{)}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{0}}$

For the problem (A), the${\mathrm{j}}^{\mathrm{th}}$ component of$\mathrm{A}\stackrel{→}{\mathrm{v}}$ will be,

${\mathrm{Av}}_{\mathrm{j}}={\mathrm{a}}_{\mathrm{j},1}{\mathrm{v}}_1+{\mathrm{a}}_{\mathrm{j},2}{\mathrm{v}}_2+...+{\mathrm{a}}_{\mathrm{j},\mathrm{n}}{\mathrm{v}}_{\mathrm{n}}={∫}_{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{j},\mathrm{n}}{\mathrm{v}}_{\mathrm{j}}={\mathrm{λ\pm ¹}}_{\mathrm{j}}$

Let ${\mathrm{v}}_1=\max \left\{{\mathrm{v}}_1,...,{\mathrm{v}}_{\mathrm{n}}\right\}$. Then,

$$\begin{gathered} {\mathrm{λ\pm ¹}}_{\mathrm{i}}=\underset{\underset{\mathrm{j}≠\mathrm{i}}{\mathrm{j}=1}}{\overset{\mathrm{n}}{∫}}{\mathrm{a}}_{\mathrm{i},\mathrm{j}}{\mathrm{v}}_{\mathrm{j}}+{\mathrm{a}}_{\mathrm{i},\mathrm{i}}{\mathrm{v}}_{\mathrm{i}} \\ \left(\mathrm{λ}-{\mathrm{a}}_{\mathrm{i},\mathrm{i}}\right){\mathrm{v}}_{\mathrm{i}}=\underset{\underset{\mathrm{j}≠\mathrm{i}}{\mathrm{j}=1}}{\overset{\mathrm{n}}{∫}}{\mathrm{a}}_{\mathrm{i},\mathrm{j}}{\mathrm{v}}_{\mathrm{j}}≤\underset{\underset{\mathrm{j}≠\mathrm{i}}{\mathrm{j}=1}}{\overset{\mathrm{n}}{∫}}{\mathrm{a}}_{\mathrm{i},\mathrm{i}}{\mathrm{v}}_{\mathrm{i}} \\ \left(\mathrm{λ}-{\mathrm{a}}_{\mathrm{i},\mathrm{i}}\right)≤\underset{\underset{\mathrm{j}≠\mathrm{i}}{\mathrm{j}=1}}{\overset{\mathrm{n}}{∫}}{\mathrm{a}}_{\mathrm{i},\mathrm{j}} \\ \mathrm{λ}=\underset{\underset{\mathrm{j}≠\mathrm{i}}{\mathrm{j}=1}}{\overset{\mathrm{n}}{∫}}{\mathrm{a}}_{\mathrm{i},\mathrm{j}}+{\mathrm{a}}_{\mathrm{i},\mathrm{i}}={∫}_{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{i},\mathrm{j}}=1 \end{gathered}$$

For the problem (B) , the entire calculation will be same as in the part (A) and the only difference is that to replace_{${\mathrm{v}}_{\mathrm{i}}→\left|{\mathrm{v}}_{\mathrm{i}}\right|$}

Suppose$\mathrm{λ}=-1$is an eigenvalue of A, with its corresponding eigenvector v. Then for $\left(\mathrm{A}+\mathrm{I}\right)\mathrm{v}$,

$$\begin{gathered} 0=\left({\mathrm{a}}_{11}+1\right){\mathrm{v}}_1+{\mathrm{a}}_{1,2}{\mathrm{v}}_2+...+{\mathrm{a}}_{1,\mathrm{n}}{\mathrm{v}}_{\mathrm{n}} \\ ={\mathrm{a}}_{2,1}{\mathrm{v}}_1+({\mathrm{a}}_{2,2}+1){\mathrm{v}}_2+...+{\mathrm{a}}_{2,\mathrm{n}}{\mathrm{v}}_{\mathrm{n}} \\ =... \\ ={\mathrm{a}}_{\mathrm{n},1}{\mathrm{v}}_{∂1}+{\mathrm{a}}_{\mathrm{n},2}{\mathrm{v}}_2+...+\left({\mathrm{a}}_{\mathrm{n}.\mathrm{n}}+1\right){\mathrm{v}}_{\mathrm{n}} \end{gathered}$$

This is only achievable, when ${\mathrm{v}}_1={\mathrm{v}}_2=...={\mathrm{v}}_{\mathrm{n}}=0$, but this means that $Av=0≠-v$. Therefore -1 cannot be an eigenvalue of A

By having $\mathrm{λ}=1$as an eigenvalue of A, we can have,

$\mathrm{Ax}=0$

$$\begin{gathered} \left({\mathrm{a}}_{1,1}-1\right){\mathrm{x}}_1+{\mathrm{a}}_{1,2}{\mathrm{x}}_2+...+{\mathrm{a}}_{1,\mathrm{n}}{\mathrm{x}}_{\mathrm{n}}=0⇒{\mathrm{a}}_{1,2}({\mathrm{x}}_2-{\mathrm{x}}_1)+...+{\mathrm{a}}_{1,\mathrm{n}}({\mathrm{x}}_{\mathrm{n}}-{\mathrm{x}}_1)=0 \\ {\mathrm{a}}_{2,1}{\mathrm{x}}_1+({\mathrm{a}}_{2,2}-1){\mathrm{x}}_2+...+{\mathrm{a}}_{2,\mathrm{n}}{\mathrm{x}}_{\mathrm{n}}=0⇒{\mathrm{a}}_{2,2}({\mathrm{x}}_1-{\mathrm{x}}_2)+...+{\mathrm{a}}_{1,\mathrm{n}}({\mathrm{x}}_{\mathrm{n}}-{\mathrm{x}}_2)=0 \\ ... \\ {\mathrm{a}}_{\mathrm{n},1}{\mathrm{x}}_1+{\mathrm{a}}_{\mathrm{n},2}{\mathrm{x}}_2+...+({\mathrm{a}}_{\mathrm{n},\mathrm{n}+1}{\mathrm{x}}_{\mathrm{n}}=0⇒{\mathrm{a}}_{\mathrm{n},2}({\mathrm{x}}_1-{\mathrm{x}}_{\mathrm{n}})+...+{\mathrm{a}}_{1,\mathrm{n}}({\mathrm{x}}_{\mathrm{n}-1}-{\mathrm{x}}_{\mathrm{n}}\left)\right)=0 \\ {\mathrm{x}}_1={\mathrm{x}}_2=...={\mathrm{x}}_{\mathrm{n}} \end{gathered}$$

Therefore,

$\mathrm{x}=\left[\begin{array}{c}\mathrm{c}\\ \mathrm{c}\\ â‹\circledR \\ \mathrm{c}\end{array}\right]$

Thus the eigenvalue of matrix A is $\mathrm{λ}=\underset{\underset{\mathrm{j}≠\mathrm{i}}{\mathrm{j}=1}}{\overset{\mathrm{n}}{∫}}{\mathrm{a}}_{\mathrm{i},\mathrm{j}}+{\mathrm{a}}_{\mathrm{i},\mathrm{i}}={∫}_{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{a}}_{1,\mathrm{j}}=1$and we have shown that $\mathrm{λ}=-1$fails to be an eigenvalue of A, and show that the eigenvector with eigenvalue 1 are the vector of the form

$\left[\begin{array}{c}c\\ c\\ â‹\circledR \\ c\end{array}\right]$where c is nonzero