Achar os autovalores e autovetores do operador linear do R3 dado por F(1,0,0)=(2,0,0), F(0,1,0)=(2,1,2) e F(0,0,1)=(3,2,1).

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Para encontrar os autovalores do operador dado, reralizaremos os cálculos abaixo:

\(\begin{align} & A=\left[ \begin{matrix} 2 & 0 & 0 \\ 2 & 1 & 2 \\ 3 & 2 & 1 \\ \end{matrix} \right] \\ & I=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & \det \left( \lambda I-A \right)=0 \\ & \det \left( \left[ \begin{matrix} 2-\lambda & 0 & 0 \\ 2 & 1-\lambda & 2 \\ 3 & 2 & 1-\lambda \\ \end{matrix} \right] \right)=0 \\ & {{\lambda }_{1}}=-1 \\ & {{\lambda }_{2}}=2 \\ & {{\lambda }_{3}}=3 \\ \end{align}\ \)

Calcularemos agora os autovetores desse operador linear:

\(\begin{align} & A-{{\lambda }_{1}}\times E=\left( \begin{matrix} 3 & 0 & 0 \\ 2 & 2 & 2 \\ 3 & 2 & 2 \\ \end{matrix} \right) \\ & {{v}_{1}}=\left( \begin{matrix} 0 \\ -1 \\ 1 \\ \end{matrix} \right) \\ & \\ & A-{{\lambda }_{2}}\times E=\left( \begin{matrix} 0 & 0 & 0 \\ 2 & -1 & 2 \\ 3 & 2 & -1 \\ \end{matrix} \right) \\ & {{v}_{2}}=\left( \begin{matrix} \frac{-3}{7} \\ \frac{8}{7} \\ 1 \\ \end{matrix} \right) \\ & \\ & A-{{\lambda }_{3}}\times E=\left( \begin{matrix} -1 & 0 & 0 \\ 2 & -2 & 2 \\ 3 & 2 & -2 \\ \end{matrix} \right) \\ & {{v}_{3}}=\left( \begin{matrix} 0 \\ 1 \\ 1 \\ \end{matrix} \right) \\ \end{align} \)

Portanrto, os autovalores serão: \(\boxed{{\lambda _1} = - 1,{\lambda _2} = 2{\text{ e }}{\lambda _3} = 3}\) e os autovetores serão:

\(\boxed{{v_1} = \left( {\begin{array}{ccccccccccccccc} 0 \\ { - 1} \\ 1 \end{array}} \right),{v_2} = \left( {\begin{array}{ccccccccccccccc} {\frac{{ - 3}}{7}} \\ {\frac{8}{7}} \\ 1 \end{array}} \right){\text{ e }}{v_3} = \left( {\begin{array}{ccccccccccccccc} 0 \\ 1 \\ 1 \end{array}} \right)}\)