Question

encontre os autovalores e autovetores de A^15, sendo A [-1 -2 -21 2 1-1 -1 0 ]

Answer 1

$\displaystyle A= \left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]$
perceba que:
$A^2=\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$
e que
$A^3=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right] =\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]$
então:
$A^{15}=\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]$
Sabemos que $A\vec{v}=\lambda\vec{v}\implies A\vec{v}-\lambda\vec{v}=0\implies \vec{v}(A-\lambda I)=0$
então:
$\displaystyle i)~~~~\vec{v}\left(\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]- \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right] \right)=0\\\\ii)~~~\vec{v}\left(\left[\begin{array}{ccc}-1-\lambda&-2&-2\\1&2-\lambda&1\\-1&-1&-\lambda\end{array}\right] \right)=0$
calcular determinante da matriz
$\displaystyle i)~~~~\det\left[\begin{array}{ccc}-1-\lambda&-2&-2\\1&2-\lambda&1\\-1&-1&-\lambda\end{array}\right] \\\\ii)~~~\det(A-\lambda I)=-(\lambda-1)^2(\lambda +1)=-\lambda^3+\lambda^2+\lambda-1$
esse determinante é a equação característica, suas raízes são os autovetores da matriz dada, como estava na forma reduzida a gente percebe que 1, 1 e -1 são raízes desse polinômio então:

$\displaystyle i)~~~~-(\lambda-1)(\lambda-1)(\lambda+1)\implies \boxed{\begin{cases} \lambda_1=1\\\lambda_2=1\\\lambda_3=-1\end{cases}}$
encontramos os autovalores da matriz dada

para calcular os autovetores basta lembrarmos de 
$\vec{v}A=\lambda\vec{v}$
ou seja:
$\vec{v}= \left[\begin{array}{c}a\\b\\c\end{array}\right]$
$\displaystyle i)~~~~\vec{v}(A-\lambda I)=0\\\\~~~~~\text{para }\lambda_1~e~\lambda_2\\\\ii)~~~\left[\begin{array}{c}a\\b\\c\end{array}\right]\left(\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]-  \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\right)=\vec{0}\\\\iii)~~~\left[\begin{array}{c}a\\b\\c\end{array}\right]\left[\begin{array}{ccc}-2&-2&-2\\1&1&1\\-1&-1&0\end{array}\right]=\vec{0}\\\\iv)~~~\text{reduzir}~~ (escalon amento) : }~\left[\begin{array}{c}a\\b\\c\end{array}\right]  \left[\begin{array}{ccc}1&1&1\\0&0&0\\0&0&0\end{array}\right]=\vec{0}\\\\v)~~~a+b+c=0\implies a=-b-c\\\\vi)~~~\left[\begin{array}{c}-b-c\\b\\c\end{array}\right]=\left[\begin{array}{c}-b\\b\\0\end{array}\right]+\left[\begin{array}{c}-c\\0\\c\end{array}\right], ~e~c,b\neq0\\\\vii)~~~\vec{v}_1=\left[\begin{array}{c}-1\\1\\0\end{array}\right]~~~~~~~~\vec{v}_2=\left[\begin{array}{c}-1\\0\\1\end{array}\right]$
e finalmente
$\displaystyle i)~~~~\vec{v}_3=  \left[\begin{array}{c}x\\y\\z\end{array}\right] \\\\ii)~~~\left[\begin{array}{c}x\\y\\z\end{array}\right]\left(\left[\begin{array}{ccc}-1&-2&-2\\1&2&1\\-1&-1&0\end{array}\right]+ \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\right)=\vec{0}\\\\iii)~~\left[\begin{array}{c}x\\y\\z\end{array}\right]\left[\begin{array}{ccc}0&-2&-2\\1&3&1\\-1&-1&1\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right] \\\\iv)~~~\text{reduzir matriz:}\left[\begin{array}{c}x\\y\\z\end{array}\right]\left[\begin{array}{ccc}1&0&-2\\0&1&1\\0&0&0\end{array}\right]=  \left[\begin{array}{ccc}0\\0\\0\end{array}\right] \\\\v)~~~\begin{cases} x-2z=0\\y+z=0\end{cases}\implies \begin{cases} x=2z\\y=-z\end{cases}\\\\vi)~~~    \left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}2z\\-z\\z\end{array}\right],~z\neq0\implies z=1\\\\v)~~~\vec{v}_3=  \left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

logo os autovetores dessa matriz são:
$\boxed{\boxed{ \left[\begin{array}{c}-1\\0\\1\end{array}\right] , \left[\begin{array}{c}-1\\1\\0\end{array}\right] , \left[\begin{array}{c}2\\-1\\1\end{array}\right] }}$


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