Seja:
A=\left[\begin{array}{cc} -15 & 4 \\\\ -56 & 17 \end{array}\right]
Na base
\hat{\mathbf{i}}=(1,0) e
\hat{\mathbf{j}}=(0,1) .
a) Encontre
D diagonal e
B tal que
A=B D B^{-1} .
Esse é o processo de diagonalização:
\begin{aligned} D & =\left[\begin{array}{cc} \lambda_{1} & 0 \\\\ 0 & \lambda_{2} \end{array}\right] \\\ B & =\left[\begin{array}{ll} x_{1} & x_{2} \\\ y_{1} & y_{2} \end{array}\right] \end{aligned}
Onde
\lambda_{i} é um autovalor de
A e
x_{i}, y_{i} são as cordenadas do autovetor associado a
\lambda_{i} . Para os autovalores de
A :
\begin{gathered} \left|\begin{array}{cc} -15-\lambda & 4 \\\\ -56 & 17-\lambda \end{array}\right|=0 \\\ \therefore(-15-\lambda)(17-\lambda)+224=0 \end{gathered}
Resolvendo a equação do segundo grau em
\lambda :
\begin{aligned} & \lambda_{1}=1+4 \sqrt{2} \\\ & \lambda_{2}=1-4 \sqrt{2} \end{aligned}
Para os autovetores de
A :
\begin{gathered} \left\{\begin{array}{l} {\left[\begin{array}{cc} -15 & 4 \\\\ -56 & 17 \end{array}\right]\left[\begin{array}{l} x_{1} \\\ y_{1} \end{array}\right]=\lambda_{1}\left[\begin{array}{l} x_{1} \\\ y_{1} \end{array}\right]} \\\ {\left[\begin{array}{cc} -15 & 4 \\\\ -56 & 17 \end{array}\right]\left[\begin{array}{l} x_{2} \\\ y_{2} \end{array}\right]=\lambda_{2}\left[\begin{array}{l} x_{2} \\\ y_{2} \end{array}\right]} \end{array}\right. \\\ \underset{\text { Subst. }}{\Longrightarrow}\left\{\begin{array}{l} -15 x_{1}+4 y_{1}=(1+4 \sqrt{2}) x_{1} \\\\ -56 x_{1}+17 y_{1}=(1+4 \sqrt{2}) y_{1} \\\\ -15 x_{2}+4 y_{2}=(1-4 \sqrt{2}) x_{2} \\\\ -56 x_{2}+17 y_{2}=(1-4 \sqrt{2}) y_{2} \end{array}\right. \end{gathered}
Escolhendo
x_{1}, x_{2}=1 (existem infinitos autovetores, eles são determinados apenas por uma direção) e resolvendo para
y_{1}, y_{2} :
\begin{gathered} \left\{\begin{aligned} -15+4 y_{1} & =(1+4 \sqrt{2}) \\\\ -15+4 y_{2} & =(1-4 \sqrt{2}) \end{aligned}\right. \\\ \therefore\left\{\begin{array}{l} y_{1}= \\\ y_{2}= \\\ = \end{array}+\sqrt{2}-\sqrt{2}\right. \end{gathered}
Assim:
\begin{aligned} & D=\left[\begin{array}{cc} 1+4 \sqrt{2} & 0 \\\\ 0 & 1-4 \sqrt{2} \end{array}\right] \\\ & B=\left[\begin{array}{cc} 1 & 1 \\\\ 4+\sqrt{2} & 4-\sqrt{2} \end{array}\right] \end{aligned}
b) Encontre
A na base
\mathbf{e}_{1}=(1,2) e
\mathbf{e}_{2}=(-1,4) .
Seja:
M:=\left[\begin{array}{cc} 1 & -1 \\\\ 2 & 4 \end{array}\right]
Assim,
A na base e é dada por
M^{-1} A M .
Computando a inversa de
M :
\begin{aligned} & {\left[\begin{array}{ll} a & c \\\ b & d \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\\\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\\\ 0 & 1 \end{array}\right] } \\\ \therefore & {\left[\begin{array}{ll} a+2 c & -a+4 c \\\ b+2 d & -b+4 d \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\\\ 0 & 1 \end{array}\right] } \end{aligned}
Igualando as componentes:
\left\{\begin{array}{l} a+2 c=1 \\\ b+2 d=0 \\\\ -a+4 c=0 \\\\ -b+4 d=1 \end{array}\right.
Resolvendo o sistema:
M^{-1}=\left[\begin{array}{cc} \frac{2}{3} & \frac{1}{6} \\\\ -\frac{1}{3} & \frac{1}{6} \end{array}\right]
Assim:
A_{\mathbf{e}}=\left[\begin{array}{cc} -\frac{25}{3} & \frac{124}{3} \\\\ -\frac{4}{3} & \frac{31}{3} \end{array}\right]