Determine, se existir, os autovalores e autovetores das matrizes: A=\left(\begin{array}{ccc}0 & 2 & -2 \\\\ 2 & 0 & -2 \\\\ -2 & -2 & 4\end{array}\right), B=\left(\begin{array}{ccc}0 & 0 & 1 \\\\ 0 & 3 & 0 \\\\ -1 & 0 & 0\end{array}\right), C=\left(\begin{array}{cccc}1 & 2 & 3 & 4 \\\\ 0 & 2 & 0 & 0 \\\\ 0 & 0 & 2 & 1 \\\\ 0 & 0 & 0 & -1\end{array}\right), D=\left(\begin{array}{ccc}-\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\\\ 0 & 0 & -1\end{array}\right), E=\left(\begin{array}{cccc}0 & -2 & 0 & 2 \\\\ -2 & 0 & 2 & 0 \\\\ 0 & 2 & 0 & -2 \\\\ 2 & 0 & -2 & 0\end{array}\right), F=\left(\begin{array}{cc}1 & -1 \\\\ 1 & 1\end{array}\right).

Passo 1

O enunciado pede os autovalores e autovetores das matrizes
A=\begin{pmatrix}0&2&-2\\\ 2&0&-2\\\ -2&-2&4\end{pmatrix},\quad B=\begin{pmatrix}0&0&1\\\ 0&3&0\\\ -1&0&0\end{pmatrix},\quad C=\begin{pmatrix}1&2&3&4\\\ 0&2&0&0\\\ 0&0&2&1\\\ 0&0&0&-1\end{pmatrix},
D=\begin{pmatrix}-\tfrac12&-\tfrac12&-\tfrac12\\\ -\tfrac12&-\tfrac12&\tfrac12\\\ 0&0&-1\end{pmatrix},\quad E=\begin{pmatrix}0&-2&0&2\\\ -2&0&2&0\\\ 0&2&0&-2\\\ 2&0&-2&0\end{pmatrix},\quad F=\begin{pmatrix}1&-1\\\ 1&1\end{pmatrix}.
Vamos usar para cada caso o polinômio característico \det(M-\lambda I)=0 e então resolver (M-\lambda I)v=0 para obter os autovetores.

Passo 2

Para A calculamos
\det(A-\lambda I) =\det\begin{pmatrix}-\lambda&2&-2\\\ 2&-\lambda&-2\\\ -2&-2&4-\lambda\end{pmatrix} =\lambda(\lambda-6)(\lambda+2).
Logo os autovalores são
\lambda_1=0,\quad \lambda_2=6,\quad \lambda_3=-2.
Resolvendo (A-\lambda I)v=0 obtemos:
\lambda=0:\;v=(1,1,1), \qquad \lambda=6:\;v=(1,1,-2), \qquad \lambda=-2:\;v=(1,-1,0).

Passo 3

Para B o polinômio característico é
\det(B-\lambda I)=-(\lambda-3)(\lambda^2+1)=0,
daí
\lambda_1=3,\quad \lambda_2=i,\quad \lambda_3=-i.
Os autovetores são dados por
\lambda=3:\;v=(0,1,0), \quad \lambda=i:\;v=(1,0,i), \quad \lambda=-i:\;v=(1,0,-i).

Passo 4

Para C, sendo triangular, os autovalores são as entradas da diagonal:
\lambda_1=1,\;\lambda_{2,3}=2,\;\lambda_4=-1.
Resolvendo (C-\lambda I)v=0 achamos:
\lambda=1:\;v=(1,0,0,0),
\lambda=2:\;v_1=(2,1,0,0),\quad v_2=(3,0,1,0),
\lambda=-1:\;v=( -9,0,-2,6 ).

Passo 5

Para D obtemos
\det(D-\lambda I)=-\lambda(\lambda+1)^2=0
logo \lambda_1=0, \lambda_{2,3}=-1.
Resolvendo:
\lambda=0:\;v=(1,-1,0), \quad \lambda=-1:\;v_1=(1,1,0),\;v_2=(1,0,1).

Passo 6

Para E (matriz simétrica) o polinômio característico dá
\lambda_1=4,\;\lambda_2=-4,\;\lambda_{3,4}=0.
Autovetores normalizados (até escala) são
\lambda=4:\;v=(1,-1,-1,1), \quad \lambda=-4:\;v=(1,1,-1,-1), \quad \lambda=0:\;v_1=(1,0,1,0),\;v_2=(0,1,0,1).

Passo 7

Para F o polinômio característico é
\det\begin{pmatrix}1-\lambda&-1\\\ 1&1-\lambda\end{pmatrix} =(1-\lambda)^2+1=0,
logo
\lambda_{1,2}=1\pm i.
Os autovetores são
\lambda=1+i:\;v=(1,-i), \qquad \lambda=1-i:\;v=(1,i).

Resposta

A lista completa de autovalores e autovetores (até escala) é:
(a) Para
A: \begin{cases} \lambda=0,\;v=(1,1,1),\\\ \lambda=6,\;v=(1,1,-2),\\\ \lambda=-2,\;v=(1,-1,0). \end{cases}
(b) Para
B: \begin{cases} \lambda=3,\;v=(0,1,0),\\\ \lambda=i,\;v=(1,0,i),\\\ \lambda=-i,\;v=(1,0,-i). \end{cases}
(c) Para
C: \begin{cases} \lambda=1,\;v=(1,0,0,0),\\\ \lambda=2,\;v_1=(2,1,0,0),\;v_2=(3,0,1,0),\\\ \lambda=-1,\;v=(-9,0,-2,6). \end{cases}
(d) Para
D: \begin{cases} \lambda=0,\;v=(1,-1,0),\\\ \lambda=-1,\;v_1=(1,1,0),\;v_2=(1,0,1). \end{cases}
(e) Para
E: \begin{cases} \lambda=4,\;v=(1,-1,-1,1),\\\ \lambda=-4,\;v=(1,1,-1,-1),\\\ \lambda=0,\;v_1=(1,0,1,0),\;v_2=(0,1,0,1). \end{cases}
(f) Para
F: \begin{cases} \lambda=1+i,\;v=(1,-i),\\\ \lambda=1-i,\;v=(1,i). \end{cases}

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