\[
A_1=\begin{pmatrix}
0&0&1\\
1&0&-1\\
0&1&1
\end{pmatrix},\quad
A_2=\begin{pmatrix}
1&-3&3\\
3&-5&3\\
6&-6&4
\end{pmatrix},\quad
A_3=\begin{pmatrix}
-3&1&-1\\
7&5&-1\\
-6&6&-2
\end{pmatrix}.
\]

\[
\underline{A_1}
\]

\[
\det(\lambda I-A_1)
=
\begin{vmatrix}
\lambda&0&-1\\
-1&\lambda&1\\
0&-1&\lambda-1
\end{vmatrix}
=
\lambda^3-\lambda^2+\lambda-1
=
(\lambda-1)(\lambda^2+1).
\]

不可完全分解

\[
\boxed{A_1\text{ 不可對角化}}
\]

\[
\underline{A_2}
\]

\[
\det(\lambda I-A_2)
=
\begin{vmatrix}
\lambda-1&3&-3\\
-3&\lambda+5&-3\\
-6&6&\lambda-4
\end{vmatrix}
=
\lambda^3-12\lambda-16
=
(\lambda-4)(\lambda+2)^2.
\]

\[
\therefore \text{特徵值為 }\lambda=4,\ -2\text{（代數重數為 }2\text{）}.
\]

\[
A_2-4I=
\begin{pmatrix}
-3&-3&3\\
3&-9&3\\
6&-6&0
\end{pmatrix}
\overset{\text{reduced echelon form}}{\longrightarrow}
\begin{pmatrix}
1&0&-\frac12\\
0&1&-\frac12\\
0&0&0
\end{pmatrix}.
\]

\[
E_4=\operatorname{span}\left\{
\begin{pmatrix}
1\\
1\\
2
\end{pmatrix}
\right\},\qquad gm(4)=1=am(4).
\]

\[
A_2+2I=
\begin{pmatrix}
3&-3&3\\
3&-3&3\\
6&-6&6
\end{pmatrix}
\overset{\text{reduced echelon form}}{\longrightarrow}
\begin{pmatrix}
1&-1&1\\
0&0&0\\
0&0&0
\end{pmatrix}.
\]

\[
E_{-2}
=
\operatorname{span}\left\{
\begin{pmatrix}
1\\
1\\
0
\end{pmatrix},
\begin{pmatrix}
-1\\
0\\
1
\end{pmatrix}
\right\},\qquad gm(-2)=2=am(-2).
\]

\[
\therefore A_2\text{ 可對角化。}
\]

\[
P=
\begin{pmatrix}
1&1&-1\\
1&1&0\\
2&0&1
\end{pmatrix},
\qquad
D=
\begin{pmatrix}
4&0&0\\
0&-2&0\\
0&0&-2
\end{pmatrix}.
\]

\[
P^{-1}A_2P=D.
\]

\[
\boxed{
A_2\text{ 可對角化},\ 
P=
\begin{pmatrix}
1&1&-1\\
1&1&0\\
2&0&1
\end{pmatrix},
\ 
D=
\begin{pmatrix}
4&0&0\\
0&-2&0\\
0&0&-2
\end{pmatrix}
}
\]

\[
\underline{A_3}
\]

\[
A_3=
\begin{pmatrix}
-3&1&-1\\
-7&5&-1\\
-6&6&-2
\end{pmatrix}.
\]
\[
\det(\lambda I-A_3)=\lambda^3-12\lambda-16=(\lambda-4)(\lambda+2)^2.
\]
\[
\therefore \text{ 特徵值為 } \lambda=4,-2,\quad am(4)=1,\ am(-2)=2.
\]

\[
A_3+2I=
\begin{pmatrix}
-1&1&-1\\
-7&7&-1\\
-6&6&0
\end{pmatrix}
\overset{\text{reduced echelon form}}{\longrightarrow}
\begin{pmatrix}
1&-1&0\\
0&0&1\\
0&0&0
\end{pmatrix}.
\]
\[
E_{-2}=\operatorname{span}\left\{
\begin{pmatrix}
1\\
1\\
0
\end{pmatrix}
\right\},
\qquad gm(-2)=1.
\]

\[
gm(-2)=1<2=am(-2).
\]
\[
\boxed{A_3\text{ 不可對角化}.}
\]

關於 $A_1 = \begin{bmatrix}0&0&1\\1&0&-1\\0&1&1\end{bmatrix}$$$

在談 cyclic space 時，也有談考慮的基底下的表現矩陣，並由此表現矩陣來談其 characteristic polynomial。上述矩陣即為此種形式，所以可以直接寫下 $p_{A_1}(t) = (-1)^3(t^3-t^2+t-1)=-t^3+t^2-t+1$。