∣ A − λ I ∣ = ∣ 1 − λ 0 2 0 1 − λ 0 2 0 1 − λ ∣ = 0 ( 1 − λ ) 3 − 0 − 0 − 4 ( 1 − λ ) − 0 − 0 = 0 ( 1 − λ ) ( ( 1 − λ ) 2 − 4 ) = 0 ( 1 − λ ) ( 1 − λ − 2 ) ( 1 − λ + 2 ) = 0 ( 1 − λ ) ( − 1 − λ ) ( 3 − λ ) = 0 ( λ 2 − 1 ) ( λ − 3 ) = 0 λ 1 = 1 , λ 2 = − 1 , λ 3 = 3 |A-\lambda I|=\begin{vmatrix}
1-\lambda & 0&2 \\
0& 1-\lambda&0\\
2&0&1-\lambda
\end{vmatrix}=0\\
(1-\lambda)^3-0-0-4(1-\lambda)-0-0=0\\
(1-\lambda)((1-\lambda)^2-4)=0\\
(1-\lambda )(1-\lambda-2)(1-\lambda+2)=0\\
(1-\lambda)(-1-\lambda)(3-\lambda)=0\\
(\lambda^2-1)(\lambda-3)=0\\
\lambda_1=1, \lambda_2=-1, \lambda_3=3 ∣ A − λ I ∣ = ∣ ∣ 1 − λ 0 2 0 1 − λ 0 2 0 1 − λ ∣ ∣ = 0 ( 1 − λ ) 3 − 0 − 0 − 4 ( 1 − λ ) − 0 − 0 = 0 ( 1 − λ ) (( 1 − λ ) 2 − 4 ) = 0 ( 1 − λ ) ( 1 − λ − 2 ) ( 1 − λ + 2 ) = 0 ( 1 − λ ) ( − 1 − λ ) ( 3 − λ ) = 0 ( λ 2 − 1 ) ( λ − 3 ) = 0 λ 1 = 1 , λ 2 = − 1 , λ 3 = 3
λ 1 = 1 , a 1 ⃗ = ( x , y , z ) ≠ 0 ⃗ ( A − λ 1 I ) a 1 ⃗ = 0 ⃗ ( 0 0 2 0 0 0 2 0 0 ) ⋅ ( x y z ) = ( 0 0 0 ) 2 z = 0 , 2 x = 0 x = z = 0 , y ∈ R a 1 ⃗ = ( 0 , 1 , 0 ) \lambda_1=1, \vec{a_1}=(x,y,z)\neq\vec{0}\\
(A-\lambda_1I)\vec{a_1}=\vec{0}\\
\begin{pmatrix}
0& 0&2 \\
0 & 0&0\\
2&0&0
\end{pmatrix}\cdot \begin{pmatrix}
x \\
y\\z
\end{pmatrix}=\begin{pmatrix}
0 \\
0\\0
\end{pmatrix}\\
2z=0,2x=0\\
x=z=0, y\in R\\
\vec{a_1}=(0,1,0) λ 1 = 1 , a 1 = ( x , y , z ) = 0 ( A − λ 1 I ) a 1 = 0 ⎝ ⎛ 0 0 2 0 0 0 2 0 0 ⎠ ⎞ ⋅ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ 2 z = 0 , 2 x = 0 x = z = 0 , y ∈ R a 1 = ( 0 , 1 , 0 )
λ 2 = − 1 , a 2 ⃗ = ( x , y , z ) ≠ 0 ⃗ ( A − λ 2 I ) a 2 ⃗ = 0 ⃗ ( 2 0 2 0 2 0 2 0 2 ) ⋅ ( x y z ) = ( 0 0 0 ) 2 x + 2 z = 0 , 2 y = 0 x = − z , y = 0 a 2 ⃗ = ( 1 , 0 , − 1 ) \lambda_2=-1, \vec{a_2}=(x,y,z)\neq\vec{0}\\
(A-\lambda_2I)\vec{a_2}=\vec{0}\\
\begin{pmatrix}
2& 0&2 \\
0 & 2&0\\
2&0&2
\end{pmatrix}\cdot \begin{pmatrix}
x \\
y\\z
\end{pmatrix}=\begin{pmatrix}
0 \\
0\\0
\end{pmatrix}\\
2x+2z=0,2y=0\\
x=-z, y=0\\
\vec{a_2}=(1,0,-1) λ 2 = − 1 , a 2 = ( x , y , z ) = 0 ( A − λ 2 I ) a 2 = 0 ⎝ ⎛ 2 0 2 0 2 0 2 0 2 ⎠ ⎞ ⋅ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ 2 x + 2 z = 0 , 2 y = 0 x = − z , y = 0 a 2 = ( 1 , 0 , − 1 )
λ 3 = 3 , a 3 ⃗ = ( x , y , z ) ≠ 0 ⃗ ( A − λ 3 I ) a 3 ⃗ = 0 ⃗ ( − 2 0 2 0 − 2 0 2 0 − 2 ) ⋅ ( x y z ) = ( 0 0 0 ) − 2 x + 2 z = 0 , − 2 y = 0 , 2 x − 2 z = 0 x = z = 1 , y = 0 a 3 ⃗ = ( 1 , 0 , 1 ) \lambda_3=3, \vec{a_3}=(x,y,z)\neq\vec{0}\\
(A-\lambda_3I)\vec{a_3}=\vec{0}\\
\begin{pmatrix}
-2& 0&2 \\
0 & -2&0\\
2&0&-2
\end{pmatrix}\cdot \begin{pmatrix}
x \\
y\\z
\end{pmatrix}=\begin{pmatrix}
0 \\
0\\0
\end{pmatrix}\\
-2x+2z=0,-2y=0, 2x-2z=0\\
x=z=1, y=0\\
\vec{a_3}=(1,0,1) λ 3 = 3 , a 3 = ( x , y , z ) = 0 ( A − λ 3 I ) a 3 = 0 ⎝ ⎛ − 2 0 2 0 − 2 0 2 0 − 2 ⎠ ⎞ ⋅ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ − 2 x + 2 z = 0 , − 2 y = 0 , 2 x − 2 z = 0 x = z = 1 , y = 0 a 3 = ( 1 , 0 , 1 )
a 1 ⃗ = ( 0 , 1 , 0 ) , ∣ a 1 ⃗ ∣ = 1 , a 2 ⃗ = ( 1 , 0 , − 1 ) , ∣ a 2 ⃗ ∣ = 2 , a 3 ⃗ = ( 1 , 0 , 1 ) , ∣ a 3 ⃗ ∣ = 2 P = ( 0 1 0 1 2 0 − 1 2 1 2 0 1 2 ) B = P − 1 A P B = ( 1 0 0 0 − 1 0 0 0 3 ) \vec{a_1}=(0,1,0), |\vec{a_1}|=1,\\
\vec{a_2}=(1,0,-1), |\vec{a_2}|=\sqrt2,\\
\vec{a_3}=(1,0,1),|\vec{a_3}|=\sqrt2\\
P=\begin{pmatrix}
0 & 1&0 \\
\frac{1}{\sqrt2} & 0& -\frac{1}{\sqrt2}\\
\frac{1}{\sqrt2}&0 &\frac{1}{\sqrt2}
\end{pmatrix}\\
B=P^{-1}AP\\
B=\begin{pmatrix}
1 & 0&0 \\
0 & -1&0\\
0&0&3
\end{pmatrix} a 1 = ( 0 , 1 , 0 ) , ∣ a 1 ∣ = 1 , a 2 = ( 1 , 0 , − 1 ) , ∣ a 2 ∣ = 2 , a 3 = ( 1 , 0 , 1 ) , ∣ a 3 ∣ = 2 P = ⎝ ⎛ 0 2 1 2 1 1 0 0 0 − 2 1 2 1 ⎠ ⎞ B = P − 1 A P B = ⎝ ⎛ 1 0 0 0 − 1 0 0 0 3 ⎠ ⎞
A 2 = ( 1 0 2 0 1 0 2 0 1 ) ⋅ ( 1 0 2 0 1 0 2 0 1 ) = ( 5 0 4 0 1 0 4 0 5 ) A 3 = A 2 ⋅ A = ( 5 0 4 0 1 0 4 0 5 ) ⋅ ( 1 0 2 0 1 0 2 0 1 ) = = ( 13 0 14 0 1 0 14 0 13 ) A^2=\begin{pmatrix}
1 & 0&2 \\
0&1&0\\
2&0&1
\end{pmatrix}\cdot\begin{pmatrix}
1 & 0&2 \\
0&1&0\\
2&0&1
\end{pmatrix}=\begin{pmatrix}
5 & 0&4 \\
0&1&0\\
4&0&5
\end{pmatrix}\\
A^3=A^2\cdot A=\begin{pmatrix}
5 & 0&4 \\
0&1&0\\
4&0&5
\end{pmatrix}\cdot \begin{pmatrix}
1 & 0&2 \\
0&1&0\\
2&0&1
\end{pmatrix}=\\
=\begin{pmatrix}
13 & 0&14 \\
0&1&0\\
14&0&13
\end{pmatrix} A 2 = ⎝ ⎛ 1 0 2 0 1 0 2 0 1 ⎠ ⎞ ⋅ ⎝ ⎛ 1 0 2 0 1 0 2 0 1 ⎠ ⎞ = ⎝ ⎛ 5 0 4 0 1 0 4 0 5 ⎠ ⎞ A 3 = A 2 ⋅ A = ⎝ ⎛ 5 0 4 0 1 0 4 0 5 ⎠ ⎞ ⋅ ⎝ ⎛ 1 0 2 0 1 0 2 0 1 ⎠ ⎞ = = ⎝ ⎛ 13 0 14 0 1 0 14 0 13 ⎠ ⎞
A 4 = A 3 ⋅ A = ( 13 0 14 0 1 0 14 0 13 ) ⋅ ( 1 0 2 0 1 0 2 0 1 ) = = ( 41 0 40 0 1 0 40 0 41 ) A^4=A^3\cdot A=\begin{pmatrix}
13 & 0&14 \\
0&1&0\\
14&0&13
\end{pmatrix}\cdot \begin{pmatrix}
1 & 0&2 \\
0&1&0\\
2&0&1
\end{pmatrix}=\\
=\begin{pmatrix}
41 & 0&40 \\
0&1&0\\
40&0&41
\end{pmatrix} A 4 = A 3 ⋅ A = ⎝ ⎛ 13 0 14 0 1 0 14 0 13 ⎠ ⎞ ⋅ ⎝ ⎛ 1 0 2 0 1 0 2 0 1 ⎠ ⎞ = = ⎝ ⎛ 41 0 40 0 1 0 40 0 41 ⎠ ⎞
A n = ( a 0 b 0 1 0 b 0 a ) A^n=\begin{pmatrix}
a & 0&b \\
0&1&0\\
b&0&a
\end{pmatrix} A n = ⎝ ⎛ a 0 b 0 1 0 b 0 a ⎠ ⎞
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