A = [ 1 − 1 1 2 0 3 1 4 − 1 ] A=\begin{bmatrix}
1 & -1&1 \\
2 & 0&3\\
1&4&-1
\end{bmatrix} A = ⎣ ⎡ 1 2 1 − 1 0 4 1 3 − 1 ⎦ ⎤
Begin with an initial nonzero approximation of
x 0 = [ 1 1 1 ] x_0=\begin{bmatrix}
1 \\
1\\
1
\end{bmatrix} x 0 = ⎣ ⎡ 1 1 1 ⎦ ⎤
Then obtain the following approximations
Iteration
x 1 = A x 0 = [ 1 − 1 1 2 0 3 1 4 − 1 ] [ 1 1 1 ] = [ 1 5 4 ] x_1=Ax_0=\begin{bmatrix}
1 & -1&1 \\
2 & 0&3\\
1&4&-1
\end{bmatrix}\begin{bmatrix}
1 \\
1\\
1
\end{bmatrix}
=\begin{bmatrix}
1 \\
5\\
4
\end{bmatrix} x 1 = A x 0 = ⎣ ⎡ 1 2 1 − 1 0 4 1 3 − 1 ⎦ ⎤ ⎣ ⎡ 1 1 1 ⎦ ⎤ = ⎣ ⎡ 1 5 4 ⎦ ⎤
“Scaled” Approximation
→ 1 [ 1 5 4 ] \to1\begin{bmatrix}
1 \\
5\\
4
\end{bmatrix} → 1 ⎣ ⎡ 1 5 4 ⎦ ⎤
Iteration
x 2 = A x 1 = [ 1 − 1 1 2 0 3 1 4 − 1 ] [ 1 5 4 ] = [ 0 14 17 ] x_2=Ax_1=\begin{bmatrix}
1 & -1&1 \\
2 & 0&3\\
1&4&-1
\end{bmatrix}\begin{bmatrix}
1 \\
5\\
4
\end{bmatrix}
=\begin{bmatrix}
0 \\
14\\
17
\end{bmatrix} x 2 = A x 1 = ⎣ ⎡ 1 2 1 − 1 0 4 1 3 − 1 ⎦ ⎤ ⎣ ⎡ 1 5 4 ⎦ ⎤ = ⎣ ⎡ 0 14 17 ⎦ ⎤
“Scaled” Approximation
→ 14 [ 0 1 1.2 ] \to14\begin{bmatrix}
0 \\
1\\
1.2
\end{bmatrix} → 14 ⎣ ⎡ 0 1 1.2 ⎦ ⎤
Iteration
x 3 = A x 2 = [ 1 − 1 1 2 0 3 1 4 − 1 ] [ 0 14 17 ] = [ 3 51 39 ] x_3=Ax_2=\begin{bmatrix}
1 & -1&1 \\
2 & 0&3\\
1&4&-1
\end{bmatrix}\begin{bmatrix}
0 \\
14\\
17
\end{bmatrix}
=\begin{bmatrix}
3 \\
51\\
39
\end{bmatrix} x 3 = A x 2 = ⎣ ⎡ 1 2 1 − 1 0 4 1 3 − 1 ⎦ ⎤ ⎣ ⎡ 0 14 17 ⎦ ⎤ = ⎣ ⎡ 3 51 39 ⎦ ⎤
“Scaled” Approximation
→ 3 [ 1 17 13 ] \to3\begin{bmatrix}
1 \\
17\\
13
\end{bmatrix} → 3 ⎣ ⎡ 1 17 13 ⎦ ⎤
Iteration
x 4 = A x 3 = [ 1 − 1 1 2 0 3 1 4 − 1 ] [ 3 51 39 ] = [ − 9 123 168 ] x_4=Ax_3=\begin{bmatrix}
1 & -1&1 \\
2 & 0&3\\
1&4&-1
\end{bmatrix}\begin{bmatrix}
3 \\
51\\
39
\end{bmatrix}
=\begin{bmatrix}
-9 \\
123\\
168
\end{bmatrix} x 4 = A x 3 = ⎣ ⎡ 1 2 1 − 1 0 4 1 3 − 1 ⎦ ⎤ ⎣ ⎡ 3 51 39 ⎦ ⎤ = ⎣ ⎡ − 9 123 168 ⎦ ⎤
“Scaled” Approximation
→ − 9 [ 1 − 13.7 − 18.7 ] \to-9\begin{bmatrix}
1 \\
-13.7\\
-18.7
\end{bmatrix} → − 9 ⎣ ⎡ 1 − 13.7 − 18.7 ⎦ ⎤
Note that the approximations appear to be approaching scalar multiples of
[ 1 − 14 − 19 ] \begin{bmatrix}
1 \\
-14\\
-19
\end{bmatrix} ⎣ ⎡ 1 − 14 − 19 ⎦ ⎤
With x = ( 1 , − 13.7 , − 18.7 ) x=(1,-13.7,-18.7) x = ( 1 , − 13.7 , − 18.7 ) A A A A A A A x Ax A x
A x = [ 1 − 1 1 2 0 3 1 4 − 1 ] [ 1 − 13.7 − 18.7 ] = [ − 4 − 54.1 − 35.1 ] Ax=\begin{bmatrix}
1 & -1&1 \\
2 & 0&3\\
1&4&-1
\end{bmatrix}\begin{bmatrix}
1 \\
-13.7\\
-18.7
\end{bmatrix}
=\begin{bmatrix}
-4 \\
-54.1\\
-35.1
\end{bmatrix} A x = ⎣ ⎡ 1 2 1 − 1 0 4 1 3 − 1 ⎦ ⎤ ⎣ ⎡ 1 − 13.7 − 18.7 ⎦ ⎤ = ⎣ ⎡ − 4 − 54.1 − 35.1 ⎦ ⎤
Then, because
A x ⋅ x = − 4 ⋅ 1 + ( − 54.1 ) ( − 13.7 ) + ( − 35.1 ) ( − 18.7 ) = 1393.54 x ⋅ x = 1 ⋅ 1 + ( − 13.7 ) ( − 13.7 ) + ( − 18.7 ) ( − 18.7 ) = 538.38 Ax\cdot x=-4\cdot 1+(-54.1)(-13.7)+(-35.1)(-18.7)=1393.54\\
x\cdot x=1\cdot1+(-13.7)(-13.7)+(-18.7)(-18.7)=538.38 A x ⋅ x = − 4 ⋅ 1 + ( − 54.1 ) ( − 13.7 ) + ( − 35.1 ) ( − 18.7 ) = 1393.54 x ⋅ x = 1 ⋅ 1 + ( − 13.7 ) ( − 13.7 ) + ( − 18.7 ) ( − 18.7 ) = 538.38
you can compute the Rayleigh quotient to be
λ = A x ⋅ x x ⋅ x = 1393.54 538.38 = 2.6 \lambda=\frac{Ax\cdot x}{x\cdot x}=\frac{1393.54}{538.38}=2.6 λ = x ⋅ x A x ⋅ x = 538.38 1393.54 = 2.6
which is a good approximation of the dominant eigenvalue
λ = 3 \lambda=3 λ = 3
Answer [ 1 − 14 − 19 ] , λ = 3 \begin{bmatrix}
1 \\
-14\\
-19
\end{bmatrix}, \lambda=3 ⎣ ⎡ 1 − 14 − 19 ⎦ ⎤ , λ = 3
Comments