find the eigenvalues and eigen vectors of the symmetric matrix : a= [1 1 3/1 5 1/ 3 1 1]
Answers
Step-by-step explanation:
(1) \begin{bmatrix} 9 & 3 \\ 2 & 9 \end{bmatrix}[
9
2
3
9
]
Start from forming a new matrix by subtracting λ
from the diagonal entries of the given matrix:
\left[\begin{array}{cc}9 - \lambda & 3\\2 & 9 - \lambda\end{array}\right][
9−λ
2
3
9−λ
]
The determinant of the obtained matrix is \lambda^{2} - 18 \lambda + 75λ
2
−18λ+75
Solve the equation
The roots are comes to be \lambda_{1} = 9 - \sqrt{6}λ
1
=9−
6
, \lambda_{2} = \sqrt{6} + 9λ
2
=
6
+9
These are the eigenvalues.
Next, find the eigenvectors.
a. \lambda = 9 - \sqrt{6}λ=9−
6
\left[\begin{array}{cc}9 - \lambda & 3\\2 & 9 - \lambda\end{array}\right] = \left[\begin{array}{cc}\sqrt{6} & 3\\2 & \sqrt{6}\end{array}\right][
9−λ
2
3
9−λ
]=[
6
2
3
6
]
The null space of this matrix is\left\{\left[\begin{array}{c}- \frac{\sqrt{6}}{2}\\1\end{array}\right]\right\}{[
−
2
6
1
]}
This is the eigenvector.
b. \lambda = \sqrt{6} + 9λ=
6
+9
\left[\begin{array}{cc}9 - \lambda & 3\\2 & 9 - \lambda\end{array}\right] = \left[\begin{array}{cc}- \sqrt{6} & 3\\2 & - \sqrt{6}\end{array}\right][
9−λ
2
3
9−λ
]=[
−
6
2
3
−
6
]
The null space of this matrix is \left\{\left[\begin{array}{c}\frac{\sqrt{6}}{2}\\1\end{array}\right]\right\}{[
2
6
1
]}
This is the eigenvector.
(2)
\begin{bmatrix} 2 & 0 & 1 \\ 0& 2 & 0 \\ 1 & 0 & 2 \end{bmatrix}
⎣
⎢
⎡
2
0
1
0
2
0
1
0
2
⎦
⎥
⎤
as above done we well proceed the same
Start from forming a new matrix by subtracting λ
λ from the diagonal entries of the given matrix:
\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right].
⎣
⎢
⎡
2−λ
0
1
0
2−λ
0
1
0
2−λ
⎦
⎥
⎤
.
The determinant of the obtained matrix is
- \lambda^{3} + 6 \lambda^{2} - 11 \lambda + 6−λ
3
+6λ
2
−11λ+6
Solve the equation \lambda_{1} = 3λ
1
=3 , \lambda_{2} = 2λ
2
=2 , \lambda_{3} = 1λ
3
=1
These are the eigenvalues.
Next, find the eigenvectors.
a. \lambda = 3λ=3
\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right] = \left[\begin{array}{ccc}-1 & 0 & 1\\0 & -1 & 0\\1 & 0 & -1\end{array}\right]
⎣
⎢
⎡
2−λ
0
1
0
2−λ
0
1
0
2−λ
⎦
⎥
⎤
=
⎣
⎢
⎡
−1
0
1
0
−1
0
1
0
−1
⎦
⎥
⎤
The null space of this matrix is \left\{\left[\begin{array}{c}1\\0\\1\end{array}\right]\right\}
⎩
⎪
⎨
⎪
⎧
⎣
⎢
⎡
1
0
1
⎦
⎥
⎤
⎭
⎪
⎬
⎪
⎫
This is the eigenvector.
b. \lambda = 2λ=2
\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 1\\0 & 0 & 0\\1 & 0 & 0\end{array}\right]
⎣
⎢
⎡
2−λ
0
1
0
2−λ
0
1
0
2−λ
⎦
⎥
⎤
=
⎣
⎢
⎡
0
0
1
0
0
0
1
0
0
⎦
⎥
⎤
The null space of this matrix is\left\{\left[\begin{array}{c}0\\1\\0\end{array}\right]\right\}
⎩
⎪
⎨
⎪
⎧
⎣
⎢
⎡
0
1
0
⎦
⎥
⎤
⎭
⎪
⎬
⎪
⎫
This is the eigenvector.
c. \lambda = 1λ=1
\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 0\\1 & 0 & 1\end{array}\right]
⎣
⎢
⎡
2−λ
0
1
0
2−λ
0
1
0
2−λ
⎦
⎥
⎤
=
⎣
⎢
⎡
1
0
1
0
1
0
1
0
1
⎦
⎥
⎤
The null space of this matrix is\left\{\left[\begin{array}{c}-1\\0\\1\end{array}\right]\right\}
⎩
⎪
⎨
⎪
⎧
⎣
⎢
⎡
−1
0
1
⎦
⎥
⎤
⎭
⎪
⎬
⎪
⎫
This is the eigenvector.
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