(1) [ 9 3 2 9 ] \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:
[ 9 − λ 3 2 9 − λ ] \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 λ 2 − 18 λ + 75 \lambda^{2} - 18 \lambda + 75 λ 2 − 18 λ + 75
Solve the equation
The roots are comes to be λ 1 = 9 − 6 \lambda_{1} = 9 - \sqrt{6} λ 1 = 9 − 6 λ 2 = 6 + 9 \lambda_{2} = \sqrt{6} + 9 λ 2 = 6 + 9
These are the eigenvalues.
Next, find the eigenvectors.
a. λ = 9 − 6 \lambda = 9 - \sqrt{6} λ = 9 − 6
[ 9 − λ 3 2 9 − λ ] = [ 6 3 2 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{ [ − 6 2 1 ] } \left\{\left[\begin{array}{c}- \frac{\sqrt{6}}{2}\\1\end{array}\right]\right\} { [ − 2 6 1 ] }
This is the eigenvector.
b. λ = 6 + 9 \lambda = \sqrt{6} + 9 λ = 6 + 9
[ 9 − λ 3 2 9 − λ ] = [ − 6 3 2 − 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 { [ 6 2 1 ] } \left\{\left[\begin{array}{c}\frac{\sqrt{6}}{2}\\1\end{array}\right]\right\} { [ 2 6 1 ] }
This is the eigenvector.
(2)
[ 2 0 1 0 2 0 1 0 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:
[ 2 − λ 0 1 0 2 − λ 0 1 0 2 − λ ] . \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
− λ 3 + 6 λ 2 − 11 λ + 6 - \lambda^{3} + 6 \lambda^{2} - 11 \lambda + 6 − λ 3 + 6 λ 2 − 11 λ + 6
Solve the equation λ 1 = 3 \lambda_{1} = 3 λ 1 = 3 λ 2 = 2 \lambda_{2} = 2 λ 2 = 2 λ 3 = 1 \lambda_{3} = 1 λ 3 = 1
These are the eigenvalues.
Next, find the eigenvectors.
a. λ = 3 \lambda = 3 λ = 3
[ 2 − λ 0 1 0 2 − λ 0 1 0 2 − λ ] = [ − 1 0 1 0 − 1 0 1 0 − 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 { [ 1 0 1 ] } \left\{\left[\begin{array}{c}1\\0\\1\end{array}\right]\right\} ⎩ ⎨ ⎧ ⎣ ⎡ 1 0 1 ⎦ ⎤ ⎭ ⎬ ⎫
This is the eigenvector.
b. λ = 2 \lambda = 2 λ = 2
[ 2 − λ 0 1 0 2 − λ 0 1 0 2 − λ ] = [ 0 0 1 0 0 0 1 0 0 ] \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{ [ 0 1 0 ] } \left\{\left[\begin{array}{c}0\\1\\0\end{array}\right]\right\} ⎩ ⎨ ⎧ ⎣ ⎡ 0 1 0 ⎦ ⎤ ⎭ ⎬ ⎫
This is the eigenvector.
c. λ = 1 \lambda = 1 λ = 1
[ 2 − λ 0 1 0 2 − λ 0 1 0 2 − λ ] = [ 1 0 1 0 1 0 1 0 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{ [ − 1 0 1 ] } \left\{\left[\begin{array}{c}-1\\0\\1\end{array}\right]\right\} ⎩ ⎨ ⎧ ⎣ ⎡ − 1 0 1 ⎦ ⎤ ⎭ ⎬ ⎫
This is the eigenvector.
#339914 on Dec 2023
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