Solved on Feb 07, 2024

Find the eigenvalues $\lambda_{1}, \lambda_{2}, \lambda_{3}$ of matrix $A=\left(\begin{array}{ccc}3 & 0 & 0 \\ \frac{1}{2} & 5 & 0 \\ \frac{1}{12} & \frac{1}{6} & 0\end{array}\right)$ and their corresponding eigenvectors.

STEP 1

Assumptions
1. Matrix $A$ is a lower triangular matrix.
2. The eigenvalues of a triangular matrix are the entries on its main diagonal.
3. The eigenvectors corresponding to each eigenvalue of a triangular matrix can be found by solving the system $(A - \lambda I)x = 0$ , where $\lambda$ is an eigenvalue and $I$ is the identity matrix of the same size as $A$ .

STEP 2

Identify the eigenvalues of $A$ by looking at the entries on the main diagonal.
$\lambda_{1} = 3, \lambda_{2} = 5, \lambda_{3} = 0$

STEP 3

Since the eigenvalues must be ordered such that $\lambda_{1} < \lambda_{2} < \lambda_{3}$ , we reorder them accordingly.
$\lambda_{1} = 0, \lambda_{2} = 3, \lambda_{3} = 5$

STEP 4

For the eigenvalue $\lambda_{1} = 0$ , find the eigenvectors by solving the system $(A - \lambda_{1} I)x = 0$ .

STEP 5

Substitute $\lambda_{1} = 0$ into the equation and set up the augmented matrix for the system.
$\begin{align} (A - 0 \cdot I)x &= 0 \\ \left(\begin{array}{ccc} 3 & 0 & 0 \\ \frac{1}{2} & 5 & 0 \\ \frac{1}{12} & \frac{1}{6} & 0 \end{array}\right)x &= 0 \end{align}$

STEP 6

Solve the system of equations represented by the augmented matrix.
$\begin{align} 3x_1 &= 0 \\ \frac{1}{2}x_1 + 5x_2 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 &= 0 \end{align}$

STEP 7

From the first equation, we get $x_1 = 0$ .

STEP 8

Substitute $x_1 = 0$ into the second and third equations.
$\begin{align} 5x_2 &= 0 \\ \frac{1}{6}x_2 &= 0 \end{align}$

STEP 9

From the second equation, we get $x_2 = 0$ .

STEP 10

Since $x_1 = x_2 = 0$ , the eigenvector corresponding to $\lambda_{1} = 0$ can have any value for $x_3$ . We can choose $x_3 = 1$ for simplicity.

STEP 11

The eigenvector corresponding to $\lambda_{1} = 0$ is then $(0, 0, 1)^T$ .

STEP 12

For the eigenvalue $\lambda_{2} = 3$ , find the eigenvectors by solving the system $(A - \lambda_{2} I)x = 0$ .

STEP 13

Substitute $\lambda_{2} = 3$ into the equation and set up the augmented matrix for the system.
$\begin{align} (A - 3 \cdot I)x &= 0 \\ \left(\begin{array}{ccc} 0 & 0 & 0 \\ \frac{1}{2} & 2 & 0 \\ \frac{1}{12} & \frac{1}{6} & -3 \end{array}\right)x &= 0 \end{align}$

STEP 14

Solve the system of equations represented by the augmented matrix.
$\begin{align} 0x_1 + 0x_2 + 0x_3 &= 0 \\ \frac{1}{2}x_1 + 2x_2 + 0x_3 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 - 3x_3 &= 0 \end{align}$

STEP 15

The first equation gives us no information as it is always true. We can choose $x_3 = 1$ for simplicity.

STEP 16

Substitute $x_3 = 1$ into the third equation to find $x_1$ and $x_2$ .
$\frac{1}{12}x_1 + \frac{1}{6}x_2 - 3 = 0$

STEP 17

Multiply the third equation by 12 to clear the fractions.
$x_1 + 2x_2 - 36 = 0$

STEP 18

Substitute $x_3 = 1$ into the second equation to find $x_1$ and $x_2$ .
$\frac{1}{2}x_1 + 2x_2 = 0$

STEP 19

Multiply the second equation by 2 to clear the fraction.
$x_1 + 4x_2 = 0$

STEP 20

Subtract the second equation from the third equation to solve for $x_2$ .
$-2x_2 = 36$

STEP 21

Solve for $x_2$ .
$x_2 = -18$

STEP 22

Substitute $x_2 = -18$ into the second equation to solve for $x_1$ .
$x_1 + 4(-18) = 0$

STEP 23

Solve for $x_1$ .
$x_1 = 72$

STEP 24

The eigenvector corresponding to $\lambda_{2} = 3$ is then $(72, -18, 1)^T$ .

STEP 25

For the eigenvalue $\lambda_{3} = 5$ , find the eigenvectors by solving the system $(A - \lambda_{3} I)x = 0$ .

STEP 26

Substitute $\lambda_{3} = 5$ into the equation and set up the augmented matrix for the system.
$\begin{align} (A - 5 \cdot I)x &= 0 \\ \left(\begin{array}{ccc} -2 & 0 & 0 \\ \frac{1}{2} & 0 & 0 \\ \frac{1}{12} & \frac{1}{6} & -5 \end{array}\right)x &= 0 \end{align}$

STEP 27

Solve the system of equations represented by the augmented matrix.
$\begin{align} -2x_1 &= 0 \\ \frac{1}{2}x_1 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 - 5x_3 &= 0 \end{align}$

STEP 28

From the first equation, we get $x_1 = 0$ .

STEP 29

Substitute $x_1 = 0$ into the third equation.
$\frac{1}{6}x_2 - 5x_3 = 0$

STEP 30

Since $x_1 = 0$ , we can choose $x_2 = 1$ and solve for $x_3$ .
$\frac{1}{6} - 5x_3 = 0$

STEP 31

Multiply the equation by 6 to clear the fraction.
$1 - 30x_3 = 0$

STEP 32

Solve for $x_3$ .
$x_3 = \frac{1}{30}$

STEP 33

The eigenvector corresponding to $\lambda_{3} = 5$ is then $(0, 1, \frac{1}{30})^T$ . However, we can scale this vector to have integer entries by multiplying by 30.

STEP 34

The scaled eigenvector is $(0, 30, 1)^T$ .
The solutions are: (a) $\lambda_{1}=0$ (b) $\lambda_{2}=3$ (c) $\lambda_{3}=5$ (d) A basis for the eigenspace corresponding to the smallest eigenvalue $\lambda_{1}$ is $(0,0,1)^{T}$ (e) A basis for the eigenspace corresponding to the eigenvalue $\lambda_{2}$ is $(72,-18,1)^{T}$ (f) A basis for the eigenspace corresponding to the largest eigenvalue $\lambda_{3}$ is $(0,30,1)^{T}$

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STEP 1

STEP 2

Identify the eigenvalues of AAA by looking at the entries on the main diagonal.λ1=3,λ2=5,λ3=0\lambda_{1} = 3, \lambda_{2} = 5, \lambda_{3} = 0λ1​=3,λ2​=5,λ3​=0

STEP 3

Since the eigenvalues must be ordered such that λ1<λ2<λ3\lambda_{1} < \lambda_{2} < \lambda_{3}λ1​<λ2​<λ3​, we reorder them accordingly.λ1=0,λ2=3,λ3=5\lambda_{1} = 0, \lambda_{2} = 3, \lambda_{3} = 5λ1​=0,λ2​=3,λ3​=5

STEP 4

For the eigenvalue λ1=0\lambda_{1} = 0λ1​=0, find the eigenvectors by solving the system (A−λ1I)x=0(A - \lambda_{1} I)x = 0(A−λ1​I)x=0.

STEP 5

STEP 6

Solve the system of equations represented by the augmented matrix.3x1=012x1+5x2=0112x1+16x2=0 \begin{align*} 3x_1 &= 0 \\ \frac{1}{2}x_1 + 5x_2 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 &= 0 \end{align*} 3x1​21​x1​+5x2​121​x1​+61​x2​​=0=0=0​

STEP 7

From the first equation, we get x1=0x_1 = 0x1​=0.

STEP 8

Substitute x1=0x_1 = 0x1​=0 into the second and third equations.5x2=016x2=0 \begin{align*} 5x_2 &= 0 \\ \frac{1}{6}x_2 &= 0 \end{align*} 5x2​61​x2​​=0=0​

STEP 9

From the second equation, we get x2=0x_2 = 0x2​=0.

STEP 10

Since x1=x2=0x_1 = x_2 = 0x1​=x2​=0, the eigenvector corresponding to λ1=0\lambda_{1} = 0λ1​=0 can have any value for x3x_3x3​. We can choose x3=1x_3 = 1x3​=1 for simplicity.

STEP 11

The eigenvector corresponding to λ1=0\lambda_{1} = 0λ1​=0 is then (0,0,1)T(0, 0, 1)^T(0,0,1)T.

STEP 12

For the eigenvalue λ2=3\lambda_{2} = 3λ2​=3, find the eigenvectors by solving the system (A−λ2I)x=0(A - \lambda_{2} I)x = 0(A−λ2​I)x=0.

STEP 13

STEP 14

STEP 15

The first equation gives us no information as it is always true. We can choose x3=1x_3 = 1x3​=1 for simplicity.

STEP 16

Substitute x3=1x_3 = 1x3​=1 into the third equation to find x1x_1x1​ and x2x_2x2​.112x1+16x2−3=0 \frac{1}{12}x_1 + \frac{1}{6}x_2 - 3 = 0 121​x1​+61​x2​−3=0

STEP 17

Multiply the third equation by 12 to clear the fractions.x1+2x2−36=0 x_1 + 2x_2 - 36 = 0 x1​+2x2​−36=0

STEP 18

Substitute x3=1x_3 = 1x3​=1 into the second equation to find x1x_1x1​ and x2x_2x2​.12x1+2x2=0 \frac{1}{2}x_1 + 2x_2 = 0 21​x1​+2x2​=0

STEP 19

Multiply the second equation by 2 to clear the fraction.x1+4x2=0 x_1 + 4x_2 = 0 x1​+4x2​=0

STEP 20

Subtract the second equation from the third equation to solve for x2x_2x2​.−2x2=36 -2x_2 = 36 −2x2​=36

STEP 21

Solve for x2x_2x2​.x2=−18 x_2 = -18 x2​=−18

STEP 22

Substitute x2=−18x_2 = -18x2​=−18 into the second equation to solve for x1x_1x1​.x1+4(−18)=0 x_1 + 4(-18) = 0 x1​+4(−18)=0

STEP 23

Solve for x1x_1x1​.x1=72 x_1 = 72 x1​=72

STEP 24

The eigenvector corresponding to λ2=3\lambda_{2} = 3λ2​=3 is then (72,−18,1)T(72, -18, 1)^T(72,−18,1)T.

STEP 25

For the eigenvalue λ3=5\lambda_{3} = 5λ3​=5, find the eigenvectors by solving the system (A−λ3I)x=0(A - \lambda_{3} I)x = 0(A−λ3​I)x=0.

STEP 26

STEP 27

Solve the system of equations represented by the augmented matrix.−2x1=012x1=0112x1+16x2−5x3=0 \begin{align*} -2x_1 &= 0 \\ \frac{1}{2}x_1 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 - 5x_3 &= 0 \end{align*} −2x1​21​x1​121​x1​+61​x2​−5x3​​=0=0=0​

STEP 28

From the first equation, we get x1=0x_1 = 0x1​=0.

STEP 29

Substitute x1=0x_1 = 0x1​=0 into the third equation.16x2−5x3=0 \frac{1}{6}x_2 - 5x_3 = 0 61​x2​−5x3​=0

STEP 30

Since x1=0x_1 = 0x1​=0, we can choose x2=1x_2 = 1x2​=1 and solve for x3x_3x3​.16−5x3=0 \frac{1}{6} - 5x_3 = 0 61​−5x3​=0

STEP 31

Multiply the equation by 6 to clear the fraction.1−30x3=0 1 - 30x_3 = 0 1−30x3​=0

STEP 32

Solve for x3x_3x3​.x3=130 x_3 = \frac{1}{30} x3​=301​

STEP 33

The eigenvector corresponding to λ3=5\lambda_{3} = 5λ3​=5 is then (0,1,130)T(0, 1, \frac{1}{30})^T(0,1,301​)T. However, we can scale this vector to have integer entries by multiplying by 30.

STEP 34

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Identify the eigenvalues of $A$ by looking at the entries on the main diagonal.
$\lambda_{1} = 3, \lambda_{2} = 5, \lambda_{3} = 0$

Since the eigenvalues must be ordered such that $\lambda_{1} < \lambda_{2} < \lambda_{3}$ , we reorder them accordingly.
$\lambda_{1} = 0, \lambda_{2} = 3, \lambda_{3} = 5$

For the eigenvalue $\lambda_{1} = 0$ , find the eigenvectors by solving the system $(A - \lambda_{1} I)x = 0$ .

Solve the system of equations represented by the augmented matrix.
$\begin{align} 3x_1 &= 0 \\ \frac{1}{2}x_1 + 5x_2 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 &= 0 \end{align}$

From the first equation, we get $x_1 = 0$ .

Substitute $x_1 = 0$ into the second and third equations.
$\begin{align} 5x_2 &= 0 \\ \frac{1}{6}x_2 &= 0 \end{align}$

From the second equation, we get $x_2 = 0$ .

Since $x_1 = x_2 = 0$ , the eigenvector corresponding to $\lambda_{1} = 0$ can have any value for $x_3$ . We can choose $x_3 = 1$ for simplicity.

The eigenvector corresponding to $\lambda_{1} = 0$ is then $(0, 0, 1)^T$ .

For the eigenvalue $\lambda_{2} = 3$ , find the eigenvectors by solving the system $(A - \lambda_{2} I)x = 0$ .

The first equation gives us no information as it is always true. We can choose $x_3 = 1$ for simplicity.

Substitute $x_3 = 1$ into the third equation to find $x_1$ and $x_2$ .
$\frac{1}{12}x_1 + \frac{1}{6}x_2 - 3 = 0$

Multiply the third equation by 12 to clear the fractions.
$x_1 + 2x_2 - 36 = 0$

Substitute $x_3 = 1$ into the second equation to find $x_1$ and $x_2$ .
$\frac{1}{2}x_1 + 2x_2 = 0$

Multiply the second equation by 2 to clear the fraction.
$x_1 + 4x_2 = 0$

Subtract the second equation from the third equation to solve for $x_2$ .
$-2x_2 = 36$

Solve for $x_2$ .
$x_2 = -18$

Substitute $x_2 = -18$ into the second equation to solve for $x_1$ .
$x_1 + 4(-18) = 0$

Solve for $x_1$ .
$x_1 = 72$

The eigenvector corresponding to $\lambda_{2} = 3$ is then $(72, -18, 1)^T$ .

For the eigenvalue $\lambda_{3} = 5$ , find the eigenvectors by solving the system $(A - \lambda_{3} I)x = 0$ .

Solve the system of equations represented by the augmented matrix.
$\begin{align} -2x_1 &= 0 \\ \frac{1}{2}x_1 &= 0 \\ \frac{1}{12}x_1 + \frac{1}{6}x_2 - 5x_3 &= 0 \end{align}$

From the first equation, we get $x_1 = 0$ .

Substitute $x_1 = 0$ into the third equation.
$\frac{1}{6}x_2 - 5x_3 = 0$

Since $x_1 = 0$ , we can choose $x_2 = 1$ and solve for $x_3$ .
$\frac{1}{6} - 5x_3 = 0$

Multiply the equation by 6 to clear the fraction.
$1 - 30x_3 = 0$

Solve for $x_3$ .
$x_3 = \frac{1}{30}$

The eigenvector corresponding to $\lambda_{3} = 5$ is then $(0, 1, \frac{1}{30})^T$ . However, we can scale this vector to have integer entries by multiplying by 30.