Eigenvectors for a 3x3 matrix

  • #1
3
0
Member advised that the homework template must be used.
Hi,
I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. The matrix is
(I have a ; since I can't have a space between each column. Sorry):
[20 ; -10 ; 0]
[-10 ; 30 ; 0]
[0 ; 0 ; 40]

I’ve already calculated the eigenvalues, which are: λ1=13.8196, λ2=36.18 , λ3= 40
The next step is to calculate the eigenvectors for every eigenvalue λ and it’s now I keep getting the wrong answer. Let us take λ1 for instance. We'll get:

[20-13.8192 ; -10 ; 0]
[-10 ; 30-13.8196 ; 0]
[0 ; 0 ; 40-13.8196]

According to the solution manual, we get the following eigenvector for the eigenvalue λ1 :

[ 0.85066 ]
[ 0.527 ]
[ 0 ]

I tried to reduce the matrix by following these steps:
[20-13.8192 ; -10 ; 0] [6.1808 ; -10 ; 0] [1 ; -1.6179 ; 0]
[-10 ;30-13.8196 ; 0] ~ [-10 ; -16.1808 ; 0] ~ [-1 ; 1.6181 ; 0]
[0 ; 0 ; 40-13.8196] [0 ; 0 ; 1] [0 ; 0 ; 0]

I don’t get anywhere after this since, as can be seen, I won't get the same eigenvectors. How did they get 0.85066 and 0.527 ?! If you can explain it for λ1 then I'll probably solve it for λ2 and λ3 as well.

Thanks!
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
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Hi,
I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. The matrix is
(I have a ; since I can't have a space between each column. Sorry):
[20 ; -10 ; 0]
[-10 ; 30 ; 0]
[0 ; 0 ; 40]

I’ve already calculated the eigenvalues, which are: λ1=13.8196, λ2=36.18 , λ3= 40
The next step is to calculate the eigenvectors for every eigenvalue λ and it’s now I keep getting the wrong answer. Let us take λ1 for instance. We'll get:

[20-13.8192 ; -10 ; 0]
[-10 ; 30-13.8196 ; 0]
[0 ; 0 ; 40-13.8196]

According to the solution manual, we get the following eigenvector for the eigenvalue λ1 :

[ 0.85066 ]
[ 0.527 ]
[ 0 ]

I tried to reduce the matrix by following these steps:
[20-13.8192 ; -10 ; 0] [6.1808 ; -10 ; 0] [1 ; -1.6179 ; 0]
[-10 ;30-13.8196 ; 0] ~ [-10 ; -16.1808 ; 0] ~ [-1 ; 1.6181 ; 0]
[0 ; 0 ; 40-13.8196] [0 ; 0 ; 1] [0 ; 0 ; 0]

I don’t get anywhere after this since, as can be seen, I won't get the same eigenvectors. How did they get 0.85066 and 0.527 ?! If you can explain it for λ1 then I'll probably solve it for λ2 and λ3 as well.

Thanks!

For nice formatting (not much more work than what you already did), use the LaTeX commands "\pmatrix}{ ...your matrix... } " or "begin{bmatrix} ...your matrix... end{bmatrix}". Put the whole thing between two $signs (to start the equation) and two more $signs to end the equation.

Individual elements are entered as "a11 & a12 & a13 \\ a21 & a22 & a23 \\ ... " so "&" is a column-separator, while "\\" is an end-of-row and/or start a new-row command.

In the "pmatrix" version the matrix elements are all placed within a pair of curly brackets "{..}". In the second option the words "begin" and "end" must be preceded by a backslash, so they read as "\begin " and "\end"; however, I avoided writing that pair so as to not confuse the typesetter for this message.

The first option gives you
$$A = \pmatrix{20 & -10 & 0 \\-10 & 30 & 0 \\ 0 & 0 & -40}$$
while the second one gives you
$$A = \begin{bmatrix} 20 & -10 & 0 \\-10 & 30 & 0 \\ 0 & 0 & -40 \end{bmatrix}$$

Just right-click on the formula and select "show math as Tex commands" to see the details.
 
Last edited:
  • #3
15,094
12,768
You have ##\lambda_1= 25 -\sqrt{125}\; , \;\lambda_2= 25 +\sqrt{125}\; , \;\lambda_3=-40##. Have you tried to solve the three equation systems ##A \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = \lambda_i \cdot \begin{bmatrix}x\\y\\z\end{bmatrix}## by hand?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Hi,
I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. The matrix is
(I have a ; since I can't have a space between each column. Sorry):
[20 ; -10 ; 0]
[-10 ; 30 ; 0]
[0 ; 0 ; 40]

I’ve already calculated the eigenvalues, which are: λ1=13.8196, λ2=36.18 , λ3= 40
The next step is to calculate the eigenvectors for every eigenvalue λ and it’s now I keep getting the wrong answer. Let us take λ1 for instance. We'll get:

[20-13.8192 ; -10 ; 0]
[-10 ; 30-13.8196 ; 0]
[0 ; 0 ; 40-13.8196]

According to the solution manual, we get the following eigenvector for the eigenvalue λ1 :

[ 0.85066 ]
[ 0.527 ]
[ 0 ]

I tried to reduce the matrix by following these steps:
[20-13.8192 ; -10 ; 0] [6.1808 ; -10 ; 0] [1 ; -1.6179 ; 0]
[-10 ;30-13.8196 ; 0] ~ [-10 ; -16.1808 ; 0] ~ [-1 ; 1.6181 ; 0]
[0 ; 0 ; 40-13.8196] [0 ; 0 ; 1] [0 ; 0 ; 0]

I don’t get anywhere after this since, as can be seen, I won't get the same eigenvectors. How did they get 0.85066 and 0.527 ?! If you can explain it for λ1 then I'll probably solve it for λ2 and λ3 as well.

Thanks!


For ##\lambda_1 = 25 - 5 \sqrt{5} \doteq 18.8196## the equations ##(\mathbf{A} - \lambda_1 \mathbf{I}) \mathbf{x} = \mathbf{0}## read as
$$
\begin{array}{rcl}
(-5 + 5 \sqrt{5}) x_1 -10 x_2 =0 \\
-10 x_1 + (5+5 \sqrt{5}) x_2 = 0\\
(15 + 5 \sqrt{5}) x_3 = 0
\end{array}
$$
The first two equations are equivalent, so keeping only the first one we get the general solution as
$$x_2 = \frac{1}{2}(\sqrt{5}-1) x_1, \: x_3 = 0,$$
where ##x_1## is ab arbitrary nonzero real number.

The ratio of ##x_2## to ##x_1## is ##x_2: x_1 = (1/2)(\sqrt{5}-1) \doteq 0.6180339880##. For the book's solution this ratio is .6180658176, which could be made closer to the true result by including more precision in the printed vales of the components. The norm of the book's as-printed eigenvector is 1.002376215, which is nearly equal to 1, so the book is choosing ##x_1## to make ##\sqrt{x_1^2 + x_2^2 + x_3^2} = 1,## which is one of the standard normalizations for eigenvectors.
 
  • #5
3
0
For ##\lambda_1 = 25 - 5 \sqrt{5} \doteq 18.8196## the equations ##(\mathbf{A} - \lambda_1 \mathbf{I}) \mathbf{x} = \mathbf{0}## read as
$$
\begin{array}{rcl}
(-5 + 5 \sqrt{5}) x_1 -10 x_2 =0 \\
-10 x_1 + (5+5 \sqrt{5}) x_2 = 0\\
(15 + 5 \sqrt{5}) x_3 = 0
\end{array}
$$
The first two equations are equivalent, so keeping only the first one we get the general solution as
$$x_2 = \frac{1}{2}(\sqrt{5}-1) x_1, \: x_3 = 0,$$
where ##x_1## is ab arbitrary nonzero real number.

The ratio of ##x_2## to ##x_1## is ##x_2: x_1 = (1/2)(\sqrt{5}-1) \doteq 0.6180339880##. For the book's solution this ratio is .6180658176, which could be made closer to the true result by including more precision in the printed vales of the components. The norm of the book's as-printed eigenvector is 1.002376215, which is nearly equal to 1, so the book is choosing ##x_1## to make ##\sqrt{x_1^2 + x_2^2 + x_3^2} = 1,## which is one of the standard normalizations for eigenvectors.

Thanks for explaining!
 

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