Please check my Eigenvector solutions.

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In summary, the conversation discusses finding the characteristic equations, eigenvalues, and eigenvectors of a given matrix. The initial solution provided by the speaker was incorrect, as 1 is not an eigenvalue and the characteristic equation was incorrect. After correction, the speaker also mentions that x_2 should equal t and x_1 equals 0, leading to further discussion about the correct values for lambda and the mistake in stating "let x_1=t".
  • #1
hadizainud
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Homework Statement


Find the characteristic equations, eigenvalues and eigenvector of the following matrix

Homework Equations


Eigen 2.png

The Attempt at a Solution


Eigen 1.png


Somehow somewhere I think the solution is wrong, based on online Eigenvector calculator on the web. Please do provide me actual answers and solutions. Thanks in advance!
 
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  • #2
Your first eigenvector corresponding to [itex]\lambda= 3[/itex] is correct.

However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote [itex](A- \lambda I)x= 0[/itex]:
[tex]\begin{bmatrix}2 & 0 \\ 8 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex]
which gives
[tex]\begin{bmatrix}2x_1 \\ 8x_1- 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
which then requires that [itex]2x_1= 0[/itex] and [itex]8x_1- 2x_2= 0[/itex].
From the second [itex]x_2= 4x_1[/itex] but the first says [itex]x_1= 0[/itex] so [itex]x_2= 4(0)= 0[/itex]. There is NO non-trivial vector satisfying this. 1 is NOT an eigenvalue
(I have no idea where you got "[itex]x_1+ x_2= 0[/itex]".)

The characteristic equation is given by
[tex]\left|\begin{array}{cc}3- \lambda & 0 \\ 8 & -1- \lambda\end{array}\right|= 0[/tex]
Which is, of course, simply [itex](3-\lambda)(-1- \lambda)= 0[/itex].
 
  • #3
Eigen 3.png

Take a look at this. I've corrected it. Please let this answer correct :)

Some mistake there;
Eigenvector for lambda = 1 is [0;1]

One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?
 
Last edited:
  • #4
If you compute what you think are an eigenvector and eigenvalue pair, stick them back in! They had better satisfy [itex] Ax = \lambda x[/itex] since that is, after all, the equation whose solutions you were looking for in the first place.

hadizainud said:
One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?

The variable [itex]x_2[/itex] should equal t, if that's what you're asking.
 
  • #5
hadizainud said:
View attachment 37957
Take a look at this. I've corrected it. Please let this answer correct :)

Some mistake there;
Eigenvector for lambda = 1 is [0;1]
You mean "for lambda= -1".

One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?
You had just shown that [itex]x_1= 0[/itex] so you can't say "let x_1=t".
 
  • #6
Thanks Stringy and HallsofIvy :)
 

1. How do I check my Eigenvector solutions?

To check your Eigenvector solutions, you can use the Eigenvector equation: Av = λv. Simply plug in your matrix A and the corresponding eigenvector v, and check if it satisfies the equation. If it does, then you have found a valid eigenvector solution.

2. What if my Eigenvector solutions do not satisfy the Eigenvector equation?

If your Eigenvector solutions do not satisfy the Eigenvector equation, then they are not valid. This could mean that there is an error in your calculations or that the matrix A does not have any eigenvectors.

3. Can I have multiple Eigenvector solutions for the same matrix?

Yes, it is possible to have multiple Eigenvector solutions for the same matrix. In fact, a matrix can have multiple eigenvectors with different eigenvalues.

4. How do I know if my Eigenvector solutions are correct?

One way to check if your Eigenvector solutions are correct is to verify that they satisfy the Eigenvector equation. Another way is to use a calculator or software to compute the eigenvalues and eigenvectors of the matrix and compare them to your solutions.

5. Can Eigenvector solutions be complex numbers?

Yes, Eigenvector solutions can be complex numbers. This is especially true for matrices with complex entries. In this case, the eigenvalues and eigenvectors will also be complex.

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