Please check my Eigenvector solutions.


by hadizainud
Tags: check, eigenvector, solutions
hadizainud
hadizainud is offline
#1
Aug12-11, 06:24 AM
P: 15
1. The problem statement, all variables and given/known data
Find the characteristic equations, eigenvalues and eigenvector of the following matrix

2. Relevant equations
Name:  Eigen 2.png
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3. The attempt at a solution
Click image for larger version

Name:	Eigen 1.png
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ID:	37947

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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HallsofIvy
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#2
Aug12-11, 07:15 AM
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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].
hadizainud
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#3
Aug12-11, 03:48 PM
P: 15
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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"?

stringy
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#4
Aug12-11, 07:05 PM
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Please check my Eigenvector solutions.


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.

Quote Quote by hadizainud View Post

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.
HallsofIvy
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#5
Aug12-11, 08:48 PM
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Quote Quote by hadizainud View Post
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".
hadizainud
hadizainud is offline
#6
Aug13-11, 03:24 AM
P: 15
Thanks Stringy and HallsofIvy :)


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