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Finding a vector associated with an eigenvalue

  • Thread starter Zem
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  • #1
Zem
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Find a general solution of the given system using the method (A - [tex]\lambda[/tex]I)V2 = V1.

[tex]x'_1 = 2x_1 - 5x_2, x'_2 = 4x_1 - 2x_2[/tex]
[tex]x' =
\left(\begin{array}{cc}2&-5\\4&-2\end{array}\right)[/tex]

characteristic equation:
(2 - [tex]\lambda[/tex])((-2) - [tex]\lambda[/tex]) + 20 = 0
[tex]\lambda[/tex]^2 + 16 = 0
[tex]\lambda[/tex] = 4i

Using this method:
(A - 4i[tex]\lambda[/tex])V_2 = V_1
[tex]\left(\begin{array}{cc}2-4i&-5\\4&-2-4i\end{array}\right) *[/tex] [tex]\left(\begin{array}{cc}a\\b\end{array}\right) = [/tex] [tex]\left(\begin{array}{cc}0\\0\end{array}\right) [/tex]

(2 - 4i)a - 5b = 0
4a - (-2 - 4i)b = 0

When there are no complex roots, I can set a or b = 1 to find the value of the other. And when I row reduce this in my ti89, I get

[tex]\left(\begin{array}{cc}1&i 0\\0&0 0\end{array}\right) [/tex] with a space between the i and 0 in the top row, and the two 0's in the 2nd row.

How do I find a and b in this problem? When I find their values I will have V_1.
 
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Answers and Replies

  • #2
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What is your vector space [itex]V[/itex], and what field [itex]F[/itex] is it over? For the moment I'm going to assume [itex]V = \mathbb{C}^2[/itex] and [itex]F= \mathbb{C}[/itex].

Also, it is not stated but you must mean that [itex]x_1', x_2'[/itex] are the coordinates of the image of [itex](x_1,x_2)[/itex] under the linear operator [itex]T[/itex] defined by the equations you have for [itex]x_1', x_2'[/itex]. ie:

[tex] T(x_1,x_2) = (x_1', x_2') [/tex]

(By the way, have you verified that [itex]T[/itex] is in fact linear?)

Zem said:
Using this method:
(A - 4i[tex]\lambda[/tex])V_2 = V_1
[tex]\left(\begin{array}{cc}2-4i&-5\\4&-2-4i\end{array}\right) *[/tex] [tex]\left(\begin{array}{cc}a\\b\end{array}\right) = [/tex] [tex]\left(\begin{array}{cc}0\\0\end{array}\right) [/tex]

(2 - 4i)a - 5b = 0
4a - (-2 - 4i)b = 0
There is a sign error here, this should be:

(2 - 4i)a - 5b = 0
4a + (-2 - 4i)b = 0

When there are no complex roots, I can set a or b = 1 to find the value of the other. And when I row reduce this in my ti89, I get

How do I find a and b in this problem? When I find their values I will have V_1.
You don't need a calculator to do this. Simply choose one of the above two equations and solve for one of the variables in terms of the other. As a check, plug that value into the other equation (you should get an identity). You'll end up with an equation for a in terms of b (or vice versa). Choose a value for a or b that makes (a, b) non-zero, and voila, you have an eigenvector for your linear operator corresponding to the eigenvalue [itex]\lambda = 4i[/itex].

edit: Re-reading your question, it asks for a *general* solution, so you can leave your answer in terms of a or b, just be sure to include the proper restrictions on a or b so that your answer gives only the non-zero solutions.
 
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  • #3
Zem
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Originally Posted by nocturnal
You don't need a calculator to do this. Simply choose one of the above two equations and solve for one of the variables in terms of the other. As a check, plug that value into the other equation (you should get an identity). You'll end up with an equation for a in terms of b (or vice versa). Choose a value for a or b that makes (a, b) non-zero, and voila, you have an eigenvector for your linear operator corresponding to the eigenvalue .
My problem was that I was stuck on assigning b = 1, but it appears much better to make b = a's coefficient.
When I set [tex]b = (2-4i), a = 5[/tex], so
[tex]V_1 = \left(\begin{array}{cc}5\\2-4i\end{array}\right)[/tex]
:cool:
 
  • #4
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Did you see the edit at the bottom of my post? The problem states that its looking for the *general* solution, so you wanto to find the set of all such vectors that satisfy this equation.
 
  • #5
Zem
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Yes, I was just stuck in the middle of the problem. Now all I have to do is put the real and complex parts of the solutions together to get the general solutions of x_1 and x_2. Piece of cake.

I am curious, though, what you were saying about [tex]T[/tex] and [tex]F[/tex]. Is that stuff for linear algebra? We did not cover anything regarding vectors over fields or [tex]T[/tex]. Perhaps I should have mentioned that this was an ODE homework problem.
Thanks!
 
  • #6
113
1
Zem said:
I am curious, though, what you were saying about [tex]T[/tex] and [tex]F[/tex]. Is that stuff for linear algebra? We did not cover anything regarding vectors over fields or [tex]T[/tex]. Perhaps I should have mentioned that this was an ODE homework problem.
Thanks!
Yes, those are terms from Linear Algebra. Essentially a vector space is a set of "vectors" satisfying certain properties. The vectors can be acted on by "scalars" (elements of a field - another set satisfying certain properties). A Linear Operator is a function satisfying certain properties that maps a vector space into itself. It is in the context of Linear Operators that we talk about eigenvectors and eigenvalues. If you take Linear Algebra they will go into great detail about these concepts.
 
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