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Help with Eigenvectors

  • Thread starter hex.halo
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1. Homework Statement

Find the eigenvalues and correspointing eigenvectors of the matrix:

[1,1;1,1]

2. Homework Equations



3. The Attempt at a Solution

I can solve the determinant to get the eigenvalues: e1=2, e2=0, and from here I am supposed to sub these values back into my matrix and find the eigenvectors, right? I'm not sure how to perform this process though...
 

Answers and Replies

tiny-tim
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Find the eigenvalues and correspointing eigenvectors of the matrix [1,1;1,1]

I can solve the determinant to get the eigenvalues: e1=2, e2=0.
Hi hex.halo! :smile:

The standard way (which you must be able to do): just write [1,1;1,1](a,b) = (0,0), and solve; and the same for [1,1;1,1](a,b) = (2a,2b). :smile:

(And the shortcut in this case: the eigenvectors for M + a.I are the same as for M, for any matrix M and any number a … so just look for the eigenvectors of [0,1;1,0], which you can probably guess! :smile:)
 
HallsofIvy
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Some people (on this board) use a "complicated method" of putting the eigenvalues into the matrix [itex]A-\lambda I[/itex] and row reducing but I much prefer tiny-tims method: any eigenvector, by definition, must satisfy [itex]Av= \lambda v[/itex]. Set up that equation and solve for v.
For example, with eigenvalue 2, you must solve
[tex]\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y\end{array}\right]= \left[\begin{array}{c} 2x \\ 2y\end{array}\right][/itex]
You will NOT be able to solve for both x and y, only for y, say, as a function of x because any multiple of an eigenvector is also an eigenvector. Choose any convenient value of x, calculate the corresponding y, and any eigenvector is a multiple of that.
 
tiny-tim
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¿ "board" ?​

oh … do you mean "planck" … ? :biggrin:

(:confused: I take two!)
 
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So, you're telling me that I need to solve y in terms of x from that (very nice) matrix set up, then I can sub in ANY number for x that I want to get my eigenvector and have the right answer? Wouldn't this result in an infinite number of correct eigenvectors? Actually, don't worry about that, I imagine i'll have that explained to me at some point. What I'd really like to know is, for any question, am I right to sub in 1 for my x value then just put a scalar multiple out the front of the matrix to represent all the multiples and that would be correct?
 
HallsofIvy
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Yes, of course! If v is an eigenvector corresponding to eigenvalue [itex]\lambda[/itex], then so is any number times v. The set of all eigenvectors corresponding to a given eigenvalue forms a subspace! That always has an infinite number of vectors.

Yes, if you choose any one of the vectors in the (one-dimensional) subspace, any other can be written as a multiple of it. You can choose x (or y and then solve for x) to be any number you please. "1" seems like a simple choice!

It is possible that the set of eigenvectors corresponding to an eigenvalue is a subspace of dimension larger than 1. For a 2 by 2 that would of course be all of R2. In that case that the matrix is larger than 2 by 2, with a single eigenvalue having "eigenspace" of dimension greater than 1, solving the equation [itex]Ax= \lambda x[/itex] would give an equation where other variables depend on two (or more) of the variables. In that case, taking them to be 1 and 0, then 0 and 1 should give you independent basis vectors for the subspace.
 

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