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Eigenvectors/Eigenvalues watch

1. Hello, the question I have is to find the eigenvalues and corresponding eigenvectors for the matrix I

The (correct) eigenvalues I get are 4, 4 and 8.

For the eigenvalues 4 and 4, the corresponding eigenvector I get is (1,-sqrt{3}, 1)

and for the eigenvalue 8 I get (0, 1/sqrt{3}, 1).

The answer (according to wolframalpha) says that there are actually 3 eigenvectors which are
(1,0,0)
(0,-sqrt{3},1)
(0,1/sqrt{3},1).

Can someone explain to me what is going on here? How can there be two different eigenvectors for the same eigenvalue?

EDIT: Apologies, I got a bit lazy with the LaTeX
2. (Original post by KeyFingot)
Hello, the question I have is to find the eigenvalues and corresponding eigenvectors for the matrix I

The (correct) eigenvalues I get are 4, 4 and 8.

For the eigenvalues 4 and 4, the corresponding eigenvector I get is (1,-sqrt{3}, 1)

and for the eigenvalue 8 I get (0, 1/sqrt{3}, 1).

The answer (according to wolframalpha) says that there are actually 3 eigenvectors which are
(1,0,0)
(0,-sqrt{3},1)
(0,1/sqrt{3},1).

Can someone explain to me what is going on here? How can there be two different eigenvectors for the same eigenvalue?

EDIT: Apologies, I got a bit lazy with the LaTeX
when you work through the process, you find that there are an infinite number of eigenvector.
So lets say you found you eigenvectors using the charastic equation det(M-Ix)
x=eigen value

To find the eigenvector for the eigenvalue
You wanna solve (M-Ix)v=0 where v is the eigenvector. This is obtain from Mv=xv
Mv-Ixv=0
v(M-Ix)=0
I is the identity Matrix. its needed since you cannot do matrix-vector

so if you solve (M-Ix)v=0 expanding you see that you get a system of linear equations.
3 equations with 3 variables. which represent 3 planes. For 3 planes,they all either intersect at one point (unique solution), intersect along a line (infinite solution) or form a triangular prism (no solutions)

(M-Ix) is not invertible, so there is no unique solutions, by definition. If it were, the only solution would the 0 vector, which is trivial.
So if solutions exist it means, the 3 planes intersection along a line. Each point on the line is an eigenvector. hence infinitely many eigenvectors
3. (Original post by KeyFingot)
Can someone explain to me what is going on here? How can there be two different eigenvectors for the same eigenvalue?
You're thinking about it the opposite way round to me If you think of an enlargement of scale factor 2 in the x, y and z directions - so A = {{2,0,0},{0,2,0},{0,0,2}} - you see there is one eigenvalue (2) with three different eigenvectors (the three which make up the standard basis). The eigenvalue is simply measuring how much "stretch" there is along the eigenvector's direction; if there are two eigenvectors with the same "stretch" along them, then the two eigenvalues will be the same.
So the way I think about it is that there are two eigenvectors which happen to have the same eigenvalue, rather than my impression of your view which was "an eigenvalue with two corresponding eigenvectors". (In different contexts, it might be helpful to think of it that way round; eigenstuff is very versatile so it crops up everywhere!)

ETA: That is, the eigenvectors are {1,0,0} with eigenvalue 4, {0,-sqrt3, 1} with eigenvalue 4, and {0,1/sqrt(3),1} with eigenvalue 8, assuming I haven't mistranscribed something.
4. (Original post by Smaug123)
You're thinking about it the opposite way round to me If you think of an enlargement of scale factor 2 in the x, y and z directions - so A = {{2,0,0},{0,2,0},{0,0,2}} - you see there is one eigenvalue (2) with three different eigenvectors (the three which make up the standard basis). The eigenvalue is simply measuring how much "stretch" there is along the eigenvector's direction; if there are two eigenvectors with the same "stretch" along them, then the two eigenvalues will be the same.
So the way I think about it is that there are two eigenvectors which happen to have the same eigenvalue, rather than my impression of your view which was "an eigenvalue with two corresponding eigenvectors". (In different contexts, it might be helpful to think of it that way round; eigenstuff is very versatile so it crops up everywhere!)

ETA: That is, the eigenvectors are {1,0,0} with eigenvalue 4, {0,-sqrt3, 1} with eigenvalue 4, and {0,1/sqrt(3),1} with eigenvalue 8, assuming I haven't mistranscribed something.
Thanks for the explanation! How do I go about finding the 3rd eigenvector then?
5. A single eigenvalue can have more than a 1-dimensional space of eigenvectors.

When working out eigenvectors for an eigenvalue you can end up with a system of equations which essentially boil down to (for example)

and this actually has a 3 dimensional space of eigenvectors. Why, because well you could set , or as an eigenvector for that eigenvalue, and all three of these vectors are linearly independent.
6. (Original post by KeyFingot)
How can there be two different eigenvectors for the same eigenvalue?
Usually, this doesn't happen. However, if an eigenvalue appears more than once when you're solving the characteristic equation (called degeneracy if you like big fancy words), sometimes it will have more than one corresponding eigenvector.

For the eigenvalues 4 and 4, the corresponding eigenvector I get is (1,-sqrt{3}, 1)

and for the eigenvalue 8 I get (0, 1/sqrt{3}, 1).

The answer (according to wolframalpha) says that there are actually 3 eigenvectors which are
(1,0,0)
(0,-sqrt{3},1)
(0,1/sqrt{3},1).
To help understand what's going on here, try adding together the first two eigenvectors Wolfram Alpha gives you - you'll get the eigenvector you found. In fact, any linear combination of the first two Wolfram results will also be an eigenvector corresponding to eigenvalue 4.

There isn't actually a unique solution, any two linearly independent eigenvectors with eigenvalue 4 will work. Ignore the Wolfram results for now, and find a second eigenvector to go with (1,-sqrt{3}, 1).
7. (Original post by tommm)
Usually, this doesn't happen. However, if an eigenvalue appears more than once when you're solving the characteristic equation (called degeneracy if you like big fancy words), sometimes it will have more than one corresponding eigenvector.

To help understand what's going on here, try adding together the first two eigenvectors Wolfram Alpha gives you - you'll get the eigenvector you found. In fact, any linear combination of the first two Wolfram results will also be an eigenvector corresponding to eigenvalue 4.

There isn't actually a unique solution, any two linearly independent eigenvectors with eigenvalue 4 will work. Ignore the Wolfram results for now, and find a second eigenvector to go with (1,-sqrt{3}, 1).
I can't ignore the third eigenvector though, I'm told in the question that I need to find the three eigenvectors ...
8. (Original post by KeyFingot)
I can't ignore the third eigenvector though, I'm told in the question that I need to find the three eigenvectors ...
I just meant to ignore what Wolfram Alpha had given you and find your own third eigenvector, sorry if I didn't make that clear! When you're finding the eigenvector with eigenvalue 4, you should find two possibilities.
9. It helps if you try to solve it like your doing an AX=0 system to get the null space.

You will ultimately get the solutions in parametric form which makes it very easy to spot the eigenvectors out.
10. (Original post by tommm)
I just meant to ignore what Wolfram Alpha had given you and find your own third eigenvector, sorry if I didn't make that clear! When you're finding the eigenvector with eigenvalue 4, you should find two possibilities.
Oh right I see! Yeah I found that from the repeated eigenvalue (4) that there are two ways to satisfy the equations. I had two sets of simultaneous equations; one equation only x dependent and the other only y&z dependent, so I was able to get the two eigenvectors after thinking more carefully about the solutions. Thanks!

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