Calculating eigenvalues

[boneill3]

HI all,

I'm having a bit of trouble calculating the eigenvector of the following matrix.

3 -2

1 -1

I've found the the eigen values of 2.4 and -0.41 these are rounded off.

using y^2 - 2y -1

I have two questions
1 : Can I use aprroximate eigenvalues to compute the eigen Vector?

and two: If so I'm trying to calulate it by the following say for eigenvalue 2.4


3a -2b = 2.4a
a -b = 2.4b

Im getting the vector of

-2.4
-4.8

I'm not sure how to check if this is right or not

Kindest regards
Brendan

[NickThissen]

The eigenvalues you've found are correct, although I think you should not round them off.

Using the quadratic formula you can easily find the eigenvalues to be 1+sqrt(2) and 1-sqrt(2).

I think you can still find the approximate eigenvectors if you use approximate eigenvalues, but in a simple case like this I would always recommend you to use the exact values. If this is a homework problem then I am pretty sure the answers are required to be exact.


On the the eigenvectors, I don't know how you tried to find the eigenvector, I have never seen that way before. It might be alright but I simply don't know (linear algebra is not my best subject hehe).

I have always been taught to find the eigenvalues by substituting the eigenvalues in the matrix (yI - A).

For example, suppose you have your matrix A:

3  -2
1  -1

Then (yI - A) is:

y-3  2
-1  y+1

(This is ofcourse the matrix you use to find the eigenvalues y in the first place).
You then find the characteristic polynomial and find both eigenvalues, 1+sqrt(2) and 1-sqrt(2).

You then substitute them for y, for example let's try y = 1-sqrt(2):

-2-sqrt(2)      2
  -1        2-sqrt(2)

Putting this in reduced row echelon form we get:

  1      -2/[2+sqrt(2)]
  0            0

From this we can see that the eigenvector [x1, x2] must be:
x1 - 2/[2+sqrt(2)] x2 = 0
x1 = 2/[2+sqrt(2)] x2

Putting x2 = some variable t, we find the eigenvector to be:

[  2/[2+sqrt(2)]  ,    1  ]

or equivalently (divide by (2+sqrt(2)) and by 2):

[   1     ,    1 + 1/2 sqrt(2)  ]

In the same way we find the other eigenvalue (corresponding to y = 1 + sqrt(2)) to be:

[   1    ,   1 - 1/2 sqrt(2)  ]

Lastly, I don't think your eigenvector is correct. It should be a multiple of

[   1    ,   1 - 1/2 sqrt(2)  ]

but it is not, is it?

[jemidiah]

First your questions:
1: yes, though only approximately. As long as you use close approximations of the eigenvalues, because the system is linear (and therefore not chaotic) the eigenvectors you get will be a similarly close approximation.

2: you check if an eigenvector is "right" or not by going back to the definition of an eigenvector. That is, given your matrix A, an eigenvalue y, and an associated eigenvector v, you check if A*v = y*v. If you've used approximations, then depending on exactly what you approximated you should check if that equality approximately holds (it may hold exactly, it depends on what you do). Note that the equation Av=yv I've written is identical to the system of equations you've written (with v=<a b>), I've just used matrix notation.


The method the you used to calculate eigenvectors is technically correct, though I think you forgot to multiply the 2.4's by a and b [both equations are true only if you ignore the a and b on the right hand sides]. That method degenerates into Nick's version when you try to solve the system of equations--he just wrote it in matrix form.

I find it very helpful to think of eigenvalues as the values of y for which the matrix (A-yI) is not full-rank. You then have several ways of finding the necessary values of y: the traditional way, using the characteristic polynomial, or using intuition to tweak y into a value that yields linearly dependent rows (or equivalently columns).

For example, consider the matrix

1 1
1 1

Then A-yI is

1-y 1
1   1-y

Using the traditional method, the characteristic polynomial is (1-y)(1-y)-1 = 1-2y+y²-1 = y²-2y = y(y-2). This has roots with y=0,2.

Using the non-full-rank method, we try to find two values of y for which A-yI is not full-rank. Clearly the original matrix, A-0I, is not full-rank, so y=0 must be an eigenvalue. Next, the matrix

-1  1
 1 -1

clearly has linearly dependent rows. This matrix is given by y=2 [that is, the matrix is A-2I], so y=2 is the other eigenvalue. This method can be significantly faster than the brute force method of the characteristic polynomial, but isn't nearly as reliable. However I enjoy using it because it makes a sort of game out of (contrived) eigenvalue problems that have nice small integer eigenvalues.


As for finding an eigenvector given an eigenvalue, Nick's method should work great, though again I enjoy a more intuitionistic method. By definition of an eigenvector v given an eigenvalue y and a matrix A, we know Av = vy(I), so Av-vIy = 0, or (A-Iy)v = 0 [the zero vector]. That is, v is by definition in the null space [aka kernel] of A-Iy, and v is not the zero vector.

(Aside) Note that the existence such a vector v which gets mapped to 0 is enough to make A-Iy non-invertible, which is enough to ensure that the rows (and columns) of A-Iy cannot all be linearly independent, which means that A-Iy is not full-rank, which is the fact that we used above to find y in the first place. If no eigenvalue exists for a particular A-Ix matrix, then no non-zero vector gets mapped to the 0 vector by the A-Ix matrix, so that A-Ix is invertible, and the determinant of A-Ix is non-zero. This is the fact we use to say that the roots of the characteristic polynomial are the only values which have any associated eigenvectors.

Back to what I was saying, since v is in the kernel of A-Iy, and any non-eigenvector is not in the kernel of A-Iy [since then Au != Iyu for a non-eigenvector u], non-zero vectors v are eigenvectors of A-Iy if and only if v is in the kernel of A-Iy. That is, a vector v != 0 is an eigenvector if and only if A-yI maps v to the zero vector. Using my above matrix, we need to find eigenvectors A = the 2x2 matrix with all 1's using A-0I and A-2I.

The matrix A-0I,

1 1
1 1

maps <1, -1> to 0, since 1*1+1*-1 = 0. Thus <1, -1> is an eigenvector of A, and (0, <1, -1>) is an eigenpair [an ordered pair of an eigenvalue and an associated eigenvector].

The matrix A-2I,

-1  1
 1 -1

maps <1, 1> to 0, since -1*1+1*1 = 0 and 1*1+-1*1 = 0. Thus <1, 1> is an eigenvector of A and (2, <1, 1>) is an eigenpair.

This intuitive method was much, much, much faster than the brute force method of solving a system of equations in the manner Nick did in his previous post, for this particular problem. Again, however, this method only works well for contrived problems which have simple, generally integer- or half integer-valued coefficients. I again find it more fun to break the monotony of silly eigenvector/value problems, though if you're looking for a mechanistic, computational method, the one Nick described is definitely the way to go.

[David_B]

I put the matrix into an online calculator:

Eigenvalue/Eigenvector Calculator

Eigenvalue 1: 2.414213562373095
Corresponding Eigenvector: (0.9596829822606673, 0.2810846377148203)

Eigenvalue 2: -0.41421356237309503
Corresponding Eigenvector: (0.5054494651244236, 0.8628562094610168)

[NickThissen]

If you look closely you can see that is the same as I got.

The eigenvalues are just the values of 1 +/- 1/2 sqrt(2).

The eigenvectors are the same, but for some reason (probably the way they are calculated with a computer, I don't know how) they are multiplied by some factor.

If you look at the last eigenvector for example, if you divide 1 by 0.5054494651244236 you get approximately:
1.97842...

If you divide 1 + 1/2 sqrt(2) by 0.8628562094610168 this also yields:
1.97842...

This means that if you multiply this eigenvector by 1.97842 (which leaves it a valid eigenvector) you get the same eigenvector I had.