# Linear Algebra - Eigenvectors

1. Aug 1, 2010

### ganondorf29

1. The problem statement, all variables and given/known data
A = [2,1,2;2,1,2;2,1,2]

Find the Eigenvectors of A

2. Relevant equations

3. The attempt at a solution

First I found the eigenvalues of A
$$det(A - \lambda I) = 0$$

$$\lambda = 0,5$$

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A-5I

[-3,1,2;2,-4,2;2,1,-3] --(rref)--> [1,0,-1;0,1,1;0,0,0]

x1 = x3
x2 = -x3

v1 = [1;-1;1]

Is this the right eigenvector for A-5I?
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A-0I

[2,1,2;2,1,2;2,1,2] --(rref)--> [1,1/2,1;0,0,0;0,0,0]

x1 + x2/2 + x3 = 0

I'm not sure what to do in order to find the eigenvector for A-0I

2. Aug 1, 2010

### lanedance

the characteristic equation gives you a bit more help than that
$$A = \begin{pmatrix} 2 & 1 & 2 \\ 2 & 1 & 2 \\ 2 & 1 & 2 \end{pmatrix}$$

$$A - \lambda I = \begin{pmatrix} 2- \lambda & 1 & 2 \\ 2 & 1- \lambda & 2 \\ 2 & 1 & 2- \lambda \end{pmatrix}$$

taking the determinant simplifies to
$$\lambda^2(5 -\lambda)$$

giving eigenvalues:
5 with algebraic multiplicity of 1
0 with algebraic multiplicity of 2

the goemtric multiplicity (number of linearly independent eignevectors) is always less than or equal to the algebraic multiplicity

this means there will be one eignevector corresponding to [itex] \lambda = 5 [\itex] and upto 2 corresponding to [itex] \lambda = 0 [\itex]

Last edited: Aug 1, 2010
3. Aug 1, 2010

### lanedance

to check if its right try multiplying the vector by the matrix

$$A.v_1 = \begin{pmatrix} 2 & 1 & 2 \\2 & 1 & 2 \\2 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \\-1 \\1 \end{pmatrix} = \begin{pmatrix}3 \\3 \\3 \end{pmatrix} \neq 5 v_1$$

so this is not correct

4. Aug 1, 2010

### ganondorf29

For A-5I
x1 = x3
x2 = x3

so [1;1;1] is the eigenvector for $$\lambda = 5$$

[2,1,2;2,1,2;2,1,2] * [1;1;1] = [5;5;5]

But I'm still not sure how to find the eigenvectors for A-0I

5. Aug 1, 2010

### vela

Staff Emeritus
You have x1 + x2/2 + x3 = 0. You can solve for any one of the variables in terms of the other two, so you have two degrees of freedom. So let x2=s and x3=t. Express the solutions in the form (x1, x2, x3) = s(some vector) + t(some vector). Those two vectors will be the eigenvectors.

6. Aug 1, 2010

### lanedance

ollowing on from Vela
x1 + x2/2 + x3 = 0
is the equaton of a plane through the origin, which can be spanned by any two linearly independent vectors lying in that plane, solving for x2 in term so of x1 & x3, gives
x2 = -2(x1 + x3)

note vela suggested solving for x1 first, which is fine, but i like the symmetry soving for x2

then let x1=s, x3=t and follow on, note this is equivalent to solving the 2 cases below
a) x1=1, x3=0
b) x1=0, x3=1
whih may be easier to see first time round