Repeated Eigenvalues

hex.halo

1. The problem statement, all variables and given/known data

Obtain the eigenvalues and corrosponding eigenvectors for the matrix: [2,2,1;1,3,1;1,2,2]

2. Relevant equations



3. The attempt at a solution

I can solve for the eigenvalues, 5, 1, and 1

I can solve the eigenvalue 5 for the eigenvector B[1;1;1]

Yet somehow, the 1 eigenvalues are supposed to give me two different eigenvectors. Unfortunately, this makes no sense to me, which is why i'm here...

Can someone please explain to me how to finish solving this question?

HallsofIvy

Since 1 is an eigenvaue, the equation
[tex]\left[\begin{array}{ccc}2 & 2 & 1\\1 & 3 & 1\\ 1 & 2 & 2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\right]= \left[\begin{array}{c}x \\ y \\ z\end{array}\right][/tex]
must have a non-trivial solution. Multiplying that out and looking at individual rows, we must have 2x+ 2y+ z= x, x+ 3y+ z= y, x+ 2y+ 2z= z. Because 1 is an eigenvalue, those are not independent and must have an infinite number of solutions. In fact, in this particular problem, the three equations those reduce to, x+ 2y+ z= 0, x+ 2y+ z= 0, and x+ 2y+ z= 0, are all the same equation. You can solve for z as a function of x and y: z= -x- 2y. That means you can choose any numbers you like for x and y, then solve for z to get an eigenvalue. Can you choose them so that you get two independent vectors? (I always like to use simple numbers like 0 and 1.)

Pere Callahan

I call your matrix B. Then I presume you solved [itex]\det(B-\lambda\mathbb{I}_{3\times 3})=0[/itex] to find the eigenvalues [itex]\lambda[/itex]. You found [itex]\lambda_1=5 \text{ and } \lambda_2=1[/itex].

Once you have done that I further assume you tried to find the eigenvector [itex]v_i[/itex] associated to the eigenvector [itex]\lambda_i[/itex] by solving [itex](B-\lambda_i\mathbb{I}_{3\times 3})v_i=0[/itex]. For your eigenvalue one you will find that this equation leads to a two-dimensional solution space for [itex]v_2[/itex], the so called eigenspace. ANY vector in this two-dimensional space will be an eigenvector with eigenvalue one.