# Repeated Eigenvalues

by hex.halo
Tags: eigenvalues, repeated
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Since 1 is an eigenvaue, the equation $$\left[\begin{array}{ccc}2 & 2 & 1\\1 & 3 & 1\\ 1 & 2 & 2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\right]= \left[\begin{array}{c}x \\ y \\ z\end{array}\right]$$ 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.)
 P: 587 I call your matrix B. Then I presume you solved $\det(B-\lambda\mathbb{I}_{3\times 3})=0$ to find the eigenvalues $\lambda$. You found $\lambda_1=5 \text{ and } \lambda_2=1$. Once you have done that I further assume you tried to find the eigenvector $v_i$ associated to the eigenvector $\lambda_i$ by solving $(B-\lambda_i\mathbb{I}_{3\times 3})v_i=0$. For your eigenvalue one you will find that this equation leads to a two-dimensional solution space for $v_2$, the so called eigenspace. ANY vector in this two-dimensional space will be an eigenvector with eigenvalue one.