# Eigenvector of matrix of equal numbers

For matrix the matrix

$$A = \begin{bmatrix} 3&1&1\\ 1&3&1\\ 1&1&3\\ \end{bmatrix}$$

with eigenvalues $\lambda_1=5$, $\lambda_2=2$, $\lambda_3=2$, I am trying to find the corresponding eigenvector corresponding to the eigenvalue 2. I got

$$(A - 2I_3) = \begin{bmatrix} 1&1&1\\ 1&1&1\\ 1&1&1\\ \end{bmatrix}$$

Reducing it (row reduced echelon form), I get:

$$\left[ \begin{array} {ccc|c} 1&1&1&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{array}\right]$$

Ending up with $x_1 + x_2 + x_3 = 0$. How would I find the eigenvector from there? Usually, I end up getting two equations and it's easy from there. How would you do it with one?

• For instance, $x_2$ and $x_3$ are free, but $x_1=-x_2-x_3$. This is two-dimensional. A basis is $(1,-1,0)$ and $(1,0,-1)$. – Julien Apr 28 '13 at 23:30
• So to make it in terms of $x_1$, you set $x_1$ to 1 and set once $x_2$ to zero, and in the other vector, $x_3$ to zero (the other would be -1 as the equation $x_1 = -x_2-x_3$ shows)? – darksky Apr 28 '13 at 23:33
• Not really. Say $x_2=s$ and $x_3=t$ are free. Then $x_1=-x_2-x_3=-s-t$. So $x=(x_1,x_2,x_3)=(-s-t,s,t)=s(-1,1,0)+t(-1,0,1)$. This proves that the solution set is spanned by the latter two vectors. As they are independent, they constitute a basis. Of the eigenspace. – Julien Apr 28 '13 at 23:37
• I still don't get how you got (-1, 1, 0) and (-1, 0, -1). I know it's the right answer, but I how do you get it from x1+x2+x3 = 0? – darksky Apr 29 '13 at 1:30

For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal) eigenvectors for the $n$ by $n$ case can be read from the columns of $$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$ In your case, the upper left 3 by 3 corner. If you want the result orthonormal you need to divide each column by a square root of something appropriate. I have displayed the 10 by 10 version, notice how the diagonal numbers go up to 9 = 10 - 1.
You have a two-dimensional eigenspace, so there should be two linearly independent eigenvectors, that each satisfy $x_1+x_2+x_3=0$ for you to choose.
More details: Every eigenvalue has an associated eigenspace. The dimension of this eigenspace is called the geometric multiplicity of the eigenvalue, while the multiplicity of the eigenvalue as a root of the characteristic polynomial is called the algebraic multiplicity of the eigenvalue. These two constants satisfy $1\le GM(\lambda)\le AM(\lambda)$. For the problem at hand, $\lambda=2$, and $GM(\lambda)=AM(\lambda)=2$. The eigenspace corresponding to $\lambda=2$ is a two-dimensional vector space. Every vector in it satisfies $x_1+x_2+x_3=0$. What you're trying to find is a basis for this vector space. @Julien gave an example in the comments above of such a basis, but it's not unique. Any two linearly independent vectors that satisfy the equation will serve as a basis. For example, $(3,-1,-2)$ and $(5,-10,5)$.
• I understand all of this. I still don't get how you'd get to (-1, 1, 0) and (-1, 0, -1) from $x_1 + x_2 + x_3 = 0$ – darksky Apr 29 '13 at 1:29
• It's arbitrary. How do you find a solution to $x_1+x_2=0$? You can pick $(1,-1)$ or $(3,-3)$ or $(-4,4)$. – vadim123 Apr 29 '13 at 1:49