What are the eigenvectors for lambda = -1?

[Anabelle37]

Homework Statement

find the general solution to x'=Ax; where A is a 3x3matrix: A=[0 1 1; 1 0 1; 1 1 0]

Homework Equations

det(A-lambda*I)=0

The Attempt at a Solution

i found the eigenvalues to be 2, -1, -1.
for lambda=2 i found the corresponding eigenvector to be a 3x1 martrix v_1 = [1;1;1]

For lambda=-1 i am having trouble finding the eigenvectors. my A-lamda*I matrix is now a 3x3 matrix containing all 1's. how do I find the eigenvector?

 

[tiny-tim]

Hi Anabelle37!

(have a lambda: λ )

… for lambda=2 i found the corresponding eigenvector to be a 3x1 martrix v_1 = [1;1;1]

So any other eigenvector msut be perpendicular to [1;1;1] …

does that help? ​

 

[HallsofIvy]

Think about the basic definition of "eigenvalue" and "eigenvector". If -1 is an eigenvalue for this A, there must be a non-zero vector, v, such that Av= -v.

That is
\begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= -\begin{bmatrix}x \\ y \\ z\end{bmatrix}

which means we must have y+ z= -x, x+ z= -y, and x+ y= -z. Those are all the same as x+ y+ z= 0 so that z= -x- y. Any vector in the eigenspace corresponding to eigenvalue -1 is of the form <x, y, -x- y>= <x, 0, -x>+ <0 , y , -y>= x<1, 0, -1>+ y<0, 1, -1>.