Why is MATLAB giving me these weird eigenvectors?

Question

I am doing a problem which involves finding the eigenvalues and eigenvectors of the matrix $$ M=\begin{bmatrix} 0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0 \end{bmatrix}$$

When I compute by hand, I get eigenvalues of 1, -1/2, -1/2 with corresponding eigenvectors $\begin{bmatrix}1\\1\\1\end{bmatrix},\begin{bmatrix}-1\\1\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\end{bmatrix}$, but when I try to compute in MATLAB I get

>>[A B] = eig(M)

A =

   -0.7152    0.3938    0.5774
    0.0166   -0.8163    0.5774
    0.6987    0.4225    0.5774


B =

   -0.5000         0         0
         0   -0.5000         0
         0         0    1.0000

The last eigenvector makes sense, but the first two aren't scalar multiples of the ones I got, which I figured they should be. Why are these also eigenvectors?

Answer 1

You have a repeated eigenvalue. The vectors $$\begin{bmatrix}-1\\1\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\end{bmatrix}$$ form a basis for the eigenspace associated to the eigenvalue $-\frac{1}{2}$. If you check the first two eigenvectors that matlab has given you then you will find that they are also a basis for this space. At least they would be if they were calculated exactly. There's going to be some rounding error.

Answer 2

The eigenvectors are normalized to have a norm of 1 and are chosen to be orthogonal. They will be unique (up to scalar multiplication) only if they have different eigenvalues.

Here you have two eigenvectors that share the same eigenvalue.

Answer 3

Try and convince yourself $M=A*B*A'$. This is because of the repeated eigenvalue $(-1/2)$. Thus the eigenspace corresponding to this eigenvalue has a dimension of $2$. Thus it can have any two orthogonal vectors from that space as the basis. You found one of them, MATLAB found a different one.