A differentail equations exercise asks for the eigenvectors of $$\begin{bmatrix} 3&2&4\\ 2&0&2\\ 4&2&3 \end{bmatrix}.$$
The answer in the back of the book gives $$\begin{bmatrix} 2\\ 1\\ 2 \end{bmatrix}, \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}, \begin{bmatrix} 1\\ -4\\ 1 \end{bmatrix};$$ the first two have eigenvalues $-1$, the second has eigenvalue $8$. Mathematica confirms this:
A = {{3, 2, 4}, {2, 0, 2}, {4, 2, 3}};
v2 = {{2, 1, 2}, {1, 0, -1}, {1, -4, 1}}
Table[A.v2[[i]], {i, 1, 3}]
{{2, 1, 2}, {1, 0, -1}, {1, -4, 1}}
{{16, 8, 16}, {-1, 0, 1}, {-1, 4, -1}}
However, Mathematica gives the eigenvectors as $$\begin{bmatrix} 2\\ 1\\ 2 \end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -2\\ 0 \end{bmatrix};$$ and then confirms them as eigenvectors:
A = {{3, 2, 4}, {2, 0, 2}, {4, 2, 3}};
v = Eigenvectors[A]
Table[A.v[[i]], {i, 1, 3}]
{{2, 1, 2}, {-1, 0, 1}, {-1, 2, 0}}
{{16, 8, 16}, {1, 0, -1}, {1, -2, 0}}
Why is this happening?
Simple answer:
Any scalar multiple $\mu$ of an eigenvector $v$ with eigenvalue $\lambda$ is an eigenvector: $$A(\mu v) = \mu Av = \mu \lambda v = \lambda(\mu v).$$
More complete answer:
As Will Jagy points out in the comments, any linear combination of eigenvectors with the same eigenvalue $\lambda$ with be an eigenvector for $\lambda$:
$$A\sum_{i = 1}^{n}\mu_{i}v_{i} = \sum_{i = 1}^{n}\mu_{i} Av_{i} = \sum_{i = 1}^{n}\mu_{i}\lambda v_{i} = \lambda\sum_{I = 1}^{n}\mu_{i}v_{i}.$$