Finding a basis of eigenvectors For a linear operator $T$ on $V$ find the eigenvalues of $T$ and an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix:
$V$=$R^3$, $T(a,b,c)$= $(7a-4b+10c,4a-3b+8c,-2a+b-2c)$.
I solved this question, and got that, the eigenvalues are $-1,1,2$ and 
the basis $\beta$ = {$(1,2,0),(1,4,1),(-2,0,1)$}.
But, my book gives a different answer for $\beta$, i.e.
$\beta$ = {$(1,2,0),(1,-1,-1),(2,0,-1)$}.
Is my answer also correct ? What i want to know is, whether this basis $\beta$ for $V$ can be uniquely determined ?
I think that there could be many choices for $\beta$, as even the question says "$an$" ordered basis and not "$the$" ordered basis. Am i correct ?
 A: There is no canonical choice for a basis of eigenvectors. For instance, if $(1,1,1)$ is an eigenvector, then also $(a,a,a)$ (for $a\ne0$) is, and there's no rule that makes $(1,1,1)$ preferable to $(2,2,2)$.
Your matrix is
$$
\begin{bmatrix}
7 & -4 & 10 \\
4 & -3 & 8 \\
-2 & 1 & -2
\end{bmatrix}
$$
It's readily checked that


*

*$(1,2,0)$ is an eigenvector for the eigenvalue $-1$;

*$(1,-1,-1)$ is an eigenvector for the eigenvalue $1$;

*$(2,0,-1)$ is an eigenvector for the eigenvalue $2$.


There's no canonical choice, so using $(-2,0,1)$ is as good as using $(4,0,-2)$ or $(2,0,-1)$.
However, having made the checks, your vector $(1,4,1)$ cannot be an eigenvector: if it were, it would be a scalar multiple of one of the preceding vectors, which it isn't.
Indeed
$$
\begin{bmatrix}
7 & -4 & 10 \\
4 & -3 & 8 \\
-2 & 1 & -2
\end{bmatrix}\,
\begin{bmatrix}
1\\4\\1
\end{bmatrix}=
\begin{bmatrix}
1\\0\\0\end{bmatrix}
$$
If I had to grade your test, I'd consider this a serious mistake, because you have a way to check your computations, namely that the vectors you found are indeed eigenvectors.
