I have the following problem, I have to find the spectral basis ( I hope I am translating this correctly) for the following endomorphism ( homomorphism ):
$\ R^3 -> R^3 f(x,y,z)=( (x,6y,2z),(2y),(x,4y,2z)) $
I created the associated matrix and found the eigenvalues
$$ \begin{bmatrix} 1 & 6 & 2 \\ 0 & 2 & 0 \\ 1 & 4 & 2 \\ \end{bmatrix} $$
I tried to find det(A-$\lambda$I)
$$
\begin{bmatrix}
1-\lambda & 6 & 2 \\
0 & 2-\lambda & 0 \\
1 & 4 & 2-\lambda \\
\end{bmatrix}
$$
And from this I found the roots of the eigenvalues as $\lambda=0,2,3$
Now I tried finding the basis by re-writing the matrices with every $\lambda$ value $$\lambda = 0$$ $$ \begin{bmatrix} 1 & 6 & 2 \\ 0 & 2 & 0 \\ 1 & 4 & 2 \\ \end{bmatrix}$$
The equivalent linear solution is :
$$ \begin{cases} x+6y+2z=0\\ 2y=0\\ x+4y+2z=0\\ \end{cases} $$ Since $ y=0 $ I have attributed to z a parametric solution, and my result is: $$ \begin{cases} x=-2s\\ y=0\\ z=s\\ \end{cases} $$ If I create a vector from that I get : $$ s*\begin{pmatrix} -2\\ 0\\1\\ \end{pmatrix} $$ Doing the same calculations for 2 and 3 I get: $$\lambda=3$$ $ \begin{bmatrix} -2 & 6 & 2 \\ 0 & -1 & 0 \\ 1 & 4 & -1 \\ \end{bmatrix}$ $ \begin{cases} -2x+2z=0\\ y=0\\ z=s\\ \end{cases}$ Vector: $ s*\begin{pmatrix} 1\\ 0\\1\\ \end{pmatrix} $
$$\lambda=2$$ $ \begin{bmatrix} -1 & 6 & 2 \\ 0 & 0 & 0 \\ 1 & 4 & 0 \\ \end{bmatrix}$ $ \begin{cases} x=-4s\\ y=s\\ z=-5s\\ \end{cases}$ Vector: $ s*\begin{pmatrix} -4\\ 1\\-5\\ \end{pmatrix} $
And so I find $ B_1=\{-2,0,1\}$ $ B_3=\{1,0,1\}$ that are correct basis for the endomorphism, however for $\lambda = 2$ I get the basis $ B_2=\{-4,1,-5\}$ that is exactly half of what I should get, I am unsure weather I have proceeded correctly, but seeing as I have 2/3 correct results, I don't know if it's something I'm doing wrong or I'm just going insane, I would appreciate some insight, thank you !
You answer is fine. If $\mathbf v$ is an eigenvector with eigenvalue $\lambda$, then so is $c\mathbf v$ for $c\ne0$. Unless there’s some convention about which eigenvectors to choose out of all of the possibilities, your answer is just as correct as the one you’re comparing against.