Let $T:R_3[x] \to R_3[x]$ Linear transformation such that:
$$ T(ax^2 + bx + c) = (a+b+c)x^2 + (2a + 2b + 2c)x + a+b-c $$
I want to find eigenvalues for $T$.
Therefore I looked at the representing matrix $[T]$, my calculations, brought me to this:
$$ [T] = \begin{bmatrix}1&1&1 \\ 2&2&2 \\ 1&1&-1\end{bmatrix} $$
Characteristic polynomial:
$|\lambda I - [T]| = \begin{vmatrix}\lambda - 1 & -1 & -1 \\ -2&\lambda-2&-2 \\ -1&-1&\lambda+1\end{vmatrix} = \begin{vmatrix}\lambda& -1 & -1 \\ 0&\lambda-3& -3 \\ 0&-1 &\lambda+1\end{vmatrix} = \lambda \begin{vmatrix}\lambda-3&-3 \\ -1& \lambda+1\end{vmatrix} = \lambda ((\lambda - 3)(\lambda + 1) - (-1)(-3)) = \lambda(\lambda^2 + \lambda - 3\lambda - 3 - 3) = \lambda(\lambda^2 - 2\lambda -6)$
I think I'm wrong, I keep calculating and I think wrong answers.
- Is there is a way so that I will know if my eigenvalues are correct?
- Is there a better way to find the eigenvalues so I will not continue calculating and be wrong?
Your computations are fine. It follows from them that the eigenvalues of $T$ are $0$ and $1\pm\sqrt7$. In order to check that this is correct, solve, for each $\lambda\in\left\{0,1+\sqrt7,1-\sqrt7\right\}$, the equation $T(P)=\lambda P$. If the only solution that you get is $P=0$, then you made a mistake. Otherwise, everyting is fine.