# Finding characteristic polynomial and eigenvalues of a linear transformation

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.

1. Is there is a way so that I will know if my eigenvalues are correct?
2. 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.
• What do you mean by $P$? A general vector $P = (a,b,c| a,b,c \in R )$? – Alon Jan 18 at 23:08
• No. An element of $\mathbb R_3[x]$. The map $T$ is a map from $\mathbb R_3[x]$ into itself, right?! – José Carlos Santos Jan 18 at 23:10
• Yea, a $P = (ax^2+bx+c), a,b,c \in R$ is an element in $R_3[x]$ no? Namely, $P \in R_3[x]$ – Alon Jan 18 at 23:11