0
$\begingroup$

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?
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What do you mean by $P$? A general vector $P = (a,b,c| a,b,c \in R )$? $\endgroup$ – Alon Jan 18 at 23:08
  • $\begingroup$ No. An element of $\mathbb R_3[x]$. The map $T$ is a map from $\mathbb R_3[x]$ into itself, right?! $\endgroup$ – José Carlos Santos Jan 18 at 23:10
  • $\begingroup$ Yea, a $P = (ax^2+bx+c), a,b,c \in R$ is an element in $R_3[x]$ no? Namely, $P \in R_3[x]$ $\endgroup$ – Alon Jan 18 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.