Compute eigenvalues and eigenvectors problem I really don't know how solve this problem:
Let $V$ be the space of real functions spanned by $\cos(x)$, $\cos(2x)$ and $\cos(3x)$. Let $T\in\mathcal{L}(V,V)$ con $T(\cos(x)) = 3\cos(x) + 2\cos(2x) - \cos(3x)$, $\;T(\cos(2x)) = 3\cos(2x) + \cos(3x)$ and $T(\cos(3x)) = \cos(3x)$. Find the eigenvalues and eigenvectors of $T$.
Please, I need help
Thanxs in advance.
 A: The matrix representation of your linear transformation can be written as:
$$ A:= \left(\begin{array}{ccc}3 & 2 & -1 \\0 & 3 & 1 \\0 & 0 & 1\end{array}\right) $$
Since the matrix is triangular the characteristic polynomial can be easily read off by looking at $ \det( xI-A)$. Can you proceed now with what you need to show? 
A: It's clear that
$$T\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)\ =\ \underbrace{\left(\begin{array}{ccc}3 & 2 & -1 \\0 & 3 & 1 \\0 & 0 & 1\end{array}\right)}_{A}\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)$$
where $A$ have the eigenvalues 1, 3 and 3. Later,


*

*You have that $T(\cos(3x)) = \cos(3x)$, and then $1$ is an eigenvalue of $T$ with eigenvector $v_1 = \cos(3x)$.

*Now, if $v_2 = \alpha\cos(x) + \beta\cos(2x) + \gamma\cos(3x)$ is an eigenvector of $T$ for eigenvector $3$, then
\begin{eqnarray*}
T(v_2) & = & 3v_2,\\
\alpha T(\cos(x)) + \beta T(\cos(2x)) + \gamma T(\cos(3x)) & = & 3\alpha\cos(x) + 3\beta\cos(2x) + 3\gamma\cos(3x),\\
3\alpha\cos(x) + (2\alpha + 3\beta)\cos(2x) - (\alpha - \beta - \gamma)\cos(3x) & = & 3\alpha\cos(x) + 3\beta\cos(2x) + 3\gamma\cos(3x),\\
2\alpha\cos(2x) - (\alpha - \beta + 2\gamma)\cos(3x) & = & 0.
\end{eqnarray*}
Now, $\cos(2x)$ and $\cos(3x)$ are linear independent, then you have
$2\alpha = 0 \Rightarrow \alpha = 0$ and $\alpha - \beta +2\gamma = 0 \Rightarrow \beta = 2\gamma$. Therefore
$$v_2\ =\ \gamma\left[2\cos(2x) + \cos(3x)\right],\quad \gamma \in \mathbb{K}$$
is an eigenvector for 3.

