Eigenvalues of operator f. If $V=\operatorname{span}( x(t), y(t), z(t) )$ where, these three functions are real, given as $x(t)=1, y(t)=\cos(t)$ and $s(t)=\sin(t)$
Linear operator $f:V \mapsto V$ is given as $(f(v))(t)=v(t+\frac{\pi}{4})$.I want to find eigenvalues of this linear operator.
In order to find eigenvalues of this linear  operator, i need to find it's matrix with respect to some basis, but i don't know how to choose basis for vector space defined this way. how to do that when i have this kind of vector spaces?
 A: Look at how the linear operator affects the basis.
$$(f(x))(t)=x\left(t+\frac{\pi}{4}\right)=1=1\cdot x(t)+0\cdot y(t)+0\cdot z(t)$$
$$(f(y))(t)=\cos\left(t+\frac{\pi}{4}\right)=\cos t\cos\frac{\pi}{4}-\sin t\sin\frac{\pi}{4}=\frac{\sqrt 2}{2}\cos t-\frac{\sqrt 2}{2}\sin t \\ =0\cdot x(t)+\frac{\sqrt 2}{2}y(t)-\frac{\sqrt 2}{2}z(t)$$
$$(f(z))(t)=\sin\left(t+\frac{\pi}{4}\right)=\sin t\cos\frac{\pi}{4}+\cos t\sin\frac{\pi}{4}=\frac{\sqrt 2}{2}\sin t+\frac{\sqrt 2}{2}\cos t \\ =0\cdot x(t)+\frac{\sqrt 2}{2}y(t)+\frac{\sqrt 2}{2}z(t)$$
Each of these results corresponds to a column in the matrix. Do you see how to build the matrix now?
A: Hint:
I suppose that $V$ is a vector space over the real numbers.
Note that $\{1,\cos t, \sin t\}$ are linearly independent so you can use these as a basis.
A vector $v$ of $V$ can be represented, in the basis $\{1,\cos t, \sin t\}$, as.
$$
v=a\cdot 1+b\cdot \cos t + c \sin t \qquad a,b,c \in \mathbb{R}
$$
and the given transformation acts as:
$$
f(v)=a+c \cos(t+\frac{\pi}{4})+c\sin(t+\frac{\pi}{4}) = a\cdot 1+\frac{\sqrt{2}}{2}(b+c)\cos t + \frac{\sqrt{2}}{2}(c-b) \sin t
$$
This means that :
$$
f \left( 
\begin{bmatrix}
a\\b\\c
\end{bmatrix}
\right)=
\begin{bmatrix}
a\\\frac{\sqrt{2}}{2}b+\frac{\sqrt{2}}{2}c\\-\frac{\sqrt{2}}{2}b+\frac{\sqrt{2}}{2}c
\end{bmatrix}
$$
so the matrix that represents the transformation is:
$$
\begin{bmatrix}
1&0&0\\
0&\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\
0&-\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}
\end{bmatrix}
$$
And remember that the eigenvalues of a linear transformation are independent from the basis used to represent it as a matrix.
