Vector space with basis What does it mean by the statement that $U$ is the vector space with basis $\{e^t, e^{-t}\}$? 
I also want to know whether the standard basis applies to this case. 
Thanks!
 A: A basis for a linear space is a family of linearly independent vectors spanning the linear space. In the given example the vectors of $U$ have the form
$$\lambda_1 e^t+\lambda_2 e^{-t},\quad \lambda_1,\lambda_2\in \Bbb R$$
A: Most probably the ambient space is the space of continuous (or differentiable) functions on $\mathbb{R}$ or an interval thereof.
The notation is abbreviated, probably what is meant is

consider the function $f$ defined by $f(t)=e^t$ and the function $g$ defined by $g(t)=e^{-t}$, belonging to $C^\infty(\mathbb{R})$, the space of infinitely differentiable functions on $\mathbb{R}$ (or other space);  consider $U$ the subspace of $C(\mathbb{R})$ spanned by $\{f,g\}$ and show this set is a basis.

Thus $U$ consists of all functions $\alpha f+\beta g$, that is, of the form
$$
t\mapsto \alpha f(t)+\beta g(t)=\alpha e^{t}+\beta e^{-t}
$$
The two vectors are linearly independent, because, if $\alpha f+\beta g=0$ (the zero constant function), then
$$
\alpha f(t)+\beta g(t)=0
$$
for all $t\in \mathbb{R}$ and, in particular, this holds for $t=0$ and $t=1$, so
$$
\begin{cases}
\alpha+\beta=0\\
\alpha e+\beta e^{-1}=0
\end{cases}
$$
from which it follows that $\alpha=\beta=0$, can you see why?
Now that we know it's a basis, we can consider the differentiation operator $T$, which sends $U$ to itself, because $Tf=f$ and $Tg=-g$. The matrix of the subordinate operator on $U$ is thus
\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}
