Linear dependence of these functions? How can I check if these three functions (which belong to vector space $R^R$) are linearly dependent:
$$e^{2x}, e^{3x}, x$$
If I take $\alpha, \beta, \gamma ∈ R$ and write the linear combination as:
$$\alpha e^{2x}+\beta e^{3x}+\gamma x = 0$$
How can I know if the statement is only correct if all $\alpha, \beta$ and $\gamma$ are $zero$?
 A: The workhorse for problems like this is the Wronskian. Put
\begin{align*}
f(x) &= e^{2x} & g(x) &= e^{3x} & h(x) &= x
\end{align*}
and define
$$
W(x)=
\begin{bmatrix}
f(x) & g(x) & h(x) \\
f^\prime(x) &g^\prime(x) & h^\prime(x) \\
f^{\prime\prime}(x) & g^{\prime\prime}(x) & h^{\prime\prime}(x)
\end{bmatrix}
=
\begin{bmatrix}
e^{2x} & e^{3x} & x \\
2e^{2x} &3e^{3x} & 1 \\
4e^{2x} & 9e^{3x} & 0
\end{bmatrix}
$$
If there exists an $x$ such that $\det W(x)\neq 0$, then $\{f,g,h\}$ is linearly independent. In our case, note that
$$
\det W(0)=\det\begin{bmatrix}1&1&0\\2&3&1\\ 4&9&0\end{bmatrix}=-5
$$
Hence our functions are linearly independent.
A: In general, functions $f, g, h \in C^2(\mathbb R)$ are linearly independent if for real constants $a, b, c \in \mathbb R$
$$af + bg + ch = 0, \text{ the zero function } \Rightarrow a = b= c = 0$$
Note that $af + bg + ch = 0$ implies $af' + bg' + ch' = 0$ and $af'' + bg'' + ch'' = 0$. These three equations imply $a = b = c = 0$ if the determinant
$$W(f,g,h) = \left| \begin{matrix} f & g & h \\ f' & g' & h' \\ f'' & g'' & h'' \end{matrix} \right|$$
is non-zero. This determinant is called the Wronksian, hence the $W$. In other words, the $3 \times 3$ matrix $\left( \begin{matrix} f & g & h \\ f' & g' & h' \\ f'' & g'' & h'' \end{matrix} \right)$ has trivial kernel.
In your case,
$$\begin{align} W(e^{2x}, e^{3x}, x) & = \left| \begin{matrix} e^{2x} &  e^{3x} &  x \\ 2e^{2x} &  3e^{3x} &  1 \\ 4e^{2x} &  9e^{3x} &  0 \end{matrix} \right| 
\\
& = e^{5x} \left| \begin{matrix} 1 &  1 &  x \\ 2 &  3 &  1 \\ 4 &  9 &  0 \end{matrix} \right|
\\
& = e^{5x}(6x - 5) \\ & \neq 0 \quad\quad\quad\quad \text{ , as a function }
\end{align}$$
A: If the statement holds, it is true for all $x$.  Hence, it suffices to take four values of $x$ (say, $1,2,3,4$) which give a linear system in four equations, that is inconsistent.  
