Show that the following functions are linearly independent.

Question

Show that the following functions are linearly independent. $e^x, (e^2)^x, x(x-1)$

Any help would be greatly appreciated.

Answer 1

Show that if $a, b, c\in\mathbb R$ and $$f(x) = ae^x + b(e^2)^x + cx(x-1) \equiv 0 \qquad \forall\ x\in\mathbb R$$ then necessarily $a=b=c=0$.

To do this, you can simply select some three values $x_1, x_2, x_3$ and look at the resulting system

$$\pmatrix{0\\0\\0} = \pmatrix{f(x_1) \\ f(x_2) \\ f(x_3)} = A\pmatrix{a\\b\\c}$$

A nice choice for $x_1, x_2$ and $x_3$ in this particular case is $0,1$ and $2$. $A$ is the matrix

$$A = \pmatrix{e^{x_1} & e^{2x_1} & x_1(x_1-1) \\ e^{x_2} & e^{2x_2} & x_2(x_2-1)\\ e^{x_3} & e^{2x_3} & x_3(x_3-1)}$$

Answer 2

I guess over the real numbers, so suppose $\;a,b,c\in\Bbb R\;$ are such that

$$ae^x+be^{2x}+cx(x-1)=0$$

The above is a functional equality and thus it is true for any $\;x\;$ in the common domain of definition, which is the whole real line. Thus, for example. we have:

$$\begin{align*}&x=0\;:\;\implies ae^0+be^0+c\cdot0=0\implies a+b=0\\{}\\&x=1\;:\;\implies ae^1+be^2+c\cdot 0=0\implies a+be=0\end{align*}$$

and etc. Continue a little as above and deduce it must be $\;a=b=c=0\;$ and thus the functions are lin. ind.

Answer 3

Consider the equation $$\lambda_1e^x+\lambda_2e^{2x}+\lambda_3x(x-1)=0.$$

Pluging in the values $x=0,1/2$ and $x=1$ you get the three equations $$ \lambda_1+\lambda_2=0,\hspace{.3cm}\lambda_1\sqrt{e}+\lambda_2e+\lambda_3/4=0,\hspace{.3cm}\lambda_1e+\lambda_2e^2=0, $$ which you can write in matrix form as $A\lambda=0$, where $\lambda=(\lambda_1,\lambda_2,\lambda_3)^t$ and $$ A=\left(\begin{array}{lcr}1&1&0\\ \sqrt{e}&e&1/4\\e&e^2&0\end{array}\right). $$ The determinant of $A$ is $1/4(e-e^2)\neq0$ thus $A\lambda=0$ can't have more solutions than $\lambda=0$, and so $\lambda_1,\lambda_2,\lambda_3=0$ proving by definition that $e^2,e^{2x}$ and $x(x-1)$ are linear independent.