I have $n$ lambdas, which are all different real and positive numbers, where:
$\lambda_1 < \lambda_2 < \cdots < \lambda_n$.
I then have to show that these functions are linearly independent:
$$e^{\lambda_1 t}, e^{\lambda_2 t}, \ldots, e^{\lambda_n t}$$
So I guess what I want to show is that the only solution to this equations:
$$c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}+ \ldots +c_n e^{\lambda_n t}=0, \quad \text{for all } t$$
is that all constants are zero.
I am not entirely sure how to do this when it is for $n$ lambdas – I did it earlier for just 3 lambdas, where I differentiated the function, so I have tried to use the same approach.
I wanted to get $n$ equations with $n$ unknown, so I differentiated the function $n-1$ times, and then I chose to look at the case where $t = 0$, so I have something looking like this:
$$c_1+c_2+ \ldots +c_n = 0$$
$$c_1 \lambda_1+c_2 \lambda_2+ \ldots +c_n \lambda_n = 0$$
$$ \cdots $$
$$c_1 \lambda_1^{n-1}+c_2 \lambda_2^{n-1}+ \ldots +\lambda_n^{n-1} = 0$$
Then I put this into a matrix $Ac = 0$, and now I want to show that the $\det(A)$ doesn't equal zero, so that the only solution is c = 0, but I'm not quite sure how to do this, or whether there is an easier way to do it ?
Best Answer
Define $f_j \colon \mathbf{R} \to \mathbf{R}$ by $$ f_j(t) = e^{\lambda_j t}.$$ Let $V$ denote the vector space of infinitely-differentiable functions from $\mathbf{R}$ to $\mathbf{R}$. Define $D \colon V \to V$ by $$Df = f'.$$ Note that $$Df_j = \lambda_j f_j$$ for $j = 1, \dots n$. Thus $f_1, \dots, f_n$ is a list of eigenvectors of $D$ corresponding to distinct eigenvalues. Now use the theorem that a list of eigenvectors corresponding to distinct eigenvalues is linearly independent, completing the proof that $f_1, \dots, f_n$ is a linearly independent list.