# Thread: Linear independence of exponential functions

1. ## Linear independence of exponential functions

I'm honestly a bit baffled by this problem...

Let a_1, ... , a_n = distinct numbers that do not equal 0. show that e^(a_1*t), ... , e^(a_n*t) are linearly independent over the numbers.

It says hint: suppose c_1*e^(a_1*t) + ... + c_n*e^(a_n*t) = 0 for constants c_1, ..., c_n, differentiate n-1 times. The determinant of the coefficients of the system of linear equations should be 0 (Why?)

I tried the hint, but I don't see how I can prove that the determinant of the coefficients have to equal 0.. Help!

2. This is called the Wronskian. You might try reading about it first. Be sure to come back if you have questions about it!

Wronskian - Wikipedia, the free encyclopedia

3. Look at the simple case of $e^{x}$ and $e^{2x}$. If $C_1e^x+ C_2e^{2x}= 0$ then, differentiating, $C_1e^xf+ 2C_2e^{2x}= 0$. Set x= 0 in both equations: $C_1+ C_2= 0$ and $C_1+ 2C_2= 0$. Those are clearly independent equations and have only $C_1= C_2= 0$ as solution.

But I think you are misunderstanding the hint. In this simple example, the determinant of the coefficients is $\left|\begin{array}{cc}1 & 1 \\ 1 & 2\end{array}\right|= 1$ not 0.

They are saying that if any system of equations has more than one solution, then the determinant of the coefficients must be 0 (if it were not, the matrix of determinants would have an inverse). The point here is to show that the determinant of the coefficients is NOT 0 so there is only the trivial solution.

In general you have
$\left|\begin{array}{ccccc}1 & 1 & 1 & \cdot\cdot\cdot & 1\\a_1 & a_2 & a_3 & \cdot\cdot\cdot & a_n \\ a_1^2 & a_2^2 & a_3^2 & \cdot\cdot\cdot & a_n^2 \\\cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & \cdot\cdot\cdot & a_n^{n-1}\end{array}\right|$
and you want to show that, as long as all of the numbers $a_i$ are distinct, that is NOT 0.

4. Originally Posted by HallsofIvy
Look at the simple case of $e^{x}$ and $e^{2x}$. If $C_1e^x+ C_2e^{2x}= 0$ then, differentiating, $C_1e^xf+ 2C_2e^{2x}= 0$. Set x= 0 in both equations: $C_1+ C_2= 0$ and $C_1+ 2C_2= 0$. Those are clearly independent equations and have only $C_1= C_2= 0$ as solution.

But I think you are misunderstanding the hint. In this simple example, the determinant of the coefficients is $\left|\begin{array}{cc}1 & 1 \\ 1 & 2\end{array}\right|= 1$ not 0.

They are saying that if any system of equations has more than one solution, then the determinant of the coefficients must be 0 (if it were not, the matrix of determinants would have an inverse). The point here is to show that the determinant of the coefficients is NOT 0 so there is only the trivial solution.

In general you have
$\left|\begin{array}{ccccc}1 & 1 & 1 & \cdot\cdot\cdot & 1\\a_1 & a_2 & a_3 & \cdot\cdot\cdot & a_n \\ a_1^2 & a_2^2 & a_3^2 & \cdot\cdot\cdot & a_n^2 \\\cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & \cdot\cdot\cdot & a_n^{n-1}\end{array}\right|$
and you want to show that, as long as all of the numbers $a_i$ are distinct, that is NOT 0.
This isn't a necessarily simple task. May I suggest you consider $A^T$. Then, things might look a little more familiar

5. Out of context, consider:

$D\in \textrm{End}\;\mathcal{C}^{\infty}(\mathbb{R},\mat hbb{R}),\;D(x(t))=x'(t)$

Then, $e^{a_1t},\ldots,e^{a_nt}$ are eigenvectors associated respectively to the pairwise distinct eigenvalues $a_1,\ldots,a_n$.

So, $e^{a_1t},\ldots,e^{a_nt}$ are linearly independent.

Regards.

6. Sweet! Thank you so much. Makes sense now

7. ## Re: Linear independence of exponential functions

Hi and how are you
I have the following function:
Let z be a complex number. Is the functions (n^z) are linearly independent for all n natural. If this is true can you indicate a reference for that.
This follows from the problem at hand. If $n > 1$
$n^z = e^{z\log n}$
Here $a_n = \log n$. For a countable set of functions to be independent, every finite subset of functions must be independent (definition). By the problem above we see that this is true, so the $n^z$ are all independent for natural $n>1$.