# 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 $\displaystyle e^{x}$ and $\displaystyle e^{2x}$. If $\displaystyle C_1e^x+ C_2e^{2x}= 0$ then, differentiating, $\displaystyle C_1e^xf+ 2C_2e^{2x}= 0$. Set x= 0 in both equations: $\displaystyle C_1+ C_2= 0$ and $\displaystyle C_1+ 2C_2= 0$. Those are clearly independent equations and have only $\displaystyle C_1= C_2= 0$ as solution.

But I think you are misunderstanding the hint. In this simple example, the determinant of the coefficients is $\displaystyle \left|\begin{array}{cc}1 & 1 \\ 1 & 2\end{array}\right|= 1$ 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
$\displaystyle \left|\begin{array}{ccccc}1 & 1 & 1 & \cdot\cdot\cdot & 1\\a_1 & a_2 & a_3 & \cdot\cdot\cdot & a_n \\ a_1^2 & a_2^2 & a_3^2 & \cdot\cdot\cdot & a_n^2 \\\cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & \cdot\cdot\cdot & a_n^{n-1}\end{array}\right|$
and you want to show that, as long as all of the numbers $\displaystyle a_i$ are distinct, that is NOT 0.

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

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

5. Out of context, consider:

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

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

So, $\displaystyle e^{a_1t},\ldots,e^{a_nt}$ 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.
Thank you in advance.

8. ## Re: Linear independence of exponential functions

This follows from the problem at hand. If $\displaystyle n > 1$

$\displaystyle n^z = e^{z\log n}$

Here $\displaystyle a_n = \log n$. 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 $\displaystyle n^z$ are all independent for natural $\displaystyle n>1$.

9. ## Re: Linear independence of exponential functions

Thank you very much for your answer. Can you suggest a reference such as book or a paper about this topic. It is very imporatnt to me to see the refernce cited in my reaserch.