Linear independence of exponential functions

**limddavid** · November 14th 2010, 08:17 PM

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!

**roninpro** · November 14th 2010, 08:24 PM

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

**HallsofIvy** · November 15th 2010, 04:13 AM

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.

**Drexel28** · November 15th 2010, 02:02 PM

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

**FernandoRevilla** · November 15th 2010, 03:26 PM

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.

**limddavid** · November 15th 2010, 03:53 PM

Sweet! Thank you so much. Makes sense now

**zeraoulia** · August 27th 2012, 11:06 AM

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.

**Vlasev** · August 27th 2012, 03:52 PM

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 $n>1$ .

**zeraoulia** · August 28th 2012, 07:22 AM

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.

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