Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $\left\{ a_{i}\right\} _{i=0}^{n}\subset\mathbb{R}$ which are distinct, show that $\left\{ e^{a_{i}t}\right\} \subset C^{0}\left(\mathbb{R},\mathbb{R}\right)$, form a linearly independent set of functions.

Any tips on how to go about this proof. I tried a working from the definition of an exponential and combining sums but that didn't seem to get me anywhere. I saw a tip on the internet that said write it in the form

$\mu_{1}e^{a_{1}t}+\dots+\mu_{n}e^{a_{n}t}=0$ to try to show $\mu_{1}=\dots=\mu_{n}=0$ considering each term of the left hand side must be positive, but I can't get my head around that because while I understand $e^{x}>0\forall x\in\mathbb{R}$ I cannot see why $\mu_{i}$ must be positive in any case. I have thought about differentiating but that doesn't seem to help. The question did originally ask for a "rigourous" proof but I'll take any hints right now and the provided the solution of 'is obvious' is most unhelpful to me.

Any input would be fantastic. Thank you.

share|improve this question
4  
Differentiating does in fact help a lot; you can compute the Wronskian and show that it's nonzero. Alternately divide by the largest term and let t tend to infinity. –  Qiaochu Yuan Feb 21 '11 at 20:45
    
Hint: take the $\max_i a_i$, take out the exponential corresponding to it with distributivity, assuming that its coefficient is non-zero (if it's zero, take the next biggest $a_i$). Reason about the linear combination. –  Raskolnikov Feb 21 '11 at 20:46

4 Answers 4

up vote 4 down vote accepted

If you put $t=0$, the sum of the $\mu_i$ is zero, so at least one of them is negative (unless they are all zero).

Intuitively, assume $a_n$ is greater than all of the other $a$'s. Then if $t$ gets very large, that term will dominate over all the others, so $\mu_n$ must be zero. Then argue the same about the next largest.

share|improve this answer
3  
One way to see this quickly is to divide by $e^{a_nt}$ and then set $t\to\infty$. –  Andres Caicedo Feb 21 '11 at 21:37
    
Thank you very much to both of you. Much appreciated! –  Tim Green Feb 22 '11 at 0:08

HINT $\rm\ e^{a_0\:t} = c_1\ e^{a_1\:t}\:+\:\cdots\:+ c_n\ e^{a_n\:t}\ \Rightarrow\ e^{a_0\:t}\ $ is killed by $\rm\ D-a_0\ $ and by $\rm\ (D-a_1)\:\cdots\:(D-a_n)\: $. But applying the latter to $\rm\ f = e^{a_0\:t}\ $ yields $\rm\ (a_0-a_1)\cdots (a_0-a_n)\ f \ne 0\ $ since $\rm\ a_{\:i}\ne a_{\:0}\:$ for $\rm\: i\ne 0\:$.

In other words, they are unequal because their characteristic polynomials share no roots.

share|improve this answer
1  
Regretfully I don't understand what or where $D$ is in this context nor your solution though I have no doubt someone else will have it help them in future. Thanks all the same. –  Tim Green Feb 22 '11 at 0:09
1  
D is $\frac{d}{dt}$, so applying $\frac{d}{dt}-a_0$ to $e^{a_0t}$ gives zero. –  Ross Millikan Feb 22 '11 at 0:24
    
What is the characteristic polynomial of an element of a vector space (of functions)? –  Marc van Leeuwen Oct 21 at 4:41

Suppose that for some reason you do not want to differentiate or let $t$ tend to $\infty$. Evaluating $\mu_1e^{a_1t}+\cdots+\mu_ne^{a_nt}$ at $t=0,1,\ldots,n-1$ yields the system of equations

$\begin{matrix} \mu_1 & + & \mu_2 & + & \cdots & + & \mu_n &=&0\\ \mu_1e^{a_1} &+& \mu_2e^{a_2} &+& \cdots &+& \mu_ne^{a_n}&=&0\\ \mu_1\left(e^{a_1}\right)^2 &+& \mu_2\left(e^{a_2}\right)^2 &+& \cdots &+& \mu_n\left(e^{a_n}\right)^2&=&0\\ \vdots& &\vdots & &\vdots& &\vdots & &\vdots\\ \mu_1\left(e^{a_1}\right)^{n-1} &+& \mu_2\left(e^{a_2}\right)^{n-1} &+& \cdots &+& \mu_n\left(e^{a_n}\right)^{n-1}&=&0, \end{matrix}$

or in other words,

$\begin{bmatrix} 1&1&\cdots&1\\ e^{a_1}&e^{a_2}&\cdots&e^{a_n}\\ \left(e^{a_1}\right)^2&\left(e^{a_2}\right)^2&\cdots&\left(e^{a_n}\right)^2\\ \vdots&\vdots&\ddots&\vdots\\ \left(e^{a_1}\right)^{n-1}&\left(e^{a_2}\right)^{n-1}&\cdots&\left(e^{a_n}\right)^{n-1} \end{bmatrix} \begin{bmatrix} \mu_1\\ \mu_2\\ \vdots\\ \mu_n \end{bmatrix} =\vec 0.$

Because the $e^{a_i}$ are distinct, the Vandermonde matrix $A$ above is invertible, and therefore the unique solution $\vec x$ to $A\vec x=\vec 0$ is $\vec x=\vec 0$. Thus all $\mu_i$ are zero.

Similar reasoning would apply if you evaluated at any arithmetic progression of $n$ terms. Alternatively, evaluating the $0^\text{th}$ through $(n-1)^\text{st}$ derivatives at $0$ (or another point) would give a Vandermonde system with $a_i$ in place of $e^{a_i}$. The latter approach is finding the Wronskian, as referred to in Qiaochu's comment.

share|improve this answer

Although the answer by Ross Millikan is probably the easiest elementary approach, the answer by Bill Dubuque points at a more profound reason that these exponential functions must be linearly independent functions: they are eigenvectors (eigenfunctions) of the differentiation operation $D:f\mapsto f'$ for distinct eigenvalues $a_1,\ldots,a_n$. It is therefore an instance of the fundamental fact that eigenspaces for distinct eigenvalues always form a direct sum. The essence of the argument can be formulated without any advanced language as follows.

We can prove the linear independence by induction of the number $n$ of distinct exponentials involved; the cases $n\leq1$ are trivial (an exponential function is not the zero function). Then by the induction hypothesis one can assume $e^{a_1x},\ldots,e^{a_{n-1}x}$ to be linearly independent. Now if $e^{a_1x},\ldots,e^{a_nx}$ were linearly dependent, the dependency relation must involve the final exponential $e^{a_nx}$ with nonzero coefficient, and therefore (after division by the coefficient) allow that function to be expressed as linear combination of $e^{a_1x},\ldots,e^{a_{n-1}x}$: $$ c_1e^{a_1x}+\ldots+c_{n-1}e^{a_{n-1}x}=e^{a_nx} $$ Now (restricting to the subspace of differentiable functions, where all our exponentials obviously live), the operator $D-a_nI: f\mapsto f'-a_nf$ has the property of annihilating the final exponential function $f=e^{a_nx}$, but multiplying all other exponentials by a nonzero constant (namely $a_i-a_n$ in the case of $f=e^{a_ix}$). Moreover this operator is linear so it can be applied term-by-term; application to both sides of our identity turns it into $$ c_1(a_1-a_n)e^{a_1x}+\ldots+c_{n-1}(a_{n-1}-a_n)e^{a_{n-1}x}=(a_n-a_n)e^{a_nx}=0. $$ But by the (induction) hypothesis of linear independence this can only be true if all the coefficients $c_i(a_i-a_n)$ on the left are zero, which means that all $c_i$ are zero. But in view of our original expression that is absurd. So $e^{a_1x},\ldots,e^{a_nx}$ cannot be linearly dependent, completing the induction step and the proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.