How to show that $e^{a_1 z} ,...,e^{a_n z}$ is Linearly independent over $\mathbb{C}$

Question

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1. What the problem asks

We are given distinct complex numbers

[ a_{1},a_{2},\dots ,a_{n}\in\mathbb C ,\qquad a_i\neq a_j\;(i\neq j), ]

and we consider the functions

[ f_i(z)=e^{a_i z}\qquad (i=1,\dots ,n) ]

defined on the whole complex plane.
The task is to prove that these (n) functions are linearly independent over (\mathbb C); i.e.

[ c_{1}e^{a_{1}z}+c_{2}e^{a_{2}z}+\dots +c_{n}e^{a_{n}z}=0\quad\text{for all }z\in\mathbb C \Longrightarrow c_{1}=c_{2}= \dots =c_{n}=0 . ]

In other words, the only linear combination of the exponentials that vanishes identically is the trivial one.

2. Full proof

We give two proofs; you may keep the one you find most convenient.

Proof A – Using the Wronskian (determinant of derivatives)

For (k\ge 0) let

[ f_i^{(k)}(z)=\frac{d^{\,k}}{dz^{k}}e^{a_i z}=a_i^{\,k}e^{a_i z}. ]

Form the (n\times n) Wronskian matrix

[ W(z)=\Bigl[f_i^{(j-1)}(z)\Bigr]_{\,i,j=1}^{n} =\begin{pmatrix} e^{a_1z} & e^{a_2z} & \dots & e^{a_nz}\[2pt] a_1e^{a_1z}& a_2e^{a_2z}& \dots & a_ne^{a_nz}\[2pt] \vdots & \vdots & & \vdots\[2pt] a_1^{\,n-1}e^{a_1z}& a_2^{\,n-1}e^{a_2z}&\dots & a_n^{\,n-1}e^{a_nz} \end{pmatrix}. ]

Factor out the exponentials from each column:

[ W(z)=\bigl(\operatorname{diag}(e^{a_1z},\dots ,e^{a_nz})\bigr)\, V,\qquad V=\begin{pmatrix} 1&1&\dots &1
a_1&a_2&\dots &a_n
\vdots&\vdots&&\vdots
a_1^{\,n-1}&a_2^{\,n-1}&\dots &a_n^{\,n-1} \end{pmatrix}. ]

The matrix (V) is the Vandermonde matrix built from the numbers (a_1,\dots ,a_n).
Its determinant is well‑known:

[ \det V=\prod_{1\le i<j\le n}(a_j-a_i). ]

Because the (a_i) are distinct, every factor ((a_j-a_i)) is non‑zero, hence

[ \det V\neq 0 . ]

Now

[ \det W(z)=\Bigl(\prod_{i=1}^{n}e^{a_i z}\Bigr)\det V =\Bigl(\prod_{i=1}^{n}e^{a_i z}\Bigr)\, \prod_{1\le i<j\le n}(a_j-a_i)\neq 0 ]

for every (z\in\mathbb C) (the product of exponentials is never zero).
Thus the Wronskian of the functions (e^{a_1z},\dots ,e^{a_nz}) never vanishes.

A standard result from the theory of linear ordinary differential equations (or directly from the definition of linear independence for analytic functions) says:

If the Wronskian of a set of analytic functions is not identically zero on a domain, the functions are linearly independent on that domain.

Consequently the exponentials are linearly independent on (\mathbb C).

Proof B – Using derivatives at a single point (often (z=0))

Assume a linear relation

[ c_{1}e^{a_{1}z}+c_{2}e^{a_{2}z}+\dots +c_{n}e^{a_{n}z}=0\qquad(\forall z\in\mathbb C). ]

Differentiate (k) times and then set (z=0).
Because (\frac{d^{k}}{dz^{k}}e^{a_i z}=a_i^{k}e^{a_i z}), we obtain for every (k\ge 0)

[ c_{1}a_{1}^{k}+c_{2}a_{2}^{k}+ \dots +c_{n}a_{n}^{k}=0 . \tag{1} ]

Take the first (n) equations, i.e. for (k=0,1,\dots ,n-1).
Written in matrix form they read

[ \begin{pmatrix} 1&1&\dots &1
a_{1}&a_{2}&\dots &a_{n}
\vdots &\vdots &&\vdots
a_{1}^{\,n-1}&a_{2}^{\,n-1}&\dots &a_{n}^{\,n-1} \end{pmatrix} \begin{pmatrix} c_{1}\c_{2}\ \vdots \c_{n} \end{pmatrix} = \begin{pmatrix} 0\0\ \vdots \0 \end{pmatrix}. ]

Again the matrix on the left is the Vandermonde matrix (V).
Since the (a_i) are pairwise distinct, (\det V\neq 0); therefore the only solution of this homogeneous linear system is

[ c_{1}=c_{2}= \dots =c_{n}=0 . ]

Thus the assumed linear relation forces every coefficient to be zero, proving linear independence.

Both arguments rely on the fact that the Vandermonde determinant is non‑zero precisely when the parameters (a_{1},\dots ,a_{n}) are distinct.

3. Final statement

For distinct complex numbers (a_{1},\dots ,a_{n}) the functions

[ e^{a_{1}z},\;e^{a_{2}z},\dots ,e^{a_{n}z}\qquad (z\in\mathbb C) ]

are linearly independent over (\mathbb C).
Equivalently, if

[ c_{1}e^{a_{1}z}+c_{2}e^{a_{2}z}+\dots +c_{n}e^{a_{n}z}\equiv 0, ]

then necessarily (c_{1}=c_{2}= \dots =c_{n}=0).

4. Common mistakes to avoid

Mistake	Why it is wrong	How to correct it
Assuming the relation holds only for some values of (z)	Linear independence concerns equality for all (z) in the domain. A relation that vanishes at finitely many points does not imply dependence.	Explicitly state that the equality holds for every complex number (z).
Trying to “cancel” exponentials (e.g. dividing by (e^{a_1z}))	The cancellation is illegal when the exponentials have different exponents; you cannot factor a common term out of a sum.	Work with derivatives or the Wronskian, or evaluate the identity at several points, rather than attempting to factor out an exponential.
Using the Vandermonde determinant without justification	One must recognise that the matrix formed by the coefficients after differentiating is exactly a Vandermonde matrix, and recall its determinant formula.	Write down the matrix explicitly, point out it is a Vandermonde matrix, and quote (\det V=\prod_{i<j}(a_j-a_i)).
Confusing linear independence over (\mathbb C) with pointwise independence	Linear independence means no global linear combination is identically zero, not that the functions take different values at each point.	Emphasise the definition: a linear combination that vanishes identically (as a function) forces all coefficients to be zero.
Neglecting the case where some (a_i) coincide	If two exponents are equal, the corresponding exponentials are the same function, so they are automatically dependent.	The hypothesis “(a_i) are distinct” is essential; make it clear that the proof uses (\det V\neq 0) which fails when two (a_i) coincide.

Keeping these points in mind will help you construct a rigorous and clean argument.

Original question: How to show that $e^{a_1 z} ,…,e^{a_n z}$ is Linearly independent over $\mathbb{C}$ on Mathematics Stack Exchange, licensed CC BY-SA.