函数除以公因子是否影响线性无关性？含朗斯基行列式应用疑问

Great question—this is such a clever shortcut, and it’s totally mathematically valid! Let’s break this down step by step, starting with your specific example and then generalizing to the core question.

First: Your Example Checks Out #

You’re right that dividing $e^{4x}$, $xe^{4x}$, and $x^2e{4x}$ by $e^{4x}$ (which is never zero on $(-\infty, \infty)$) gives the simpler set ${1, x, x^2}$, and verifying their linear independence is way easier. Here’s why this works:

Recall the definition of linear independence for functions: A set of functions ${f_1(x), f_2(x), f_3(x)}$ is linearly independent on an interval if the only constants $c_1, c_2, c_3$ that satisfy

$c_1f_1(x) + c_2f_2(x) + c_3f_3(x) = 0$ for all $x$ in the interval

are $c_1 = c_2 = c_3 = 0$.

For your original set:
Suppose there exist constants $c_1, c_2, c_3$ (not all zero) such that
$$c_1e^{4x} + c_2xe^{4x} + c_3x^2e{4x} = 0 \quad \forall x \in (-\infty, \infty)$$
Since $e^{4x} \neq 0$ for every real $x$, we can divide both sides by $e^{4x}$ without changing the equality’s validity. This gives:
$$c_1 + c_2x + c_3x^2 = 0 \quad \forall x \in (-\infty, \infty)$$
But we know the polynomial set ${1, x, x^2}$ is linearly independent—there’s no non-trivial combination of constants that makes this quadratic zero everywhere. The only solution is $c_1 = c_2 = c_3 = 0$, which contradicts our initial assumption that the constants weren’t all zero. Therefore, the original set ${e^{4x}, xe^{4x}, x^2e{4x}}$ must be linearly independent.

Generalizing: When Does Dividing by a Common Term Work? #

The key here is that the common term ($e^{4x}$ in your case) is never zero on the entire interval of interest. Here’s the general rule:

• If you have a set of functions ${f_1(x), f_2(x), ..., f_n(x)}$ and a function $g(x)$ that is non-zero for all $x$ in your interval, then the set ${f_1(x)/g(x), f_2(x)/g(x), ..., f_n(x)/g(x)}$ is linearly independent if and only if the original set is linearly independent.

Why? Because dividing by $g(x)$ is equivalent to multiplying each function by $1/g(x)$—and since $g(x)$ is never zero, $1/g(x)$ is a well-defined, non-zero function across the interval. This operation is a linear isomorphism on the function space (it’s invertible, with inverse multiplying by $g(x)$), and linear isomorphisms preserve linear independence (and linear dependence, for that matter).

Using Wronskians to Confirm #

You mentioned calculating the Wronskian, and this shortcut works here too. The Wronskian of the original set is:
$$W(e^{4x}, xe^{4x}, x^2e{4x})(x) = e^{4x} \cdot e^{4x} \cdot e^{4x} \cdot W(1, x, x^2)(x)$$
Since $e^{12x} \neq 0$ for all $x$, the Wronskian of the original set is zero if and only if the Wronskian of ${1, x, x^2}$ is zero. The Wronskian of the polynomial set is the Vandermonde determinant:
$$W(1, x, x^2)(x) = \begin{vmatrix} 1 & x & x^2 \ 0 & 1 & 2x \ 0 & 0 & 2 \end{vmatrix} = 2$$
Which is never zero, so the original set’s Wronskian is also never zero—confirming linear independence.

内容的提问来源于stack exchange，提问作者Benjamin Xu