I'm in an intro to DE class. On one of the quizzes, the professor asked us to determine whether the set of functions $\ x, x-1,$ and$\ x+3$ are linearly independent or linearly dependent on $\ (- \infty, \infty) $. The homework assignment that preceded the quiz contained exercises made this seem like a straightforward process: Take the Wronskian of the functions and if it equaled zero for any value of the independent variable on I or if it was identically zero, then the functions are linearly dependent.
The correct answer on the quiz was that these functions are linearly dependent.
When I saw the problem on the quiz, I went about it mindlessly. I computed the Wronskian and, obviously, it turned out to be zero. But, then I began to think about what it meant to be linearly independent.
How can these functions be constant multiples of each other if one function does not keep a proportional distance to another? Let $\ y_1=x $ and $\ y_2=x+3 $, and let $\ x_1 = 1$ and $\ x_2=4 $. For $\ x_1$, $\ \frac{y_1}{y_2}= \frac{1}{4} $, but for $\ x_2$, $\ \frac{y_1}{y_2}= \frac{4}{5} $.
What happens when these functions pass through the$\ x$-axis? Say there were constant coefficients $\ a, b,$ and $\ c$ that formed a linear combination of these functions such that the linear combination equalled zero for all$\ x$ (except where the function is zero, I suppose) in I. When one of these functions crossed the$\ x$-axis, wouldn't at least one of the constants be undetermined?
Or, looking at it differently, take our $\ y_1$ and $\ y_2$ and let $\ x = 0$. How could you express $\ y_2$ as a multiple of $\ y_1$? More clearly, when does $\ a \cdot 0 =4$ ? And if the slack were to be picked up by the third function, wouldn't its constant multiple have to change?
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$\begingroup$Maybe you should revisit the definition of linear dependence in a general vector space (functions are vectors in this context):
A list of vectors $(v_i)_{i=1}^n$ is linearly dependent in a $\mathbb K$-vector space iff there exists a vector $\lambda\in\mathbb K^n$ with $\lambda \ne 0$ and $$\sum_{i=1}^n \lambda_i v_i = 0$$
In your case the natural vector space is the space of real-valued functions. Find a nontrivial solution to $$\lambda_1 x + \lambda_2 (x-1) + \lambda_3 (x+3) = 0 \quad \forall x\in\mathbb R$$ And you are done. The obvious choice is $\lambda = (-4,3,1)^T$.
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