# Are two functions linearly dependent iff their Wronskian is $0$?

The fact that linear dependence guarantees that $$W=0$$ is easy to prove. What about the converse? According to Wikipedia its not necessarily true. But, what is the problem is this proof?

$$W=y_1y_2'-y_2y_1'=0\implies\frac{y_1y_2'-y_2y_1'}{y_1^2}=d(\frac{y_2}{y_1})=0\implies\frac{y_2}{y_1}=k$$

Which guarantees linear dependence. Why is this not what Wikipedia agrees with? Is the converse generally true as claimed in the proof above?

• Thou shalt not divide by zero. – Angina Seng Apr 22 at 9:29

That's fine if $$y_1$$ never vanishes. That's why your attempted proof doesn't work if $$y_1(x)=x^2$$ and $$y_2(x)=|x|x$$, which is the standard counterexample.

• Why does $x,2x$ work? – tatan Apr 22 at 9:37
• I don't understand your question. – José Carlos Santos Apr 22 at 9:38
• You are saying that $W=0$ implies linear dependence is fine if$y_1$never vanishes. If $y_1=x$ and $y_2=2x$, then $y_1$ vanishes but the $W=0$ and they are linearly dependent. – tatan Apr 22 at 9:41
• What I wrote was that your proof doesn't work if $y_1$ has some zero. I did not write that what you were trying to prove is false in that case. – José Carlos Santos Apr 22 at 9:43
• Okay, I understand:-) – tatan Apr 22 at 9:46

Let us assume the functions $$\ f_1,f_2,\cdots,f_n$$ which are differentiable at least $$\ (n-1)$$ times in the interval $$\ I$$.

We now consider the equation $$\ c_1f_1+c_2f_2+\cdots+c_nf_n=0$$
Differentiating successively, we get \ \begin{align} &c_1f'_1+c_2f'_2+\cdots+c_nf'_n=0\\ &c_1f''_1+c_2f''_2+\cdots+c_nf''_n=0\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &c_1f_1^{(n-1)}+c_2'f_2^{(n-1)}+\cdots+c_nf_n^{(n-1)}=0\end{align}
Now if we consider these $$\ n$$ equations as a system of $$\ n$$ equations in $$\ c_1,c_2,\cdots,c_n$$ and if the determinant of the system does not vanish, the system will have no solution except the one with each of the $$\ c_i'$$s equal to $$\ 0$$.

Thus if the Wronskian $$\ W\ne0$$, then the functions $$\ f_1,f_2,\cdots,f_n$$ are linearly independent.

$$\ \mathbf{Hence\; the\; non-vanishing\; of\; W\; is\; a\; sufficient\; condition\; that\; the \;functions\; are \;linearly\,independent.}$$

Given two functions $$\ f$$ and $$\ g$$ that are differentiable on some interval $$\ I$$.
i) If $$\ W(f,g)(x_0)\ne 0$$ for some $$\ x_0\in I$$, then $$\ f$$ and $$\ g$$ are linearly independent $$\ I$$.
ii) If $$\ f$$ and $$\ g$$ are linearly dependent, then$$\ W(f,g)(x)=0$$ for all $$\ x\in I$$.

Be careful of the fact that it does not say that if $$\ W(f,g)(x)=0$$, then $$\ f$$ and $$\ g$$ are linearly dependent. In fact, it is possible for two linearly independent functions to have a zero Wronskian.

For example, if you take $$\ f(x)=2x^2$$ and $$\ g(x)=x^4$$, you will find that $$\ W(f,g)(x)=4x^5\ne0\; \text{unless}\; x=0$$. So $$\ W$$ is not identically $$\ 0$$ here. So $$\ f$$ and $$\ g$$ are linearly independent here.