Most introductory books on Linear Algebra have a Theorem which says something like
Let $A$ be a square $n \times n$ matrix. Then the following are equivalent:
What does this mean, it simply means that if you want to check if any of these conditions is true or false, you can simply pick whichever other condition from the list and check it instead..
Your question is: Can instead of third or fourth condition, check the second? That's exactly what the Theorem says: YES.
Note: I wonder what DEQ this would be a solution for? Regardless, lets proceed.
We know that if $f(x)$ and $g(x)$ are linearly dependent on $I$ then $W(f,g)(x) = 0$ for all $x$ in the interval $I$.
Is this telling you anything about the linear dependence of the functions themselves? It does not imply that if $W(f,g)(x) = 0$ then $f(x)$ and $g(x)$ are linearly dependent. It is possible for two linearly independent functions to have a zero Wronskian!
We would analyze
$$ W(f,g) = \det\begin{bmatrix}f & g\\f' & g'\end{bmatrix} = fg' - gf'$$
- $x < 0 \rightarrow |x| = -x \rightarrow \text{Wronskian} = 0$ since $fg' - gf' = x(-1) - (-x)(1) = 0$
- $x = 0 \rightarrow |x| = 0 \rightarrow \text{Wronskian} = 0$ since $fg' - gf' = 0-0 = 0$
- $x > 0 \rightarrow |x| = x \rightarrow \text{Wronskian} = 0$ since $fg' - gf' = x(1) - (x)(1) = 0$
Since the Wronskian is zero, no conclusion can be drawn about linear independence!
For linear independence, we want to go back to the basic definitions again. We have:
- $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x \lt 0$. Thus, our equations to check for linear independence of these functions become:
$$c_1 x + c_2 x = 0~~~~ \text{for}~ x \ge 0 \\ c_1x - c_2 x = 0~~~~\text{for}~ x \lt 0$$
The only solution to this system is $c_1 = c_2 = 0 \rightarrow$ linear independence. Note that at the single point $x = 0$ does not matter.
You can also see the same argument for your second example Calculate the Wronskian of $f(t)=t|t|$ and $g(t)=t^2$ on the following intervals: $(0,+\infty)$, $(-\infty, 0)$ and $0$?
Best Answer
It's traditional mathematical English to say "$x$ vanishes" to mean $x = 0$.