linearly independent (Linear algebra) Show graphically that $y_1(x)=x^2$ and $y_2(x)=x|x|$ are linearly independent on $-\infty$ to $\infty$ but Wronskian vanishes at every point.
The Wronskian is 
$$W = \left|\begin{matrix}y_1&y_2\\y'_1&y_2'\end{matrix}\right|$$
 A: The definition linearly independent is that $ay_1 + b y_2 = 0$ only when $a=0$ and $b=0$.
Linearly dependent means the negation, that there are non-zero values of $a$ and $b$ with $ay_1 + b y_2 = 0$.
These definitions extend to not just two functions but an arbitrary number of functions.
In the particular case of two functions, the two functions are linearly dependent means that one is a non-zero multiple of another:
$a y_1 + b y_2=0$.  If $a$ and $b$ are not both zero, then neither of them will be zero.
So $-\frac{a}{b} y_1 = y_2$
One way to interpret the question is: Show graphically that $y_1(x) = x^2$ is not a (non-zero) constant multiple of $y_2(x) = x|x|$.
Do you know the relationship between graphs of functions that are multiples of one another?
A: Here is a sketch of the two functions. If they were linearly dependent, they would have approximately the same shape. Clearly they don't.

The wronskian in this case is:
$$W=x^2\times 2|x| - x|x|\times 2x = 0$$
A: Write explicitly:
$$y_2(x)=x|x|=\begin{cases}-x^2&,\;\;x<0\\{}\\\;\;\,x^2&,\;\;x\ge 0\end{cases}$$
Thus, suppose $\;ay_1(x)+by_2(x)=0\;,\;\;a,b\in\Bbb R\;$ . As this is the zero function ,we can choose any $\;x\;$ and the equality remains, yet:
$$\begin{align}&ay_1(-1)+by_2(-1)=a(-1)^2+b(-1)|-1|=&a-b=0\\{}\\
&ay_1(1)+by_2(1)=a\cdot1^2+b\cdot1\cdot|1|=&a+b=0\end{align}$$
Solving the above linear system renders $\;a=b=0\;$ .
