Linear independence and the Wronskian Suppose I have two linearly independent solution vectors \begin{bmatrix}x_1,_1(t)\\x_1,_2(t)\end{bmatrix} and \begin{bmatrix}x_2,_1(t)\\x_2,_2(t)\end{bmatrix} 
If I take the Wronskian of these 2 solution vectors, it comes out to a nonzero number since they are stated to be Linearly Independent. My question is, if you take the Wronskian of the same solution vectors but their derivative:
\begin{bmatrix}x'_{1,1(t)}&x'_{2,1(t)}\\x''_{1,1(t)}&x''_{2,1(t)}\end{bmatrix}
Would it still be linearly independent? (Would the Wronskian still be a nonzero number? 
 A: You've got a couple things going on at once here. First, you are asking about the Wronskian of two vectors of functions:
$$\mathbf{x}_1(t) = \left[\begin{array}{c} x_{1,1}(t) \\ x_{1,2}(t) \end{array}\right] \quad\text{and}\quad \mathbf{x}_2(t) = \left[\begin{array}{c} x_{2,1}(t) \\ x_{2,2}(t) \end{array}\right]$$
which is defined as:
$$W(\mathbf{x}_1,\mathbf{x}_2)(t) = \left|\begin{array}{cc} x_{1,1}(t) & x_{1,2}(t) \\ x_{1,2}(t) & x_{2,2}(t) \end{array}\right|.$$
Next you are asking about the Wronskian of two differentiable functions $x_{1,1}(t)$ and $x_{2,1}(t)$, which is defined as: 
$$W(x_{1,1},x_{2,1})(t) = \left|\begin{array}{cc} x_{1,1}(t) & x_{2,1}(t) \\ x_{1,1}'(t) & x_{2,1}'(t) \end{array}\right|.$$
So be careful there. To answer your question (for functions at least), take:
$$f(t) = t\quad\text{and}\quad g(t) = -1.$$
Then,
$$W(f,g) = \left|\begin{array}{cc} t & -1 \\ 1 & 0 \end{array}\right| = 1$$
but
$$W(f',g') = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| =0.$$
Also, this link does a great job at explaining the Wronskian.
A: Any two linearly independent constant vectors have dependent (identically zero) derivatives. Thus, the answer to your question is "not necessarily".
