Are linear signals independent if the Casoratian is $0$? Consider the signals $3^k$, $(-4)^k$, $(-1)^k$.
To compute whether they are linearly independent we must analyze their respective Casorati Matrix.
We have $\begin{bmatrix} 3^{k}&(-4)^{k}&(-1)^{k}\\3^{k+1}&(-4)^{k+1}&(-1)^{k+1}\\3^{k+2}&(-4)^{k+2}&(-1)^{k+2}\end{bmatrix}$.
I think we're just supposed to set $k=0$ and figure out whether the determinant (Casoratian) is $0$, (I'm not even sure..)
So we have $\begin{bmatrix} 1&1&1\\3&-4&-1\\9&16&1\end{bmatrix}$.
A simple thing to do is row reduce, and check the number of pivots.
So yes, it has $3$ pivots, and is invertible.
What does this mean? I see some sources saying that linear independence means that the Casoratian is $0$, others say linear dependence means the Casoratian is $0$???? 
I provided most of the body for context, but in reality, I just need the answer to: Does the Casoratian = 0 imply linear independence? (Yes or No) and hopefully an explanation would be nice. 
 A: http://mathworld.wolfram.com/Casoratian.html
Solutions of a (homogeneous) linear difference equation of the form: 
$$ x(n+k) + p_1 (n) x(n+k-1) \dots + p_k(n) x(n)=0  $$
for $n \in \mathbb N$ are linearly independent sequences iff their Casoratian (determinant of the matrix) is nonzero for $n=0$ (Zwillinger 1995).
so NO, determinant$=0$ implies linear DEPENDENCE.
In your case you want to check linear independence of three given solutions to a problem, which probably requires a linear combination of three solutions (so it is a $k=3$-th order system). We call this set the "fundamental solution set" similar to linear differential equations. Indeed we find that
$\left|\begin{array}{ccc} 1 & 1& 1\\3&-4&-1\\9&16&1\end{array} \right| = 84 \neq 0$.
The solutions are therefore independent.
so $x(n)=a 3^k + b (-4)^k + c(-1)^k $
For some parameters that depend on the initial values.
See also the lecture slides:
https://www.studydrive.net/courses/technische-universiteit-eindhoven/discrete-dynamical-systems/lectures/lecture52wa50/viewfile/634161
