Linear Independence of $v$, $Av$, and $A^2v$ Let A be a 3x3 matrix and $ v \in \Re^3 $ with $ A^3v = 0 $ but $ A^2v \neq 0 $.
Show that the vectors $ v, Av $ and $ A^2v $ are linearly independent.
(From Alan MacDonald's Linear and Geometric Algebra, Problem 3.1.5).
 A: Hint: write down the equation : $a_1v+a_2Av+a_3A^2v=0$  .Now the idea is to  multiply $A$ on both sides. What must the coefficient $a_1,a_2,a_3$ be if you multiply several times $A$ on both sides?
A: First, you might try to see what would happen if they were linearly dependent?
Certainly, you could write an equation down from that assumption.  That equation would involve a linear combination and, once you've got a linear combination, you can multiply through by $A^2$ to get the result.
Edit
Per your request, here's a more geometric way to think of it.  
First, $v$ cannot be an eigenvector of $A$.  (Can you see why?)  Thus, $v$ and $Av$ together span a 2D subspace, say $S$, of $\mathbb R^3$.
Next, we need to show that $A^2v$ does not lie in $S$.  Well, to say that $A^3v=0$, while $A^2v\neq 0$ is to say that $A^2 v$ lies in the null space of $A$. Thus, if $A^2v$ did lie in $S$, there would be some linear combination of $v$ and $Av$ that lied in the null-space of $A$.  But since $Av$ already maps to the null-space under application of $A$, we would get that $v$ must also map to the null-space of $A$, which is a contradiction.
More generally, you might think of multiplication by $A$ as folding up space, even in higher dimensions.  Multiplication by $A$ once sends the one-dimensional null-space to zero and something also maps to that null-space.  Multiplication again does it, well, again.
