Let $A$ be an $n\times n$ matrix that $A^n = 0$ but $A^{n-1}\not =0$, prove that... 
How would I do this problem? I think I know the  definitions but don't know anything past the first step. 
On a side note, is there any way to improve on doing these things as i'm so lost on every proof my teacher gives me... Most of the time I don't know what the question is asking. 
 A: Try to form a linear combination of these vectors and try to equate that to $0$, then $$\sum_{i=1}^n c_iA^{i-1}v=0$$If these vectors form a linearly independent set then $c_i$'s have to be $0$. Now premultiplying the above equation with $A$, $$\sum_{i=1}^{n-1}c_iA^i v=0$$ Observe that the last term vanished because of the given condition. So, if you go on premultipying with $A$, one by one the terms will vanish and you'll be left with $c_1A^{n-1}v=0$, but since $A^{n-1}v\ne 0$, you have $c_1=0$. Similarly, with $c_1$ gone, you can re start the above procedure and will end up with $c_2=0,$ and so on. So ultimately, the set is linearly independent. This set is very interesting and have applications in optimization algorithms. The vector space spanned by this set of vectors is called Krylov subspace.
A: Hint: a basis is a set of vectors that are linearly independent.
For your understanding of the question, if you show that $\vec{v},A\vec{v},\ldots,A^{n-1}\vec{v}$ are linearly independent vectors, then you'll have shown that they form a basis of $\mathbb{R}^n$. Then you can simply rename these vectors by calling them $\vec{v_1},\ldots,\vec{v_n}$; ie. you'll have defined the relation $A\vec{v_i}=\vec{v_{i+1}}$ for $i<n$ and $A\vec{v_n}=\vec{0}$, as requested.
A: Another hint: set $v_i=A^i v$ for $i=0,\dots,n-1$, and suppose there is a linear relation $$\sum_{i=0}^{n-1}\lambda_i v_i=0.$$ 
You show each $\lambda_i$ is $0$ by induction, multiplying this relation by $A^{n-1}$ first. 
