Proofs in linear algebra I'm pretty awful at proving linear algebra proofs, I just don't understand how you know what to do or where the information comes from.
I have some sample questions below of what I mean, I have no idea how I'm supposed to prove them.
Any help would be brilliant!


*

*Suppose $  P_n $ is the vector space of all polynomials with degree less than or equal to n.


Prove that $ \{1, x − 1, x^2 − x, x^3 − x^2, . . . , x^n − x^{n−1}\} $ is a basis for $ P_n.$


*In the vector space $ V = {f : R → R} $, prove that the set $ \{\cos 2x,\sin 5x, x^3\} $ is linearly
independent.

*Suppose $ C_{ij} $ is the $2\times3$ matrix with $1$ in the $ i,j^{th} $ entry and zero everywhere else. Prove
that $ \{C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}\} $ is a basis for $ M_{2×3}(R) $ . Also, what is the dimension of
$M_{2×3}(R)?
$


and also if you have any good tips for learning this from scratch
 A: First of all I think question number 1 has a mistake, probably you want to prove that $\{1,x-1,\dots,x^{n}-x^{n-1}\}$ is a basis for $P_{n}$. Now, for that question I hope you know that $B:=\{1,x,x^{2},\dots,x^{n}\}$ is a basis for $P_{n}$ (if not, it easy to prove), then your question is easy to prove using elementary facts on linear algebra because you only must do linear combinations between vectors of the basis B.
Question 2 it's a little harder, but I suppose that reductio ad absurdum would be a good way to start.
Question 3 follows from the isomorphism between $M_{2x3}(R)$ and $R^{2x3}$.
Generally proofs on linear algebra don't have and standard way to solve. Obviously, most of the poofs refering to prove  linear independent usually works by using reductio ad absurdum
A: As a rule, the first thing I suggest is to write down the definitions of mathematical terms.
For question 1.  Prove $v_1,v_2\dots v_n$ forms a basis.  What is a basis?
A basis is a linearly independent set of vectors that spans the space.  
Great, more words to define.
Spans the space -- every vector in the space can be formed by combination of vectors in the basis.
Linear independence --  No vector in the basis can be formed by a combination of the other vectors in the basis.  Or, $c_1 v_2 + c_2 v_2 \dots c_n v_n = 0$ if and only if $c_1...c_n = 0.$
Now you have an idea of what you need to show.
A few tricks.  Can you find a way to transform this basis to the "cannonical basis"--for polynomials that would be $1,x,x^2\dots x^n$ -- then you are done.
If you have a linearly independent set of vectors they will form a basis of some space.  You can prove that they are a basis of your target space, if you can show that they form the basis of a sub-space and than the dimension of the sub-space equals the dimension of the target space.
If you can't prove that something is true.  Try to prove that it is not true.  It might provide some insights.
Another classic technique, assume that it is not true and see if you run into a contradiction.
A: As a note for question two, if $\cos{2x},\sin{5x},$ and $x^3$ are linearly dependent as functions, then there must exist $\alpha,\beta,\delta \in \Bbb R$ such that:
$$\alpha\cos{2x}+\beta\sin{5x}+\delta x^3\equiv 0 \quad\forall x\in\mathbb{R} $$
We can show $\delta$ must be zero by considering 
$\lim\limits_{x\rightarrow \infty}\left(\frac{\alpha\cos{2x}}{x^3}+\frac{\beta\sin{5x}}{x^3}+\delta\right)=0$
To show that $\alpha$ and $\beta$ must also be zero consider multiplying the new equation:
$$\alpha\cos{2x}+\beta\sin{5x}\equiv 0 \quad\forall x\in\mathbb{R} $$
by $\cos{2x}$ and integrating from $-\pi$ to $\pi$, or similarly by $\sin{5x}$ and integrating with the same limits.
You can use that $\int_{-\pi}^{\pi} \sin(mx)\cos(nx)=0$.
