Proving a basis in linear algebra So at the moment I'm trying to go through proofs and I came across this one:
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.$
So I know that $B:=\{1,x,x^{2},\dots,x^{n}\}$ is a basis for $P_{n}$ So I don't understand how to prove this one?
 A: In finite dimensions (which is the case for you), if you know one basis $\mathcal B$, then proving another set $\mathcal L$ is a basis is easiest if you prove that 


*

*$\mathcal L$ and $\mathcal B$ have the same number of elements

*every element from $\mathcal B$ can be expressed as a linear combination of elements from $\mathbb L$.

A: Since the set you're given has $n+1$ elements, all you need to show is that it is linearly independent, because $\dim P_n=n+1$. If you consider
$$
a_0\cdot 1+a_1(x-1)+a_2(x^2-x)+\dots+a_n(x^n-x^{n-1})=0
$$
you get the matrix
$$
\begin{bmatrix}
1 & -1 & 0 & 0 & \dots & 0 & 0 & 0 \\
0 & 1 & -1 & 0 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & \dots & \dots & \dots & \dots & 0 & 1 & -1 \\
0 & \dots & \dots & \dots & \dots & 0 & 0 & 1
\end{bmatrix}
$$
which is triangular with $1$ along the main diagonal, therefore invertible.
A: Use the following principle: If an array $(a_1,a_2,\ldots, a_p)$ of vectors is linearly dependent then some $a_j$ is a linear combination of the $a_i$ with $i<j$. Now the polynomials $p_j$ in your list have strictly increasing degrees, hence none of them is a linear combination of the previous ones. It follows that the $p_j$ in your list are linearly independent, and as there are $n+1$ of them they form a basis of the $(n+1)$-dimensional space $P_n$.
Proof of said principle: If $\sum_{i=1}^p\lambda_ia_i=0$ with not all $\lambda_i=0$ there is a $\lambda_j\ne0$ with maximal $j$. For this $j$ one has
$$a_j=\sum_{i=1}^{j-1}-{\lambda_i\over\lambda_j}\>a_i\ .$$
