I am having trouble trying to solve this problem:
Check that the following vectors are bases and find the dimension of the vector space they are basis for
$S_1 = (1,0,...,0),(0,1,...,0),...,(0,0,...,1) \in K^n$
I do not know how can I put these vectors in row echelon form since I don't know how many they are or all their components.
$S_2 = 1,x,x^2,...,x^n\subset K^n$ [x]=“Polynomials of degree less than or equal to n”
How do i put these in row echelon form?
Thank you in advance.
Your first set of vectors contains $n$ vectors, where the $k$th one has zeros everywhere except in coordinate number $k$. For instance if $n=3$ we have $(1, 0, 0)$ and $(0,1,0)$ and $(0,0,1)$.
If you just write them in a matrix in the order they come, it is already in row echelon form. You just need to verify that it is a basis, which should be quite easy regardless of your preferred method of doing so.
Your second set of vectors seems ill-defined; what is $x$ and how does it give a $n$-dimensional vector?
Assuming that $x$ is an element of $K$, and the vectors you want are $(1,0,0,...)$, $(0,x,0,...)$, $(0,0,x^2,...)$ et cetera it should be almost exactly the same as the previous one.
And btw, to get it to be in $K^n$, you have to stop at $x^{n-1}$ since you included $1=x^0$.
edit:
With the new information, I interpret the second question as us being given a $K$-vector space $W$ spanned by the elements $1,x,x^2,...,x^n$.
It is indeed a $K$-vector space, with addition of two vectors in the usual sense, and multiplication of vectors with scalars (elements of $K$) also defined in the usual way.
To see the dimension, you just have to realise that $1,x,...,x^n$ are all linearly independent elements, and that their span is $n+1$-dimensional.