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In my book it says that the basis of the set of polynomials of degree at most n ie $K[x]_{≤n}$ has the basis $1,x,x^2,...,x^n$. But arent $x,x^2,...$ a linear combination of each other? This confuses me since the set isnt a set of vectors here.

NoteBook
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    The polynomials $1, \ldots, x^n$ are vectors in $K[x]_{\leq n}$. They are certainly linearly independent, as $a_0 + a_1 x + \cdots a_n x^n = 0$ implies that $a_0 = \cdots = a_n = 0$. – Travis Willse May 12 '18 at 19:02
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    The monomials $1,x,x^2,\ldots,x^n$ are linearly independent over $K$ more or less by the definition of the polynomial ring $K[x]$. – Jyrki Lahtonen May 12 '18 at 19:16
  • Anyway, looks like you are confused about what polynomials are. They are finite sums of the form $\sum_{i=0}^mc_ix^i$, $c_i\in K, m\in\Bbb{N}$. Here $x$ is an unknown, and it is not to be viewed as an element of $K$. Also, two polynomials are (by definition!!) equal if and only if all their corresponding coefficients are equal. In particular the above polynomial is the zero element of $K[x]$ if and only if all the coefficients $c_i=0$. But, this is exactly what is required for the monomials to be linearly independent over $K$. – Jyrki Lahtonen May 12 '18 at 19:22
  • Another possible source for confusion is the linearly independent OVER $K$ -part. Up to this point you should have covered abstract vector spaces over an arbitrary field. – Jyrki Lahtonen May 12 '18 at 19:24
  • And, the relation $x\cdot x-1\cdot x^2=0$ shows that $x$ and $x^2$ are linearly dependent OVER THE RING $K[x]$. In this relation the coefficients, $x$ and $-1$ are elements of the ring $K[x]$, but they are not both elements of the field of constants, so they are linearly independent over $K$. – Jyrki Lahtonen May 12 '18 at 19:27
  • @JyrkiLahtonen Ok i get that thanks. Also another small question which I dont want to make a whole thread for... For a lin. trans. $T : R^2-> R^3$ will the matrix representing T will always be a 3x2 matrix and is this in general true for any field. – NoteBook May 12 '18 at 19:34
  • In case you are troubled with the discussion I had in comments under Jeffery's answer, take a look at this thread for more information about the difference between polynomials and polynomial functions. – Jyrki Lahtonen May 12 '18 at 19:51
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    And, yes, the dimensions of the domain and the range (as vector spaces over whatever field you are working with) determines the size of a matrix representing a linear transformation. – Jyrki Lahtonen May 12 '18 at 19:53

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As said in a comment above, the polynomials $1,x,...x^n$ are linearly independent.

Maybe you have a confusion with the fact that $x^2 = x\times x$ (etc.). In this case, note that $x$ isn't a scalar so it can't be a coefficient of a linear relation.

paf
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  • So the vector represented by x , represents all the polynomials with degree 1? – NoteBook May 12 '18 at 19:06
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    No. Did I say that? The polynomials of degree 1 are the elements of the space generated by 1 and $x$ i.e. the polynomials of the form $a_0+a_1x$, as you may know. – paf May 12 '18 at 19:16