Polynomials as vector spaces? Can someone please explain how polynomials are vector spaces? I don't understand this at all. Vectors are straight, so how could a polynomial be a vector. Is it just that the coefficients are vector spaces? I'm really not understanding how polynomials tie in to linear algebra at all? I can do all the problems with them, because they are done the same way as any other vector problem, but I can't understand intuitively what I'm doing at all.
 A: When we imagine vectors, we all see arrows from our undergraduate physics lessons. And it's fine.
However, what is a vector space ? Well it's any set where you can add elements between them with proportionality factors. And you can definitely add polynomials and multiply them with reals.
I think you just need to accept that the graphical representation you like is only a way to remember the only properties that define a vector : you can add them together, and you can "extend" them by multiplication with a real (or a complex number of course...).
I've always thought that "vector space" is too complex a word to simply represent the possibility of adding together and "extending" the elements of a set.
A: A vector space has two kinds of things: vectors and scalars.  It must be possible to add two vectors, and it must be possible to multiply a vector by a scalar.  There are also some laws that the addition and the multiplication must obey, such as $s\cdot(\mathbf{a} + \mathbf{b}) = s\cdot\mathbf{a} + s\cdot\mathbf{b}$.
One example of this is to take the vectors to be $4$-tuples of real numbers, say $\langle a,b,c,d\rangle$, and the scalars to be single real numbers.  We can add two vectors (by adding them component-wise) and we can multiply a vector by a scalar, multiplying each of the components of the vector by the scalar.
Another example of this is to take the vectors to be $2\times 2$ matrices of real numbers $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ and the scalars to be single real numbers.  We can add two of these vectors using ordinary matrix addition, and we can multiply a matrix by a scalar, multiplying each of its entries by the single number.
Of course this example is exactly the same as the one in the previous paragraph.  The vectors behave exactly the same way whether we write them as $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ or as $\langle a,b,c,d\rangle$.  It doesn't matter whether the brackes are curvy or angled, or whether we write the numbers piled up or not.
Here is another example: take the vectors to be third-degree polynomials, say $ax^3 +bx^2+cx+d$, and the scalars to be real numbers.  We can add two of these vectors using ordinary polynomial addition, and we can multiply a vector by a scalar by multiplying the coefficients of the polynomial by the single number.
This example is just like the previous two.  We could abbreviate $ax^3+bx^2+cx+d$ as $\langle a,b,c,d\rangle$ (since the parts involving $x$ are aways the same, we could just agree not to write them down) or as $\begin{pmatrix}a&b\\c&d\end{pmatrix}$.  It doesn't matter if we write the four numbers in a line, or in a pile, or with $x^3$es and plus signs in between; the vectors are still behaving the same way.
So why bother to do it?  This is one of the great strengths of mathematics, to see when two different-seeming kinds of objects are actually the same, and to develop an abstract theory that applies in many different situations.  Once we develop a theory of vector spaces, we can apply that theory to all sorts of things that behave like vectors—such as polynomials—even if we don't normally think of them as vectors.  In short, by understanding polynomials as vectors, we can use the theory of vector spaces to help us solve problems that are about polynomials.
For example, in Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ I used the theory of vector spaces to show that the required polynomial actually exists, and my method for calculating that polynomial draws heavily on the theory of vector spaces; it boils down to the problem of finding the coordinates of a certain vector in a different basis.
A: A (real) vector space means, by definition any set $V$ together with operations ${+_V}: V\times V\to V$ and ${\times_V}: \mathbb R\times V \to V$, such that


*

*$+_V$ is associative and commutative and has a neutral element, and has inverses for every element of $V$.

*$\times_V$ associates with real multiplication: $a\times_V(b\times_V v)=ab\times_V v$ for all $a,b\in \mathbb R$ and $v\in V$.

*$\times_V$ distributes over $+_V$ in the sense that $a\times_V(v+_Vw)=(a\times_V v)+_V(a\times_V w)$ and $(a+b)\times_V v=(a\times_V v)+_V(b\times_V v)$for all $a,b\in \mathbb R$ and $v,w\in V$.


It happens that if you take the set of all polynomials together with addition of polynomials and multiplication of a polynomial with a number, the resulting structure satisfies these conditions. Therefore it is a vector space -- that is all there is to it.
It can be useful intuitively to visualize vectors as little arrows (or however you're used to thinking about geometric vectors), but this is just intuition which may or may not be helpful -- it is deliberately not part of the linear-algebra concept of a vector space.
A: The ring of polynomials with coefficients in a field is a vector space with basis $1, x, x^2,x^3,\ldots$. Every polynomial is a finite linear combination of the powers of $x$ and if a linear combination of powers of $x$ is 0 then all coefficients are zero (assuming $x$ is an indeterminate, not a number).
A: You are thinking of a vector as being a geometric object. They are straight lines when the space is $\mathbb{R}^n$. However vector spaces are more general than this and intuitive visualizations fall apart.
Vector spaces have is any space which has certain properties, such as closure under addition and properties of scalar multiplication.
http://en.wikipedia.org/wiki/Vector_space#Definition
To show that something is a vector space, you simply need to show that all the properties of vector spaces hold.
A: Perhaps it is best to define what a vector space is. First, they must be over a field; in this case, probably $\mathbb{R}$ or $\mathbb{C}$. For polynomials it is best to think of certain nice properties it satisfies. For example, $a(P(x) + Q(x)) = aP(x) + aQ(x)$ or $(a+b)P(x) = aP(x) + bP(x)$ would be two distributive properties it must satisfy. Look up what a vector space is for more of the axioms.
When you think of taking $w = cu+v$ with $u$ and $v$ going in the same direction, you find that $w$ goes in the same direction (up to orientation). Here you are thinking of the closed subspace of vectors on a given line. For polynomials, your "direction" might be that $P(0) = 0$. Then the direction isn't changed by adding polynomials or scaling by a real or complex number. But here your notion of "direction" that has changed. A vector space abstracts a geometric construction and so it can be impossible to explain these notions geometrically.  
A: Polynomial is just a function. It can have integer, decimal, vector, or matrix inputs. let $P(x)$ bedefined by,
$P(x) = 1 + 3x^2  = \sum{}_{i=0}^{k=2}\alpha_i x^i$
where $\alpha = [1,0,3]$. Given a matrix $A$,
$P(A) = 1 + 3A^2  = \sum{}_{i=0}^{2}\alpha_i A^i$
As it turns out, any vector $v$ with the same dimension as some square matrix* $A$ can be expressed as some polynomial P(A) so long as we have freedom to choose our basis. To see this requires some tricky albeit basic math.
Let $u$ be some basis vector we don't yet know the value of. I claim the following is valid,
$T^jv = \sum{}_{i =j}^{2}\alpha_{i-j} T^iu$
The above equation forms a triangular system of equations that can be solved for beginning with the highest index $j=2$. By 'triangular' I mean at the highest index $j=2$ you get only one term in the summation on the right side. We already know everything on the left (because we know T and v are arbitrary*).
What we are solving for are the values  $T^iu $ on the right side. For $j=2$ we can solve for $T^2u$. Then we use this in turn to move down to $j=1$ and solve $Tu$. We now know the values of $T^2u$ and $Tu$ which allows us to go down (or up depending on how you see it) to $j=0$ and solve for $u$ itself.
We then have $u_0=u, u_1= T^u, u_2 = T^2u$ as our basis vectors which we have represented v in.
But did we represent v in terms of these vectors? Yes. If you perform the above steps the equation you get for $j=0$,
$v = \sum{}_{i =0}^{2}\alpha_{i} ui
\  = \sum{}_{i =0}^{2}\alpha_{i} T^iu  
\  = P(A)u $
*These statements are not intended to be rigorous. More exact stipulations are warranted and discussion is welcome.
