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.

  • 4
    $\begingroup$ Vectors are straight LOL I'll skip the obvious jokes. A vector space is any set with an addition and scalar multiplication defined that satisfies the axioms. The idea is that we are motivated by our thoughts of vectors in Euclidean n-space, but we then abstract the key features so that we can apply our geometric insight in far more general settings. We call such abstract vectors being on the down-low. No just kidding that was just one of the many bad jokes I promised not to make. $\endgroup$ – user4894 Nov 1 '14 at 20:56
  • 3
    $\begingroup$ Why polynomials form a vector space? Because the set of polynomial with 2 operations ("+", "x") do satisfy the axioms of vector space. $\endgroup$ – SamC Nov 1 '14 at 20:59
  • 1
    $\begingroup$ A mathematical vector space is defined abstractly - I suggest you look up vector spaces online or in a text book, and check the definition. One motivating factor for looking at vector spaces is that Euclidean space fits the definition, and you can do geometry using vectors. But having abstracted the key idea from the Euclidean case, we find that there are lots of other objects which fit the definition - that's one of the reasons it is so useful. The ring of polynomials in one variable over a field is an infinite dimensional vector space. $\endgroup$ – Mark Bennet Nov 1 '14 at 21:04
  • $\begingroup$ Perhaps your own comment "because they are done the same way as any other vector problem" is the key thing here. Since there is basically no difference, why make a difference? We can expand the meaning of the word "vector", and include any type of objects that we can form linear combinations of. $\endgroup$ – Hans Lundmark Nov 1 '14 at 21:40
  • 1
    $\begingroup$ To get the best answers, I think you should write down in your question what your idea of a vector space is and what you are expecting it to do for you. This way, the replies can be tailored to proceed from your current understanding to a more comprehensive one. As it stands, people are giving definitions and axioms disregarding your own idas about what a vector space is (and piling up notions like ring and field that I think will only serve to confuse you further). $\endgroup$ – guest Nov 2 '14 at 4:22

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

  1. $+_V$ is associative and commutative and has a neutral element, and has inverses for every element of $V$.
  2. $\times_V$ associates with real multiplication: $a\times_V(b\times_V v)=ab\times_V v$ for all $a,b\in \mathbb R$ and $v\in V$.
  3. $\times_V$ 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.

  • 1
    $\begingroup$ I also think that it's important to note that the notion of vector doesn't necessarily implies the existence of a coordinate system, a metric system; it's also a bit counterintuitive because a vector space requires a field, and usually a field is like the $x$ and $y$ in a cartesian system, but a vector space, at least from the mathematical point of view, doesn't really has a mandatory coordinate system . $\endgroup$ – user2485710 Nov 1 '14 at 22:35
  • 3
    $\begingroup$ As a non-mathematician, I always saw Vectors as 'lists of numbers', or arrays if you will. Indeed, any polynomial can be expressed as a list of the coefficients for each power. $\endgroup$ – Mark Jeronimus Nov 2 '14 at 1:02
  • 1
    $\begingroup$ @Mark: Yes, that meaning is common in (among other places?) programming languages -- but that is a different generalization of geometric vectors than the one used in linear algebra. For example the set of "all functions $\mathbb R\to\mathbb R$, whether continuous or not" is a vector space too, and such a function cannot be represented simply as a list of numbers. $\endgroup$ – Henning Makholm Nov 2 '14 at 10:57
  • $\begingroup$ $1 \times_V v = v, \forall v \in V$ is missing! I don't think it can be omitted $\endgroup$ – Geralt of Rivia Oct 7 '17 at 12:32
  • $\begingroup$ Not a bad answer, but it would have been better if you had actually USED the abstract definition you gave on polynomials to show that they are indeed vector spaces $\endgroup$ – AmateurMathPirate Mar 30 '18 at 18:45

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.

  • 3
    $\begingroup$ 'Vector space' is actually a criminally misleading term. Why aren't we all using the more neutral term 'linear space' in our undergraduate courses and note that this is the same as 'vector space' but for psychological reasons we will reserve the latter term for tuples of real or complex numbers (behind the scenes: basis-dependent realizations of linear spaces)? I spend copious amounts of time scooping these preconceived notions out of my students' brains, something that could be avoided to a large extend by a simple switch in terminology. Many authors underestimate the power of terminology! $\endgroup$ – guest Nov 2 '14 at 4:09

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.


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).

  • $\begingroup$ Can you explain "the ring of polynomials over a field" a bit more in depth. $\endgroup$ – tokola Nov 1 '14 at 20:55
  • $\begingroup$ Answer edited slightly and links added. $\endgroup$ – Matt Samuel Nov 1 '14 at 20:58

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.


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.


To show that something is a vector space, you simply need to show that all the properties of vector spaces hold.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.