I was just wondering why we don't ever define multiplication of vectors as individual component multiplication. That is, why doesn't anybody ever define $\langle a_1,b_1 \rangle \cdot \langle a_2,b_2 \rangle$ to be $\langle a_1a_2, b_1b_2 \rangle$? Is the resulting vector just not geometrically interesting?

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    $\begingroup$ One of the problems with doing this is that the resulting algebraic system is not an integral domain. If we let $x=(1,0)$ and $y=(0,1)$ then $x \neq 0$ and $y \neq 0$ but $xy = 0$. This also means that $x$ and $y$ do not have an inverse. $\endgroup$ Commented Aug 23, 2012 at 15:11
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    $\begingroup$ I assume your definition of multiplication should be $\langle a_1,b_1\rangle\cdot\langle a_2,b_2\rangle=\langle a_1a_2,b_1b_2\rangle$? This is what I was reading by accident until I saw Victor's answer and got confused. Otherwise this does not obviously generalize to vector spaces of dimension more than $2$. $\endgroup$
    – mdp
    Commented Aug 23, 2012 at 15:13
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    $\begingroup$ @JamesFennell I don't think "it isn't a domain" is a very useful criterion to reject ring structures... :) $\endgroup$
    – rschwieb
    Commented Aug 23, 2012 at 15:57
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    $\begingroup$ @rschwieb - for sure! It seems to me that when this kind of multiplication is proposed it's usually an attempt to create a two dimensional version of the real numbers with the same nice properties, but it not being an integral domain breaks that. In fact, I took a Complex Analysis course where complex numbers (presented as ordered pairs) we're introduced as the solution to this problem, after it was shown that component-wise multiplication has certain "shortcomings". $\endgroup$ Commented Aug 23, 2012 at 16:11
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    $\begingroup$ @JamesFennell In that case I know what you mean, but they are only shortcomings in comparison to the complex numbers! The magic of $\mathbb{C}$ is a little hard to live up to :) Otherwise the coordinatewise product is as natural and nice as the product of groups, or the product of topological spaces... $\endgroup$
    – rschwieb
    Commented Aug 23, 2012 at 16:14

6 Answers 6


Unlike the usual operations of vector calculus, the product $\bullet$ you defined here is not covariant for Cartesian coordinate changes. This means that an equation involving $\bullet$ is not guaranteed to keep holding true if both members undergo an orthogonal coordinate change, such as a rotation of the axes.

For a 2 dimensional example, consider the following equation: \begin{equation} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \bullet \begin{bmatrix} 0 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}. \end{equation} If we rotate the plane 45° counterclockwise then \begin{align} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \to \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix},& & \begin{bmatrix} 0 \\ 1 \end{bmatrix} \to \begin{bmatrix} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix},&&\begin{bmatrix} 0 \\ 0 \end{bmatrix} \to \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \end{align} but \begin{equation}\tag{!!} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} \bullet \begin{bmatrix} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{bmatrix} \ne\begin{bmatrix} 0 \\ 0\end{bmatrix}. \end{equation} From the physicist's point of view, then, this operation is ill-posed as it should be independent of the particular coordinate system one chooses to describe physical space. This is not the case of the dot-product and the cross-product, which are independent of such a choice.

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    $\begingroup$ It may be noted that the usual dot product is not invariant under Cartesian coordinate changes either, only under coordinate changes by an orthogonal matrix. But the set of coordinate changes that behave well with respect to (=commute with) the component-wise product is very small indeed. Coordinate permutations to seem to fit the bill though. $\endgroup$ Commented Aug 26, 2012 at 11:53
  • $\begingroup$ @MarcvanLeeuwen: Yes, you're right, by "Cartesian coordinates" I mean "linear coordinates with respect to an orthonormal basis". I should have emphasized this in the text. $\endgroup$ Commented Aug 26, 2012 at 12:29
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    $\begingroup$ @MarcvanLeeuwen: A natural conjecture arising from your observation is that the biggest group of symmetries of this product $\bullet$ is the group of permutation matrices. By this I mean that if $A$ is invertible and $$(Ax)\bullet (Ay)=A(x\bullet y), \qquad \text{all}\ x, y \in \mathbb{R}^n,$$ then $A$ is a permutation matrix. I wonder if this is true. I guess it is. $\endgroup$ Commented Aug 26, 2012 at 12:43
  • $\begingroup$ Hola @GiuseppeNegro, I think this forces the group to be the identity element: cancelling $A$ from the left: $ x \cdot (Ay) = x \cdot y$ and letting $x$ be $(1,0,0,0,0...), (0,1,0,0,0,0..),...$ shows this. $\endgroup$
    – inquisitor
    Commented Dec 1, 2017 at 16:42
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    $\begingroup$ Is there something roughly similar to Hadamard product, but invariant to coordinate changes by an orthogonal matrix? $\endgroup$
    – max
    Commented Oct 5, 2020 at 6:46

This is Hadamard product, which is defined for matrices, and hence for vector columns. See Wikipedia page : Hadamard Product

  • $\begingroup$ Thanks for the answer. Why don't we do the Hadamard product in a linear algebra or calculus III course? Is it not as useful in some way? $\endgroup$
    – dan
    Commented Aug 23, 2012 at 15:30
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    $\begingroup$ I don't think it's really useful. I never used Hadamard product at all. As said on Wikipedia, Hadamard product has some applications in compression algorithms, but I'm not aware of any others applications. $\endgroup$
    – Ahriman
    Commented Aug 23, 2012 at 15:51
  • $\begingroup$ @Ahriman The product is completely natural and elementary, it's just not useful as "a product on a vector space". $\endgroup$
    – rschwieb
    Commented Aug 23, 2012 at 16:03
  • $\begingroup$ Did I say that it was not natural or elementary ? I just said that I never saw any applications of it in linear algebra. It doesn't mean that there aren't such applications. $\endgroup$
    – Ahriman
    Commented Aug 23, 2012 at 16:06
  • $\begingroup$ @Ahriman Since you said you were not aware of any other applications, I was just filling you in. The rest of my comment agrees with what you said. Don't get bent out of shape :) $\endgroup$
    – rschwieb
    Commented Aug 23, 2012 at 16:35

To elaborate yet a bit more on what Guiseppe Negro, James S. Cook and Michael Joyce have already said:

A vector is not a tuple of individual components. A vector is an element of some vector space.

When you're writing a vector as such a tuple, you're only referring to the expansion of the vector in some particular basis. But this basis is very often not even specified. Which is actually ok, because the "normal" vector operations don't in fact depend on the choice, i.e. if you transformed all you vectors to be written out in some other basis, you would have all different numbers but the same calculations would still yield correct results.

But that wouldn't work for component-wise multiplication, as was already shown. This operation simply does not work on the vectors but on their basis representation, which is only well-defined for some fixed choice of basis, which is not what you're actually interested in when studying vectors.

Of course, there are plenty of applications where you are in fact interested in tuples of numbers, but those aren't vectors then. There's nothing wrong with the Hadamard product, but it doesn't work on vectors but on matrices1. If you want to multiply components, then your objects may be called tuples or arrays or lists or whatever, but hardly vectors.

Unfortunately, many people have got this wrong, and that's why e.g. C++ programmers are blessed with2 an std::vector class that is in fact for dynamic arrays, which are even less accurately vectors than static arrays.

1Matrices suffer from a similar problem: many people use "matrix" and "linear mapping" as synonyms, but they aren't in fact the same; matrices refer to a particular basis while linear mappings need no such thing.

2Nothing against std::vector, it's great – it's just not a vector class, just like "functors" aren't in fact functors.


I think the reason you don't normally see it is just because it doesn't really have an application in linear algebra.

If you look at $\mathbb{F}^n$ as a ring instead of a vector space over $\mathbb{F}$, then what you have suggested (coordinatewise multiplication) is exactly the product ring structure of the ring $\mathbb{F}^n$. It's completely natural, and useful.

It's just not mentioned in linear algebra because you are rarely thinking of $\mathbb{F}^n$ as a ring, you are usually focused upon its vector space identity.

I think a lot of people have probably been cognitively fooled into thinking of your product "like the cross product" or "like the inner product". Those two are really useful in linear algebra, but the coordinatewise product does not compare.


I often see this product as an incorrect answer in my freshman mechanics course. If they told me they thought I wanted the direct product of $\mathbb{R}$ with itself then I suppose I would let them have their points back.

This multiplication has been on my mind lately. The algebra defined on $\mathbb{R}^2$ by this Hadamard product is equivalent to the hyperbolic numbers $\mathbb{R} \oplus j\mathbb{R}$ where $j^2=1$. Let's call your algebra $\mathcal{A}_1$ and the hyperbolic numbers $\mathcal{A}_2$

The isomorphism is given by $\Phi: \mathcal{A}_1 \rightarrow \mathcal{A}_2$ with $\Phi(a,b) = \frac{1}{2}(a+b)+\frac{1}{2}j(b-a)$. Notice that the identity for $\mathcal{A}_1$ is $(1,1)$ and $\Phi(1,1)=1$. Furthermore, $\Phi^{-1}(x+jy) = (x-y, x+y)$ which allows us to see that $\Phi^{-1}(j) = (-1,1)$. In other words, $(-1,1)$ is the "$j$" for the Hadamard product.

The geometry of $\mathcal{A}_2$ is in part exposed by thinking about $j$-multiplication:

$$ j(x+jy) = jx+j^2y = y+jx $$

Multiplication by $j$ reflects about the line $y=x$. This is obviously different than the complex numbers $\mathbb{R} \oplus i\mathbb{R}$ where multiplication by $i$ maps $(x,y)$ to $(-y,x)$. See http://en.wikipedia.org/wiki/Split-complex_number for the geometry of these hyperbolic numbers.

  • $\begingroup$ Thanks James. Do you know why this multiplication isn't taught in a linear algebra or calculus III course? $\endgroup$
    – dan
    Commented Aug 23, 2012 at 15:46
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    $\begingroup$ Well, when I teach calculus III or linear algebra I teach the dot-product in $\mathbb{R}^n$ because it allows me to elegantly create projections onto subspaces of whatever dimension I choose and I teach the cross-product in $\mathbb{R}^3$ because it allows be to elegantly select a vector perpendicular to a given pair of vectors. So far as I know I didn't need the Hadamard product to solve problems of geometry or analysis which arise in the usual discussion. There are tons and tons of these sort of constructions, we can't cover everything. $\endgroup$ Commented Aug 24, 2012 at 1:51

This is really just Giuseppe Negro's answer said differently, but your definition is flawed in the sense that it depends on the choice of a basis. So it's not really a function of the two input vectors, but rather a function of the two vectors and a choice of basis for the vector space.

In Giuseppe's example, he shows that if you expand two particularly chosen vectors in two different bases, you get two different values for your product using your definition.

Contrast that with the dot product. Let's take any two vectors $v, w$ in a vector space and compute the dot product. You open your notebook and expand $v$ and $w$ in terms of your favorite orthonormal basis $e_1, \dots, e_n$, giving $v = v_1 e_1 + \cdots + v_n e_n$ and $w = w_1 e_1 + \cdots + w_n e_n$, so you conclude that $v \cdot w = v_1 w_1 + \cdots v_n w_n$. I, on the other hand, open my notebook and expand $v$ and $w$ in my favorite orthonormal basis $f_1, \dots, f_n$, giving $v = v'_1 f_1 + \cdots + v'_n f_n$ and $w = w'_1 f_1 + \cdots + w'_n f_n$, so I conclude that $v \cdot w = v'_1 w'_1 + \cdots + v'_n w'_n$. The fact that we wind up with the same number for $v \cdot w$ (even though my $v'_i$ will not be the same as your $v_i$ and similarly for the coefficients of $w$) is what tells us that the dot product is really just a function of the vectors $v$ and $w$, not a function of their expansion coefficients in terms of some particular basis.


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