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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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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. – James Fennell Aug 23 '12 at 15:11
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 $2$. – Matthew Pressland Aug 23 '12 at 15:13
sorry about the typos, it's been corrected – dan Aug 23 '12 at 15:41
@JamesFennell I don't think "it isn't a domain" is a very useful criterion to reject ring structures... :) – rschwieb Aug 23 '12 at 15:57
@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... – rschwieb Aug 23 '12 at 16:14
up vote 17 down vote accepted

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

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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? – dan Aug 23 '12 at 15:30
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. – Ahriman Aug 23 '12 at 15:51
@Ahriman The product is completely natural and elementary, it's just not useful as "a product on a vector space". – rschwieb Aug 23 '12 at 16:03
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. – Ahriman Aug 23 '12 at 16:06
@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 :) – rschwieb Aug 23 '12 at 16:35

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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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. – Marc van Leeuwen Aug 26 '12 at 11:53
@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. – Giuseppe Negro Aug 26 '12 at 12:29
@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. – Giuseppe Negro Aug 26 '12 at 12:43

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.

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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.

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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 for the geometry of these hyperbolic numbers.

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Thanks James. Do you know why this multiplication isn't taught in a linear algebra or calculus III course? – dan Aug 23 '12 at 15:46
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. – James S. Cook Aug 24 '12 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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The definition you are suggesting for multiplication does not satisfy the rules for multiplication (at least in the algebraic ring sense). It must satisfy at least associativity and distributivity over addition (there are more, see here). You can test for yourself that $a \cdot( b \cdot c) \neq (a \cdot b) \cdot c$.

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I think associativity and distributivity do hold. If $a=\langle a_1,a_2 \rangle$, $b=\langle b_1,b_2 \rangle$ and $c=\langle c_1,c_2 \rangle$ then $$a(bc) = a( \langle b_1c_1, b_2c_2 \rangle ) = \langle a_1b_1c_1, a_2b_2c_2 \rangle = \langle a_1b_1, a_2b_2 \rangle c= (ab)c$$Because everything is done component wise all the ring axioms are satisfied. The algebraic problem, as I mention above, is that it's not an integral domain, so elements do not, in general, have an inverse. – James Fennell Aug 23 '12 at 15:15
associativity does hold – Belgi Aug 23 '12 at 15:17
The definition he provided would give $a(bc) = a( \langle b_1 b_2, c_1 c_2 \rangle = \langle a_1 a_2, b_1 b_2 c_1 c_2 \rangle$. – Victor Aug 23 '12 at 15:18
I think the confusion here is that in the text of the question, the OP refers to componentwise multiplication, but the formula given is an entirely different, non-associative, operation. – Matthew Pressland Aug 23 '12 at 15:18
This answer should probably be deleted now (somewhat unfortunately, since the asker didn't behave wrongly) – Ben Millwood Aug 23 '12 at 17:32

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