6
$\begingroup$

This question already has an answer here:

In Axler's Linear Algebra Done Right, in the introduction to vectors it says:

We could define a multiplication on $\mathbf{F}^n$ in a similar fashion [to addition], starting with two elements in $\mathbf{F}^n$ and getting another element of $\mathbf{F}^n$ by multiplying corresponding coordinates. Experience shows us that this definition is not useful for our purposes.

In what ways was this definition tested? And how did it fail? Is there some fundamental reason why this definition of multiplication is bad for a vector space?

Are there any examples of successful application this definition of multiplication?

$\endgroup$

marked as duplicate by Rahul, user91500, mrp, yo', José Carlos Santos Jun 27 '17 at 10:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ For reference, what you refer to is called the Hadamard product $\endgroup$ – Mark Jun 27 '17 at 5:47
  • 3
    $\begingroup$ You can do it if you want, but then the structure you're describing is not a vector space anymore and you're no longer doing just linear algebra. $\endgroup$ – Qiaochu Yuan Jun 27 '17 at 5:50
  • 3
    $\begingroup$ @OfekGillon it does have a unit, the all 1s vector. $\endgroup$ – Mark Jun 27 '17 at 5:52
  • $\begingroup$ What? $v=(1,1,1,...,1,1)$ is a multiplicative identity. For all u in *F**^n uv=v*u=u $\endgroup$ – D. W. Jun 27 '17 at 5:53
  • $\begingroup$ Of course! My bad! I meant a totally different thing, there are "zero dividers" $\endgroup$ – Ofek Gillon Jun 27 '17 at 5:56
3
$\begingroup$

If one equips $F^n$ with usual vector addition and component-wise multiplication, then it becomes a bona fide commutative ring called the direct product $F\times F\times\cdots\times F$ of $n$ copies of the ring (in fact field)$~F$. If you also want scalar multiplication by elements of $F$ in the picture, then this ring becomes an $F$-algebra.

So this approach is not unheard of. It just does not sit well in the context where linear algebra is studied.

One of the main characteristics of linear algebra is the high degree of symmetry that the objects it studies (namely vector spaces) possess. In the case of a finite dimensional space, one can choose almost any collection of the correct number (the dimension$~d$) of vectors, and use that as a basis to establish an isomorphism with a standard vector space $F^d$; it suffices to avoid the linearly dependent families, which (among the families of $d$ vectors) are quite rare. The fact that so many choices give essentially the same result, and therefore the space has a huge number of symmetries to itself, is really a distinguishing feature of linear algebra. If you weaken the condition of working over a field to working over a ring (even a very nice one like a PID) then you loose this overwhelming symmetry aspect (though some symmetry does remain) and this radically changes the subject. Even groups, which are the essence of symmetry itself, do not have the kind of symmetry that vector spaces have.

(On the other hand, note that working over a skew field, also known as division ring, the symmetry properties are preserved, and it is quite reasonable that a beginning course in Linear Algebra should work over a skew fields. But as far as I know nobody except Bourbaki or R. Godement actually does that.)

So here is why the component-wise product is really not useful in linear algebra: it completely breaks the symmetry property of vector spaces. Unlike nonzero vectors in $F^n$, nonzero elements of the ring $F\times F\times\cdots\times F$ come in many kinds (there are for instance invertible ones, zero divisors, and idempotents), and there are consequently very few symmetries of the ring $F\times F\times\cdots\times F$ (though permutation of the factors still count). By comparison, the introduction of a scalar product does break symmetry somewhat, but to a lesser extent. Instead of the freedom of all bases, one has to restrict to orthonormal bases to get the same results among all choices. And in addition, in some applications (notably from physics) the symmetry is already reduced in the situation one intends to study (nature singles out a specific meaningful inner product).

$\endgroup$
9
$\begingroup$

The componentwise product is occasionally useful, in fact it has a name (the Hadamard product). But it is less fundamentally important than the dot product because it does not have a nice geometric interpretation, it is not invariant under rotations and translations, it does not generalize nicely to an abstract vector space, and probably other reasons too. One reason the dot product is so useful is that we always want to write a vector as a linear combination of some other vectors, and the dot product helps us do that (at least in the case where the other vectors form an orthonormal set).

$\endgroup$
  • $\begingroup$ Wait... Isn't the L2 inner product (at least for the reals) the integral of the element wise multiplication of two functions? $\endgroup$ – D. W. Jun 27 '17 at 8:27
  • $\begingroup$ @D.W. Yes, why do you ask? $\endgroup$ – littleO Jun 27 '17 at 8:33
  • $\begingroup$ @D.W. And the standard dot product is the sum of elementwise products. Summing all the products gives it much better invariance properties, which is why it's more useful than the Hadamard product itself. (also, over complex spaces, inner product comes down to multiplying one number with conjugate of the other) $\endgroup$ – Wojowu Jun 27 '17 at 8:34
  • 2
    $\begingroup$ I think the term "Hadamard product" is used with matrices, not particularly for individual column (or row) vectors. Though of course one is allowed to reinterpret such vectors as slim matrices. $\endgroup$ – Marc van Leeuwen Jun 27 '17 at 8:43
  • $\begingroup$ @MarcvanLeeuwen Ah, good point, thanks. $\endgroup$ – littleO Jun 27 '17 at 9:03
2
$\begingroup$

Vector addition is useful because it has physical and geometric interpretations (like the combination of two or more forces in physics and the "parallelogram rule"). The dot product is useful because it is related to length and angle, through the formulas $||v||^2 = v \cdot v$ and $||u||||v||\cos \theta = u \cdot v$. The cross product is useful because it produces orthogonal vectors (and it also has a geometric interpretation -- $||u \times v||$ is the area of the parallelogram spanned by $u$ and $v$).

Does the multiplication you're describing have any interpretations like these? I would guess not.

$\endgroup$
1
$\begingroup$

The multiplication described depends on the choice of basis - and who in their right mind would work with a specific basis in a vector space?

$\endgroup$
  • $\begingroup$ The dot product also depends on a choice of basis in this sense. $\endgroup$ – Toby Bartels Jun 27 '17 at 10:13
1
$\begingroup$

There isn't really any difference between, for example, doing an algebraic calculation with $\mathbf{F}^2$ as you described and doing two separate algebraic calculations with $\mathbf{F}$.

Using $\mathbf{F}^n$ can be a nice bookkeeping tool when you need to keep track of $n$ different things at once. Or if $\mathbf{F}^n$ appears in a higher algebraic calculation, it tells us we can split the object of study apart into $n$ pieces.

But the detailed study of $\mathbf{F}^n$ doesn't really tell us anything that we didn't already know.

$\endgroup$
1
$\begingroup$

The dot product doesn't change when you rotate the vectors. Proof: It satisfies $\|a+b\|^2=\|a\|^2+2a\cdot b+\|b\|^2$, and the magnitude function is invariant under rotation.

On the other hand, this (let's call it $\otimes$) does. Proof: $(1,0)\otimes(0,1)=(0,0)$, but if you apply a forty-five degree rotation, you get $(\frac{\sqrt2}2,\frac{\sqrt2}2)\otimes (-\frac{\sqrt2}2,\frac{\sqrt2}2)=(-\frac12,\frac12)\ne(0,0)$.

$\endgroup$
  • $\begingroup$ Compare that first identity to the cosine theorem and the Pythagorean theorem, by the way $\endgroup$ – Akiva Weinberger Jun 27 '17 at 10:01
  • $\begingroup$ *law of cosines $\endgroup$ – Akiva Weinberger Jul 12 '17 at 18:22

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