Scalar product of different sized vectors

[Shadow]

The dot product of two i-sized vectors [math]\vec{A}[/math], [math]\vec{B}[/math] is defined as

 

[math]\sum_i A_iB_i[/math],

 

What if the vectors aren't equal in dimension? What is the scalar product of

 

[math]\vec{A} = (1, 3, 5)[/math] and [math]\vec{B} = (2, 4)[/math]?

[D H]

It's not defined.

[Shadow]

Which means that the cross product isn't either? But the tensor product should be, right?

 

Speaking of tensors, what is [math]\mathbf{\bar{C}}[/math] ? In the context I've seen it, it refers to a 3rd degree tensor, but I'm not sure if it has other meanings (related to tensors)

[D H]

The cross product is defined for 3-vectors period. It does not generalize directly.

The tensor product has the same constraints as the inner product. Dimensions have to match.

[ajb]

The cross product is defined for 3-vectors period. It does not generalize directly.

 

The generalisation to arbitrary dimensions is the wedge product and the Lie algebra of mulitivector fields.

[ydoaPs]

The dot product of two i-sized vectors [math]\vec{A}[/math], [math]\vec{B}[/math] is defined as

 

[math]\sum_i A_iB_i[/math],

 

What if the vectors aren't equal in dimension? What is the scalar product of

 

[math]\vec{A} = (1, 3, 5)[/math] and [math]\vec{B} = (2, 4)[/math]?

 

Doesn't [math]\vec{B} = (2, 4)[/math] imply [math]\vec{B} = (2, 4, 0)[/math]?

[D H]

The generalisation to arbitrary dimensions is the wedge product and the Lie algebra of mulitivector fields.

Not quite the same thing, though. The cross product is a vector (well, almost; it's a pseudo-vector); the wedge product is not a vector, period. The cross product as such can be defined in R³ and R⁷ only. Think of it as an offshoot of the imaginary part of the quaternion and octonion product. Why not R¹⁵, R³¹, ...? The sedenions aren't even alternative. While the Cayley–Dickson construction goes on forever, after a few iterations (reals, complex numbers, quaternions, octonions) its pretty much useless.

[Shadow]

The cross product as such can be defined in R³ and R⁷ only.

 

Shoot, forgot about that. But thanks for the reminder.

 

Why isn't the outer product defined for arbitrary dimensions? I always thought of the result as a sort of multiplication table...

 

If we had [math]\vec{A}=(2, 4)[/math] and [math]\mathbf{B}=

\left( \begin{array}{ccc}

1 & 3 & 5 \\

2 & 5 & 7 \\

4 & 7 & 9 \end{array} \right)[/math]

 

couldn't [math]\vec{A}\mathbf{B} = \mathbf{\bar{C}}[/math], where

 

 

[math]\mathbf{C}_{1ij} = \left( \begin{array}{ccc}

2 & 6 & 10 \\

4 & 10 & 14 \\

8 & 14 & 18 \end{array} \right)[/math] and [math]\mathbf{C}_{2ij} = \left( \begin{array}{ccc}

8 & 24 & 40 \\

16 & 40 & 56 \\

32 & 56 & 72 \end{array} \right)[/math]

[D H]

That's an outer product.

 

Now what are you going to do with it?

 

(In other words, what are you really asking in this thread?)

[ajb]

Not quite the same thing, though. The cross product is a vector (well, almost; it's a pseudo-vector); the wedge product is not a vector, period. The cross product as such can be defined in R³ and R⁷ only. Think of it as an offshoot of the imaginary part of the quaternion and octonion product. Why not R¹⁵, R³¹, ...? The sedenions aren't even alternative. While the Cayley–Dickson construction goes on forever, after a few iterations (reals, complex numbers, quaternions, octonions) its pretty much useless.

 

I did not say the same, I said a generalisation.

 

What is true is that the collection of multivector fields over a smooth manifold is a vector space (or if we do not include zero vectors, a module over smooth functions). (Same holds for all tensors.)

 

The fact that the wedge product holds in any dimensions makes it far more useful than the cross product.

Edited September 15, 2009 by ajb

[the tree]

[math]\vec{B} = (2, 4, 0)[/math]?

If you specify that B is a vector on the (x,y,0) plane, then yes. Which you might want to do if you were discussing, I dunno, the something on a flat surface and something airbourne. But exactly which subspace B occupies has to be specified - I think.

[ydoaPs]

If you specify that B is a vector on the (x,y,0) plane, then yes. Which you might want to do if you were discussing, I dunno, the something on a flat surface and something airbourne. But exactly which subspace B occupies has to be specified - I think.

 

Why would you be trying to multiply vectors in different co-ordinate systems?

[ajb]

Why would you be trying to multiply vectors in different co-ordinate systems?

 

Unless one of the vectors is in a subspace, which as the tree said need to be specified it is confusing. I am not sure what the original question is pointing to.

 

However, we can form the tensor product of two arbitrary vector spaces and ask questions there.

[the tree]

Why would you be trying to multiply vectors in different co-ordinate systems?

Well, that much would be silly, but finding a problem where the 2d vector isn't on the (x,y,0) plane* seems quite natural (the first thing that comes to mind is an aircraft flying with a fixed (non-zero) altitude versus a surface to air missile). Which is why it'd need to be specified to avoid confusion.

 

*osculating plane? Is that the word?

[Shadow]

The tensor product has the same constraints as the inner product. Dimensions have to match.

 

In my example dimensions didn't match...one was a 2x1 matrix, the other a 3x3 matrix.

 

My original question being answered, I was asking if the tensor product was defined for matrices (or any tensors) with different dimensions. Your first answer was no, yet what I did was an outer product, which leaves me confused...

[ajb]

In my example dimensions didn't match...one was a 2x1 matrix, the other a 3x3 matrix.

 

My original question being answered, I was asking if the tensor product was defined for matrices (or any tensors) with different dimensions. Your first answer was no, yet what I did was an outer product, which leaves me confused...

 

There is a bit of confusion here.

 

On a manifold, lets say Euclidean space of a given dimension tensors are sections of various natural bundles. Lets look at vector fields, which are sections of the tangent bundle [math]TM[/math]. The point is that the dimensions of the fibres are the same as the manifold. So, naturally vector fields are of the same "size".

 

Tensors are sections of tensor products of the tangent and cotangent bundle. So, for example matrix should be thought of as a section of [math]TM \otimes T^{*}M[/math], which thus has fibre dimension [math]n^{2}[/math] if the manifold is of dimension [math]n[/math]. So again, matrices are of the same "size".

 

But what you can do is consider more general vector bundles over the manifold and these can have fibre dimensions different to the base manifold. A good example is Lie algebra valued tensors.

 

You can then consider tensor products (and other constructions) of what ever vector bundles you like. The tensor product of two sections is locally just the standard product of the components.

 

So, you can multiply two vectors of different sizes.

 

Similar things hold for more general tensors.

Edited September 15, 2009 by ajb