4 ms·

> aren't vectors just Nx1 matrices? It's been a while since I formally studied maths, but - can you describe a situation in which this is _not_ true? (modulo t

by scubbo 6y ago

> aren't vectors just Nx1 matrices? It's been a while since I formally studied maths, but - can you describe a situation in which this is _not_ true? (modulo transposition)

dan-robertson 6y ago
Well if you think of matrices as linear maps then saying a vector is just an Nx1 matrix is like saying that a vector is just a linear map from a vector space of dimension 1 to a vector space of dimension N. It isn’t really a very useful thing to say
- tomjakubowski 6y ago
  Sure it's useful, if not earth-shattering - another way of saying it is that a vector maps scalars to vectors on the same line.
crdrost 6y ago
So, I would say that there is a canonical isomorphism between them but that they are not quite the same. So, to take a step back, I'd say that a covector is a linear function from a vector input to a scalar output (those terms themselves being somewhat fungible, the space of "vectors" needs to probably form a module over the "scalars" which must themselves be a ring—the issue is that I would as a physicist include scalar fields but you can have two nonzero fields multiply to give a zero field, so they do not obey the field axioms). Then I would say that an [m n]-tensor is technically a multilinear function from m covectors and n vectors to a scalar. This means that a [0 1]-tensor is really exactly a covector, but a [1 0]-tensor is not really exactly a vector. One can manifestly be constructed from any vector v (take any covector c and produce the scalar c(v)) and we generally have some criteria that I forget which ensures that this is not just an injection but also a bijection implying some inverse, so that these [1 0] "co-covectors" are indeed isomorphic to the vectors. And then of course this becomes much easier in a metric space where we identify a canonical bijection between vectors and covectors called the "metric", this gives a much more direct way by which co-covectors are isomorphic to vectors. Hopefully that all makes sense. The situations in which this is not true are non-metric spaces where linear maps from (linear maps from vectors to scalars) to scalars are just a subset of the "apply to this vector" instances, but other linear maps are maybe also included. Then there is a more dramatic statement which I think requires the space to be Hausdorff-paracompact or something, which is the idea that of course we have the outer product which naturally combines an [a b]-tensor and a [c d] tensor into an [a+c, b+d]-tensor ("take these vectors and covectors, give them to the first tensor, take those vectors and covectors, give them to the second tensor, then take the two scalar outputs and multiply them together), and we claim that actually an [m n]-tensor can always be realized as a finite sum of outer products of m [1 0]-tensors with n [0 1]-tensors, which allows you to do arbitrary index contractions in a geometric way.
tgb 6y ago
In an N dimensional vector space, yes, you can always choose a basis and write vectors down that way, but it turns out to often be a bad way to work. For example, picture a surface. At any point on the surface, there is a plane of vectors tangent to that surface. It's unnatural to describe the vectors there in terms of a 2x1 matrix, since that requires the choice of a basis of the tangent space. And if you try to choose a basis of the tangent space, you'd probably want it to vary continuously from one point on the surface to the next, but that's not possible to do on many surfaces, such as a sphere, thanks to the Hairy Ball Theorem. [1] [1] https://en.wikipedia.org/wiki/Hairy_ball_theorem https://en.wikipedia.org/wiki/Hairy_ball_theorem
joppy 6y ago
It’s more of a perspective and opinion thing. For example I would not say that an integer is its twos-complement representation, but I am very happy to hold onto that representation as a kind of key for the “real thing”. Similarly, in a vector space there can be abstract vectors, which exist no matter of what basis you choose to write them down. (Choosing a basis = choosing a way to encode them as a 1xN or Nx1 matrix). Often in the theory, general theorems can be proven very efficiently using these abstract vectors, but most of the time when one wants to actually do things to those vectors, picking a convenient basis and working with N coordinates is the best choice.

by scubbo 6y ago

> aren't vectors just Nx1 matrices? It's been a while since I formally studied maths, but - can you describe a situation in which this is _not_ true? (modulo transposition)

dan-robertson 6y ago

Well if you think of matrices as linear maps then saying a vector is just an Nx1 matrix is like saying that a vector is just a linear map from a vector space of dimension 1 to a vector space of dimension N. It isn’t really a very useful thing to say

tomjakubowski 6y ago
Sure it's useful, if not earth-shattering - another way of saying it is that a vector maps scalars to vectors on the same line.

crdrost 6y ago

So, I would say that there is a canonical isomorphism between them but that they are not quite the same. So, to take a step back, I'd say that a covector is a linear function from a vector input to a scalar output (those terms themselves being somewhat fungible, the space of "vectors" needs to probably form a module over the "scalars" which must themselves be a ring—the issue is that I would as a physicist include scalar fields but you can have two nonzero fields multiply to give a zero field, so they do not obey the field axioms). Then I would say that an [m n]-tensor is technically a multilinear function from m covectors and n vectors to a scalar. This means that a [0 1]-tensor is really exactly a covector, but a [1 0]-tensor is not really exactly a vector. One can manifestly be constructed from any vector v (take any covector c and produce the scalar c(v)) and we generally have some criteria that I forget which ensures that this is not just an injection but also a bijection implying some inverse, so that these [1 0] "co-covectors" are indeed isomorphic to the vectors. And then of course this becomes much easier in a metric space where we identify a canonical bijection between vectors and covectors called the "metric", this gives a much more direct way by which co-covectors are isomorphic to vectors. Hopefully that all makes sense. The situations in which this is not true are non-metric spaces where linear maps from (linear maps from vectors to scalars) to scalars are just a subset of the "apply to this vector" instances, but other linear maps are maybe also included. Then there is a more dramatic statement which I think requires the space to be Hausdorff-paracompact or something, which is the idea that of course we have the outer product which naturally combines an [a b]-tensor and a [c d] tensor into an [a+c, b+d]-tensor ("take these vectors and covectors, give them to the first tensor, take those vectors and covectors, give them to the second tensor, then take the two scalar outputs and multiply them together), and we claim that actually an [m n]-tensor can always be realized as a finite sum of outer products of m [1 0]-tensors with n [0 1]-tensors, which allows you to do arbitrary index contractions in a geometric way.

tgb 6y ago

In an N dimensional vector space, yes, you can always choose a basis and write vectors down that way, but it turns out to often be a bad way to work. For example, picture a surface. At any point on the surface, there is a plane of vectors tangent to that surface. It's unnatural to describe the vectors there in terms of a 2x1 matrix, since that requires the choice of a basis of the tangent space. And if you try to choose a basis of the tangent space, you'd probably want it to vary continuously from one point on the surface to the next, but that's not possible to do on many surfaces, such as a sphere, thanks to the Hairy Ball Theorem. [1] [1] https://en.wikipedia.org/wiki/Hairy_ball_theorem https://en.wikipedia.org/wiki/Hairy_ball_theorem

joppy 6y ago

It’s more of a perspective and opinion thing. For example I would not say that an integer is its twos-complement representation, but I am very happy to hold onto that representation as a kind of key for the “real thing”. Similarly, in a vector space there can be abstract vectors, which exist no matter of what basis you choose to write them down. (Choosing a basis = choosing a way to encode them as a 1xN or Nx1 matrix). Often in the theory, general theorems can be proven very efficiently using these abstract vectors, but most of the time when one wants to actually do things to those vectors, picking a convenient basis and working with N coordinates is the best choice.