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February 2[edit]

Relation of transformation matrices to tensors[edit]

Let $T:V\to W$ be a linear transformation from the vector space V to the vector space W, where V and W are finite-dimensional vector spaces over the same field. I know T can be identified as a member of the tensor product $V^{*}\otimes W$ . Firstly, is this the usual order in which to arrange the two factors of the tensor product? Secondly, if A is the matrix of T (which is m by n) with respect to the (ordered) basis $\mathbf {e} _{1},\mathbf {e} _{2},...,\mathbf {e} _{m}$ of W and the (ordered) basis $\mathbf {e} ^{1},\mathbf {e} ^{2},...,\mathbf {e} ^{n}$ of V^*, then does the (i, j)th entry of A correspond to the coefficient of $\mathbf {e} ^{j}\otimes \mathbf {e} _{i}$ ? In other words, do we have $T=\sum _{i=1}^{m}\sum _{j=1}^{n}a_{ij}\mathbf {e} ^{j}\otimes \mathbf {e} _{i}$ (I am deliberately not using Einstein summation notation since I'm not proficient at it)? Would we call i a contravariant index and j a covariant index?

While I understand the construction of the tensor product, I am trying to become proficient at connecting tensors with their coordinate expressions. In particular, I hope to be able to expose others who have only learned transformation matrices to the power of tensors.--Jasper Deng (talk) 02:17, 2 February 2017 (UTC)[reply]

If v*∈V* and w∈W then you can identify v*⊗w with am mapping V→W by v*⊗w(x) = v*(x)w. Note that v*(x) is a scalar, so v*(x)w scalar multiple of w. (If you write functions on the right, as in f(x)=xf, then the rule becomes x(v*⊗w)=(xv*)⊗w), i.e. the associative law.) The matrix of a linear transformation V→W is usually defined relative to a basis for V and a basis for W, say {e_j} is a basis for V and {f_i} is a basis for W. (It gets more confusing if you use the same letter e for V and W.) Define a basis {e*_j} for V* by e*_j(e_k) = δ_jk. If you compute the matrix of e*_j⊗f_i it has 1 in row i and column j, 0 elsewhere. If it works for the basis then it works for everything by linearity, so it looks like you're correct. (I admit though that I'm pretty rusty at this. I could be wrong but it seems like this is a bit backwards from the way I learned it. But I'm using the definitions given in Transformation matrix.) --RDBury (talk) 04:48, 4 February 2017 (UTC)[reply]

Computer data rates[edit]

Coming here just to make sure I've not made some silly mistake.

I'm considering getting a new Internet service plan, and before I sign up for anything, I want to have a solid sense of the speeds I'd be looking at. Do I understand rightly that uploading a 6MB image at 6mbps speed would take eight seconds, as long as the WMF servers don't experience some sort of slowdown? If I understand Data rate units correctly, 1mbps means that a 1/8MB file takes one second to upload, meaning that a 6MB file would take 48 seconds, and such a file on a 6mbps connection would take 8 seconds. However, I don't want to assume that I have this all correct — especially as I'm not sure that I can assume that these speeds scale in the way that I've suggested. Nyttend (talk) 03:43, 2 February 2017 (UTC)[reply]

@Nyttend: At the assumption that your 6 mbps is a constant, steady, rate of data transfer (which itself is already a huge simplification), then your transmission delay will be approximately what you calculated. However, there are other network delays too, which are nonnegligible - queueing delay, propagation delay (latency), processing delay (time needed for each router along the path to make its forwarding decision), to name a few. The calculation also neglects the overhead associated with headers and TCP retransmissions (though these should be small in general, no more than one part in ten for a large file).--Jasper Deng (talk) 06:12, 2 February 2017 (UTC)[reply]

@Nyttend: Also, in reality, DSL plans advertised as "6 mbps" very seldom have that bandwidth realized. I've found that 4-5 mbps is more typical on such plans.--Jasper Deng (talk) 06:29, 2 February 2017 (UTC)[reply]

Thank you, both for the calculation confirmation and for the warning. This plan isn't advertised as "6 mbps"; it's advertised as "Up to 6 mbps", and I was presuming that it wouldn't commonly be all the way there normally. I just wanted to be sure I understood the maximum speed correctly. Nyttend (talk) 12:10, 2 February 2017 (UTC)[reply]

Do note that "upto 6 Mbps" (assuming you mean Mega rather than milli!) could (mathematically and legally, even if not morally) include 1kbps, but will exclude 6.1Mbps. -- SGBailey (talk) 12:43, 2 February 2017 (UTC)[reply]

Understood :-) I've had a very different service with easily-exceeded data limits, and before I buy out the contract, I wanted to make sure that I had an accurate understanding of the statistics given in the advertisement. Thankfully, it's a fixed price, not a "price X for the first six months, and price 5X after that". I've got a laptop and a very fast network at work, so big stuff doesn't have to be done at home anyway. Nyttend (talk) 13:34, 2 February 2017 (UTC)[reply]

From my experience in France (two different landline contracts where I looked carefully the fine print), the "up to x Mbps" is actually formulated as "usually close to x Mbps" - so you have a legal recourse if they fall well short. My guess is that to advertise "up to x", the cable from your modem or equivalent has a legal obligation to be able to carry that (from a hardware and software point of view); and once it's installed, the ISP can as well let you use it since the marginal cost of a packet of data is negligible.

The real limitation is that the backbone network is not large enough to guarantee that everyone should be able to use its full download at any time (which is perfectly reasonable). For instance, I would guess your download speed is significantly delayed on Election Day if your neighbours use HD TV over the internet; in another context, I know for certain text messages sent on New Year's Eve usually arrive with a much larger latency than usual. Tigraan^{Click here to contact me} 14:35, 2 February 2017 (UTC)[reply]

Convenience link: Network performance. Tigraan^{Click here to contact me} 12:20, 2 February 2017 (UTC)[reply]