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July 26

Finding the inverse of a matrix

I was reading Gauss–Jordan_elimination, but I have no idea how it gets from step 2 to step 3 (i.e. "By performing elementary row operations on the [AI] matrix until it reaches reduced row echelon form, the following is the final result:").--115.178.29.142 (talk) 01:12, 26 July 2010 (UTC)[reply]

Have you read elementary row operation and reduced row echelon form? --Tango (talk) 01:27, 26 July 2010 (UTC)[reply]

LaTeX Symbols

I would like to check that I know the meaning of two symbols. Is it true that $\twoheadrightarrow$ is used for an surjective map and $\hookrightarrow$ is used for an injective map? Let U ⊆ V be two vector spaces. Is my use of arrows correct in the following short exact sequence?

0\twoheadrightarrow U\twoheadrightarrow V\hookrightarrow V/U\hookrightarrow 0.

— Fly by Night (talk) 19:53, 26 July 2010 (UTC)[reply]

You are right about the meanings of the arrows (although I would say a hooked arrow was an inclusion map rather than a general injection), but you might want to check the sequence again. How can you have a surjection from the zero space to another space (other than the zero space again)? And how can you have an injection from a non-zero space to the zero space? --Tango (talk) 20:04, 26 July 2010 (UTC)[reply]

I got the arrows mixed up. Thanks Tango!

0\hookrightarrow U\hookrightarrow V\twoheadrightarrow V/U\twoheadrightarrow 0.

Reading around it seems that

\hookrightarrow

for a monomorphism, i.e. an injective homomorphism. I'm not actually interested in any algebraic structure; just yet. — Fly by Night (talk) 20:08, 26 July 2010 (UTC)[reply]

It's unusual to use

\hookrightarrow

or

\twoheadrightarrow

for maps to or from 0, since any such map is automatically injective/surjective. 129.67.37.143 (talk) 22:01, 26 July 2010 (UTC)[reply]

I agree. Personally, I find this use of arrows unnecessary. You could just say that "

0\rightarrow U\rightarrow V\rightarrow V/U\rightarrow 0

" is a short exact sequence and it would be clear which maps are injections and which are surjections. PS T 02:16, 27 July 2010 (UTC)[reply]

My question was to see if I understood the arrows correctly. I gave the short exact sequence as an example. It seems that the answer to my question is that my use of arrows is correct (although not common in the given example). — Fly by Night (talk) 18:46, 27 July 2010 (UTC)[reply]

Boundary between two spatial regions

I know that in a two-dimensional space the boundary between the regions $y\geq x$ and $y<x$ is the line $x=y$ . But I am having trouble visualizing the extension to more dimensions. I believe that the boundary seperating the region $z\geq y\geq x$ from the rest of space must be a two-dimensional surface, but as I said I can't visualize it. What is this surface, and can I derive an equation for it? thanks in advance, 128.223.131.109 (talk) 22:31, 26 July 2010 (UTC)[reply]

You have two inequalities there, so you'll have two surfaces. z≥y gives you the plane z=y and y≥x gives you the plane y=x. --Tango (talk) 22:38, 26 July 2010 (UTC)[reply]

(edit conflict) The boundary as a set won't be a surface. It will be an algebraic variety. — Fly by Night (talk) 22:42, 26 July 2010 (UTC)[reply]

It will be. However, I suspect that the OP would not ask such a question if he/she knew about algebraic varieties ... PS T 02:11, 27 July 2010 (UTC)[reply]

Indeed. Often the most difficult part of answering questions on the reference desk is working out what prior knowledge the OP likely has and how best to answer the question in a way they will be able to understand. The fact that the union of two planes is a type of algebraic variety is not likely to be of any help to the OP in this case. --Tango (talk) 14:31, 27 July 2010 (UTC)[reply]

True. Actually it is a semialgebraic set. Tkuvho (talk) 14:40, 27 July 2010 (UTC)[reply]

It's both, I guess. It is an algebraic variety, since it is the zero set of the polynomial

(z-y)(y-x)

. --Tango (talk) 14:55, 27 July 2010 (UTC)[reply]

Right, I was assuming he was referring to a connected spacial region, in which case the boundary is only semi. Tkuvho (talk) 15:15, 27 July 2010 (UTC)[reply]

The region he refers to is connected, isn't it? (If not, then I need to practise visualising in 3D more...) --Tango (talk) 15:34, 27 July 2010 (UTC)[reply]

V((z-y)(y-x)) cuts the space into 4 parts. If you just take the boundary between the region that does satisfy z ≥ y ≥ x and the region that doesn't, it's shaped like a folded plane and it's not an algebraic variety. Rckrone (talk) 16:36, 27 July 2010 (UTC)[reply]

I think I get what you mean Rckrone, I can picture the planes described by Tango above, and where they would intersect. The surface I am describing is made of these planes, up to the point where they intersect. In this region, some derivatives would be singular and is this why you say it is not algebraic? Any idea of how I could plot this in mathematica? 128.223.131.109 (talk) 17:03, 27 July 2010 (UTC)[reply]

Actually, it's not algebraic because we only have portions of the planes (Rckrone and Tkuvho are absolutely right, I was being an idiot!). The whole planes, despite the singularities where they intersect, would be algebraic. (We do sometimes talk about "non-singular algebraic varieties", which would exclude cases like this.) --Tango (talk) 17:10, 27 July 2010 (UTC)[reply]

Try RegionPlot3D[x <= y <= z, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}] . -- Meni Rosenfeld (talk) 17:59, 27 July 2010 (UTC)[reply]

PST: I, too, believe that the OP wouldn't ask such a question if s/he knew about algebraic varieties. All the more reason to introduce the idea and wikify it. Why not "call a spade a spade"? — Fly by Night (talk) 18:55, 27 July 2010 (UTC)[reply]

Everyone knows what a spade is. Not everyone knows what an algebraic variety is. It just complicates things for little or no gain. --Tango (talk) 19:03, 27 July 2010 (UTC)[reply]

But if someone is going around calling a spade a hammer, then what's wrong with saying "that's a spade, not a hammer"? Surely correcting someone in a polite way, while adding a reference, is the best thing to do. I added a blue link to the article, and it is a very easy article to understand. — Fly by Night (talk) 19:14, 27 July 2010 (UTC)[reply]

Of course, it is fine to introduce the OP to basic algebraic varieties. However, that these sorts of objects are algebraic varieties (or semialgebraic varieties) is just fancy language for something that can be understood without any real machinary. Perhaps if you mentioned twisted cubics, rational normal curves, Segre varities, or more generally determinantal varieties, we would have a fun discussion where the OP would try to visualize and deduce the geometric nature of these varieties. (There are indeed some beautiful geometric properties of these varieties that you can develop with only basic linear algebra; for example, the fact that any n+1 points on the rational normal curve of degree n are linearly independent corresponds to the nonvanishing of the Vandermonde determinant. (Exercise for the OP: Suppose another such curve has this property. (Make precise "such curve".) Must it be projectively equivalent to the rational normal curve? This is an example of the sorts of easily-posed problems classical algebraic geometry has to offer.) In fact there are so many neat properties you can deduce with only visualization and basic algebra/linear algebra at your side.)

Unfortunately, at this point we would be going too far off topic. That is, I guess, the only "flaw" with algebraic geometry. It has a technical side which requires a lot of background to fully appreciate (but is beautiful), and it has a classical side which is very intuitive. Nonetheless, the true beauty of the classical side is only released once you start to move on from the basic examples of (projective) varieties such as finite sets of points, linear subspaces, hypersurfaces etc. But if you are ready to move on from these basic examples, you might as well begin studying algebraic geometry. PS T 00:56, 28 July 2010 (UTC)[reply]