Thanks, guys. Like I said earlier, this particular professor doesn't make things entirely clear during lecture, so you might see me asking questions about linear algebra stuff in here through the semester.
@Rithaniel: Good grief. Applied mathematicians, more than anyone besides differential geometers and representation theorists, need to know a lot of linear algebra.
Someone I help sometimes posted on main an exercise from his homework to find the umbilics on a (specific) tri-axial ellipsoid. I told him good luck :P
My intuition is that there are none, but I haven't thought about it.
Well, I think that I understand the definitions of vector spaces and fields, and the desire to keep things "content free" in a mathematical formulation, but I am used to the idea of building up more complex things from simpler ones - e.g. groups first, etc. But the answers I am getting imply that a scalar is a higher level concept than a field because I cannot define what I mean by scalar without first defining a field.
Oh, I see what you're thinking. You do definitely need to know what a field is, and then you're just talking about elements of that field. More generally, scalars can be thought of as elements of a ring (when you discuss modules rather than vector spaces). But fields are very basic. :)
Most math students, I would say, learn fields LONG before they learn groups, rings, etc.
I mean, we take $\Bbb R$ and $\Bbb C$ as "known" structures, pretty much, and do linear algebra over $\Bbb R$ and $\Bbb C$ without making a big deal about it. Later on, after you've learned groups, rings, etc., you might discuss vector spaces over more general fields.
I would like to be able to say that a scalar is any element in a set that I form that has addition and multiplication analogs that allow you to form a field.
I would like to be able to say that a cabbage is not a scalar. :-)
Is there any context in which a say... a set of matrices are scalars?
The set of diagonal matrices with the same entry down the diagonal (in other words, constant multiple of the identity) can be interpreted as "scalars" when the vector space is the set of (square) matrices.
My honest reaction is that you're making too big a deal out of something, but I can't tell what you're really after.
square matrices form a ring, you have addition and multiplication satisfying all the necessary axioms (but no inverses and no commutativity), so you can do modules over that
(I understand the diagonal matrix point, but that's sort of an edge case). But your earlier comment is helpful. I had thought that there was no context in which matrices could be called scalars.
And I suppose we could multiply cabbages by vectors of cabbages in the obvious way ... or maybe even some other vectors (cabbage times parsnip is another parsnip). :P
I am teaching a course of Math Methods in Physics. (I know, never let a physicist teach math). One approach to defining scalars vs. vectors vs. tensors is how they behave under coordinate transformations. But then I started to look at how a mathematician actually defines a "scalar" and it doesn't connect well to this approach.
@TedShifrin You are confusing a cabbage with the representation of a cabbage :-)
I can confuse things if I want. That's what isomorphism is for.
Yes, physicists muck up tensors badly, although I do agree that the cross product of two vectors should be a pseudo-vector and not a vector (because we suppress the exterior product, which is really what's going on).
You're just going to say that the scalars are unaffected by changes of coordinates to the underlying vector space. Why is that different?
Thank you for the discussion. If a matrix can be a scalar in some mathematical contexts then we are using the words very differently. (I think) (If it is a Cartesian matrix,)
We say that because in computer science a vector can be two column of social security numbers and grades. That set of numbers can't rotate or transform from one element into another.
Well, a vector is an element of a vector space, and we think of the vector space as one copy of itself, as opposed to multiple copies of itself and its dual.
@KieranMullen from my pov, a vector is something that can be scaled and added to other things from the same vector space. The fact that there are transformations of vector spaces is not integral (to me) for what a vector is. How do you even define a transformation of a vector space without knowing what a vector is?
So, using the Sierpiński space centered at 0 or 1 respectively as an example, taking the union of them will give the trivial topology, while their intersection give the discrete topology
However, the multitopology constructed by having all sequences starting at 1 to take the Sierpiński space center at 0 and rise versa for all sequences starting at 0 has a set of convergent sequences intermediate between those of the respective Sierpiński spaces and the discrete topology
Thus the multitopology is something stronger than a topology, of which it forms a tree with the other topologies
For those who want the real deal (which has nothing to do with the stuff I made above), read here: pdfs.semanticscholar.org/c89b/…
Here's the hard way to do the problem: inclusion-exclusion.
There are $25\choose3$ ways to choose 3 squares from the 25.
Now you have to subtract the ways that have two squares in the same row or column. There are 10 ways to choose the row/column, $5\choose2$ ways to choose the two squares in...
You will do the same thing except using permutations instead to take account of the ordering in picking the squares. The relationships between permutation and combination is the factor $r!$ which here it should be $r=3$
Thus $A$ is open if $R(\phi,0,1)$ or $R(\phi,1,0)$
Taking unions gives all the convergent nets. Since the tail of any convergent sequence has to end somewhere, we have this union being $X$. Meanwhile its intersection are all the sequences that converges simultaneously to 0 and 1. This is possible because $\tau$ is not required to be $T_1$
Hence unions, and intersections are both closed and $\tau$ defines a topology
In particular, this topology is unusual because the constant nets are not convergent
But, doesn't that get more complicated if we only have certain pieces that are indistinguishable from one another but distinguishable from others?
I.e. what if we have 3 pawns, 2 rooks, 2 bishops and we are putting them on an 8x8 board such that no two pieces of the same type are in the same row or column.
So having 2 pawns on a row is not allowed but having a pawn, rook, and bishop all in one row is allowed
Another way I can think of modelling this is to use coordinate pairs (x,y)
Then whether pieces share the same column or row reduces to the problem of how many possible ways you can pick (x1,y1), (x2,y2) such that x1=x2 or y1=y2
You do not want the squares selected to be on the same row, right? That is the same as you do not want the coordinates of the squares to share the same x values
Also since you pick squares one by one, you have three coordinate pairs (x1,y1),(x2,y2),(x3,y3).
You can also add an extra parameter to denote what piece you are handling, e.g. 1=rook, 2=pawn etc.
Thus for each piece you have a tuple (t,xi,yi)
where t is the type of the piece, and (xi,yi) is the position of that piece
Then assuming you have 3 pieces, you thus basically have the tuple (t1,t2,t3,x1,x2,x3,y1,y2,y3)
Thus your requirements translates as follows:
"you cant have 2 rooks in the same row" => (t1=t2=2 and x1=/=x2) or (t1=t3=2 and x1=/=x3) or (t2=t3=2 and x2=/=x3)
"you can have a rook and a pawn in the same row" => (t1,t2 in {1,2} t1=/=t2 and x1=x2) or (t1,t3 in {1,2} t1=/=t3 and x1=x3) or (t2,t3 in {1,2} t2=t3 and x2=x3)
I get the logic but I still don't really follow how computing this by hand would really work. How would you work this out with say a 4x4 grid and 2 rooks and 2 pawns?
Is this correct: if we just look at the rook (lets assume t = 2) then each rook position is denoted by (2, x , y) so with the first rook, you have 4 slots for x, 4 slots for y, then for the second rook you either have 4 slots for x and 3 slots for y or 3 slots for x and 4 slots for y. So you have (4x4)[(4x3) + (3x4)] = 384
No wait, you would always have 3 slots for the x value for the second rook, right?
The first rook will have 4 slots for x and 4 slots of y, yes. The second rook will only have 3 slots for x (because no two rooks share the same row) and 4 slots for y
Presumably the ordering of placing the chess pieces should not affect the conclusion, but it is a bit tedious for me to check that by hand, otherwise, you have to compute all possible permutations for each rook and pawn order. That's 4! times of the above reasoning, which the details may differ if the order matters
actually, the order will matter, placing pawns first the first rook can end up with less than 4 slots for rows and columns because pawns already occupied the board
so you have 4!=24 different terms in your sum possibly governed by some symmetry such as the column and row rules of the rooks and pawns
Product rule will probably only apply for some of the terms had the rules are something like, "no two rooks nor two pawns should occupy the same row" then you have a symmetry between rooks and pawns, and some of the 24 cases will be degenerate
The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other; thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general n queens problem of placing n non-attacking queens on an n×n chessboard, for which solutions exist for all natural numbers n with the exception of n = 2 and n = 3.
== History ==
Chess composer Max Bezzel published the eight queens puzzle in 1848. Franz Nauck published the first solutions in 1850. Nauck also extended the puzzle...
The 8 queen problem is basically a subset of the problems we tried to model above, so yes, expect it to be very hard if there is little symmetry
I realize this is a fairly basic combinatorial problem, but I'm still confused.
In how many was can 2 black and 2 white rooks be placed on a standard chessboard such that white and black don't attack each other.
The way I saw it first, there are 2 cases:
Case 1. Both white are on the same ro...
It seems there is a variant of the Kelvin-Stokes theorem stating something along the lines of $\int \mathrm d\mathbf r\times\mathbf F=\iint (\mathrm d\mathbf S\times\nabla)\times\mathbf F$. I hope it has a tidy derivation. approach0 doesn't seem to help, but I found a couple MSE posts basically deriving individual components of it. I'd like to see a reference/standard-derivation for this form. Do you think it's worth posting a reference-request question for?
I don't know what the notation means but if that's true it is going to just be a rephrasing of the general form of Stokes' theorem with differential forms to the special case of objects in R^3. This is a very common phenomenon.
I just realized addition in an elliptic curve agrees with addition of points in the Picard group of the elliptic curve. That's so surprisingly natural.
(In fact, Pic E = E x Z by some pure homological algebra arguments. The Z copy comes from the degree homomorphism Pic E -> Z)
If P, Q, R are distinct points of E, then P + Q + R = 0 means there's a line through P, Q, R in CP^2. Restrict equation of that line to E to get a meromorphic function. This is the easy direction.
Does Halmos conclude B is not a set (last page of Section **2** and last page of Section **3** in Naive Set Theory ) ? Book Link from google: https://piazza.com/class_profile/get_resource/is25bi6c6o1oh/isv14pknyni2z3
@NewStudent Is there a phrase in section 2 that you're focusing on where it sounds like that is being said? If you were already familiar with Russell's paradox or thinking about section 3, it's important to note that Halmos's B in section 2 is not the same problematic set as in alluded to in section 3.
@MikeMiller yes, but 1. That rephrasing is tidy for people not very familiar with the general form of Stokes' Theorem and 2. Whether it uses differential forms or not, I'd like to see a fairly direct derivation (maybe you could write one). I think I probably will post a question once I can gather my thoughts/references.
I'm suggesting that if you want to see a proof you should try to derive it from the general theorem and then rephrasing. I am not interested in writing such down.
I am certainly aware that the differential forms phrasing is not familiar to most Calculus students (and hence physicists, engineers, etc).
@MikeMiller thanks for your attention/response. I think I wasn't very clear that I've already proved it through those sorts of methods, I just don't have an elegant derivation.
This was really involved to write on chat, can someone have a look at this question I posted: https://math.stackexchange.com/questions/3362151/generalizing-chains-of-connected-sets-to-arbitrary-indexing-sets
@MarkS. the only intelligent remark that comes to mind for me is that, since your formula is vectorial, you can compute components by dotting with a constant vector $\vec{c}$
At which point the LHS becomes $$\int \vec{c}\cdot(d\vec{r}\times \vec{F})=\int d\vec{r}\cdot(\vec{F}\times \vec{c})$$
In which case the usual form of Stokes-Kelvin applies
does anybody know anything about quasi-nilpotent operators? ive been looking through google for any useful general result about them, they seem to be hell