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Jan
19
comment Binary operation commutative, associative, and distributive over multiplication
Nice, but the question wasn't for it to distribute over itself, but to distribute over regular multiplication. (Which this might do, I haven't checked.)
Dec
16
comment Lebesgue Measure - positive measure sets not containing intervals
In part 2., the $\alpha$ is fixed. You can have a positive measure set $A$, such that $m(A \cap I)<1$ for all intervals, but as you take smaller and smaller intervals $I_n$ "zooming in" on a part of $A$, then $m(A \cap I_n)$ arbitrarily approaches 1. Part 2. works because the $\alpha$ bounds this measure away from 1 (by some tiny amount).
Dec
4
comment Minimize the Frobenius norm of the difference of two matrices with respect to matrix: $\underset{B} {\mathrm{argmin}} \left\| A- B \right\|_F$
This is VERY nice, but I must admit I have no idea how you did it. I see what you're doing, but not how. In your very second line, how do you differentiate the Frobenius inner product without first converting it to some other notation and then trying to figure out how to convert back?
Dec
3
comment Minimize the Frobenius norm of the difference of two matrices with respect to matrix: $\underset{B} {\mathrm{argmin}} \left\| A- B \right\|_F$
@nullgeppetto Those diagonal elements will contribute the smallest square sum when their average is 0 (when they are "on average" as close to zero as they can get). Thus taking $b = \mathop{avg}\limits_{i} \{d_A(i)\}$, so that $B$ subtracts off the average of the diagonal of $A$, will minimize this case.
Dec
3
comment Minimize the Frobenius norm of the difference of two matrices with respect to matrix: $\underset{B} {\mathrm{argmin}} \left\| A- B \right\|_F$
@nullgeppetto Picture the graph of a function $d_A(i)$, where this is the $i^{th}$ element on the diagonal of $A$. Since your $B$ is diagonal, no matter how it changes, it won't affect the off diagonal elements. Since the Frobenius norm squares and then sums each element, you want $B$ to modify the diagonal of $A$ so they contribute as little as possible. When $B$ was any diagonal matrix, taking $d_B(i):=d_A(i)$ gave $d_A(i)-d_B(i)=0$, so they contributed nothing. When $d_B(i)=b$, $d_A(i)-d_B(i)$ will slide the diagonal elements of $A$ up and down equally.
Dec
2
comment Is “product” of Borel sigma algebras the Borel sigma algebra of the “product” of underlying topologies?
+Brian I don't mean to nitpick, but for clarity for later readers, the standard product $\sigma$-algebra (generated by cylinders) and the "box" $\sigma$-algebra (generated by rectangles) are still the same for countably infinite products (because you can intersect countably many cylinders to get the rectangles). If you're indexing uncountably many dimensions, every cylinder is an "infinite" rectangle, but you can not intersect uncountably many cylinders to get the smaller rectangles anymore, so the product $\sigma$-algebra is properly contained in the "box" $\sigma$-algebra.
Nov
16
comment Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
Some of us just want hints, and only look for explicit results after failing to be able to work it out. Yes they can find this online, but providing it takes away that choice.
Nov
16
comment Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
Since that's the simplest power series there is, I don't think it should be given out on homework problems. It's easy to compute.
Nov
16
comment Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
It's been a long time since I've worked with generating functions, but it seems you should be able to figure out the power series for $1/(1-z)$ and $1/(1-2z)$, figure out the constants $C$ and $D$, and then you have $A(z)$ as a linear combination of two other series, so combine them term by term.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Check the comments above, under the other Travis' answer. When you choose $j^2=-1, ij=k, ji=k$ and get a commutative set, it's isomorphic to $\mathbb C^2$.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Yep, that checks out. i $=(i,i)$, j $=(i,-i)$, and k $=(-1,1)$.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Perhaps, but any ring forms a left-algebra over itself though, right? So if they had other division rings, wouldn't they already have other noncommutative algebras? I'd assumed that he avoided commutativity because he was writing this after having figured them out, and by using one less assumption he was able to simultaneously prove that no field over a basis of 3 elements existed, and none existed even if he gave up commutativity, an objection someone would likely raise as he was giving up commutativity while jumping from 2 to 4 elements.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
From reading the text above, it seems to me that he was looking for a field larger than $\mathbb C$ and was proving that one didn't exist. Since the quaternions were the first noncommutative division algebra discovered, it seems to me that they would have helped lead to these now-well-understood properties.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Like this. I figured I'd try getting Mathematica to symbolically solve for a set of 2x2 matrices that might commute like this, and right away it reduced the equations to several possibilities. I started plugging in values and ... oh, right, I have to tell it none of them equal each other. I told it to run again and it didn't finish, so I don't have high hopes
Nov
15
comment Why is $(-1) \cdot j = j \cdot (-1)$ for quaternions?
@JamesS.Cook It's wrong to say $Q_8$ (or $\mathbb H$) comes from the cross-product. That might be the perspective of people whose experiences are based in application. Hamilton's perspective is the modern one: In order to define cross-product of vectors (a nonassociative algebra), you need a vector space, which requires a field, which requires a ring, which requires a group. $Q_8$ comes from trying to develop a group that can be turned into a field larger than $\mathbb C$. Hamilton proved this cannot be done, but the quaternions form a noncommutative division ring, the next best thing.
Nov
15
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
I have long noticed that one could define an associative (and commutative) product with $i^2=j^2 = -1$ and $k^2 = 1$, but wondered why I never heard about it. I appreciate @Micah 's comment above about getting zero divisors when trying to make a field out of a multiplicative group that contains a second element that squares to a positive real. Would this product I've specified also lead to the split quaternions, or something else? I'm guessing not since, if the split quaternions are isomorphic to the 2 by 2 matrix ring, they're also noncommutative.
Nov
13
comment What is the difference between ring homomorphism and module homomorphism?
I'm saying it wasn't a bad analogy, and examples help people who are trying to learn new concepts.
Nov
13
comment What is the difference between ring homomorphism and module homomorphism?
Clarifying your driving/flying analogy: Person $A$ and $B$ are both pilots (isomorphic under the structure: "having passed the flying exam"). Person $A$ and $C$ are both drivers (isomorphic under the structure: "having passed the driving exam"). But if person $B$ is not a driver, and person $C$ is not a pilot, these two are not isomorphic with respect to either structure.
Nov
13
comment Shorter proof of $R/I$ is a field if and only if $I$ is maximal
Your example gives a field, but I saw how $I + (x) = R$ meant that after sending $I$ to $0$, if I sent $x$ to $0$, everything collapsed, which meant $x$ was a unit in $R/I$. Further, $I$ needn't be maximal, i.e. this should work when $R/I$ is not a field. I think a more illuminating example is $I = 4\mathbb Z = (4)$. The lattice involves $(1), (2), (4)$, and $(3), (6), (12)$ "below" them. Sending $(4)$ to $(0)$, also sends $(12)$ there, meaning $(6)$ collapses to $(2)$ and $(3)$ collapses to $(1)$. $(2)$ is the only remaining maximal ideal and $3 + I$ isn't in it, so it's a unit.
Nov
12
comment What is the relationship between prime ideals and their generators?
When writing my question, I edited for simplicity, and removed everything about UFDs, but then tried to imply my question included UFDs when I said "what additional criteria are needed". Thanks for helping me understand generators better. I think over the weekend I'll type up a similar question just for UFDs and reference this one.