@TedShifrin and one other question, since the calculator doesn't accept this long number, the easiest way to determine the decimal number 2^63 in hex is to do it manually, i mean it would be probably stupid to go the way you pointed me to the relation between the levels
A student asked me what multiplication of matrices was in terms of systems of equations. It stumped me. I (probably unsuccessfully) tried to make the point that this is why passing to matrices instead of systems of equations is good.
i mean, i get this task on an exam, for example where i have limited time for multiple things, to get the value of 2^4 or 2^6 would be easy to count, but to get the 2^63 without the calculator would take a lot of time, or am i missing something?
I figured out $2^{63}$ without a calculator, @rsz. How did you use a calculator?
@Espen: Sure, it's composition of linear functions. But that's hard to understand when you're in the first week of linear algebra just discussing systems of equations :)
@Mike: Very, very disappointing. One of the problems I put on the algebra qual was to prove that commuting diagonalizable linear maps (finite dimensions, of course) are simultaneously diagonalizable. No one got it :(
Sure - but to do that one talks about transformations. It's a level beyond systems of equations. (Whence why it's natural to talk about linear transformations!)
@MikeMiller Perhaps explain it in terms of permutations? Of course, that would require decomposing a general matrix into a basis of permutation matrices... I'm not sure that would be satisfying.
@TedShifrin what is the easiest way to do that? cuz the decimal representation of the number is pretty long, so i assume there is some way to do that quickly, could you tell me?
@TedShifrin and how does that equation you wrote help me to get to the result in hexadecimal number base? i was figuring it out till now and couldn't get to the meaning
@Espen: Yes, basically that's what most people do with elementary matrices, but as I was trying to say to @Mike (who's put me on ignore) is that it's more basic than that.
linspace is a core function to MATLAB -- it's hardly cheater code. In any case, it provides you with a ground truth so you can test whatever algorithm you come up with.
@Ted How do you convince yourself the RHS is the (right-)incerse without knowing that row transformariojs are right multiplying by elementary matrices?
@Mike: I give you the proof one of my talented students told me 20 years ago (and which I've never found in any book other than mine). Look at the algorithm upside-down.
@Kaj: You prove that I failed at teaching you linear algebra. So much for my cred here.
I was also trying to playing around with $\det$ as a homomorphism $GL_2(\mathbb{Z}_p) \rightarrow \mathbb{Z}_p^\times$, but I wasn't getting anything useful...
I have no idea what that algorithm is, @user2692669, but unless I needed to write my own algorithm as a requirement, I would simply use linspace. Farting around with trying to outsmart IEEE 754 implementations is usually a fast track to pain.
I work for an company with over 15 PhDs in electrical engineering. for the past 5 minutes, I have listened to someone fail to use the microwave oven properly.
@TedShifrin Not being facetious, for a moment, actually sometimes a good way to inspire a student to think about a problem is to initially present a shitty way to do something, and hope that they come to the conclusion "there's got to be a better way." Unfortunately, 85% of students just give up anyways.
@TedShifrin Of course. It is the devil incarnate.
I learned matrix inversion by Cramers rule -- using Maple.
Let $f: X \to \mathbb{R}$ be uniformly continuous. Then for any $\epsilon > 0$ given there exists $\delta > 0$ such that $x,y \in X$,
$$|x - y| < \delta \Rightarrow |f(x) - f(y)|<\epsilon$$
In particular fix $a \in X$ arbitraly, then
$$|x - a| < \delta \implies |f(x) - f(a)| < \epsilon $$ ...
@Arkamis: Occasionally if I tell students I don't know a nice way to do something, the very good ones will take that as a challenge. But 99% give up and say that if I don't know, they certainly will never find it. ... But I reference that wonderful student (who flunked out of my university for reasons not related to mathematical talent) who gave me the right proof for matrix inverses above.
Let $f: X \to \mathbb{R}$ be uniformly continuous. Then for any $\epsilon > 0$ given there exists $\delta > 0$ such that $x,y \in X$,
$$|x - y| < \delta \Rightarrow |f(x) - f(y)|<\epsilon$$
In particular fix $a \in X$ arbitraly, then
$$|x - a| < \delta \implies |f(x) - f(a)| < \epsilon $$ ...
@TedShifrin Well, that's what category theory does, really. It organizes stuff and allows one to ask new questions. Already we see new developments about, say, Lie groupoids.
It's totally standard in precalculus and calculus that if one does not specify a particular domain, the intent is the set of all $x$ for which $f(x)$ is defined. In those settings, $x\in\Bbb R$.
because as for most humans, my brain capacity is limited. and as I learn a lot of new stuff, I also have to forget some things, there are anyway a lot of things I hope to forget, they are more of a burden
No, this is not a belated post about the moderator election results.
Instead, I noticed that Daniel Fischer has very recently topped 100K in reputation. And he accomplished that in one-year, six-months, while also contributing enormously through helpful comments on questions.
I think these feat...
Hrm, there is a user asking for answers for something which is equivalent to an exam. How come the question has received several answers and no delete votes?
I applied for job in a company abroad, and got a mail detailing a few questions. One question out of them is this. I tried to find the answers over several sites, but didn't find the exact right answer for the below question. I'm left with $10$ hrs to reply back with answer.
Find a function $F(n...
Let $f: X \to \mathbb{R}$ be uniformly continuous. Then for any $\epsilon > 0$ given there exists $\delta > 0$ such that $x,y \in X$,
$$|x - y| < \delta \Rightarrow |f(x) - f(y)|<\epsilon$$
In particular fix $a \in X$ arbitraly, then
$$|x - a| < \delta \implies |f(x) - f(a)| < \epsilon $$ ...
No, this is not a belated post about the moderator election results.
Instead, I noticed that Daniel Fischer has very recently topped 100K in reputation. And he accomplished that in one-year, six-months, while also contributing enormously through helpful comments on questions.
I think these feat...
