just a quick question. Suppose that we have two vector spaces V and W and a vector space isomorphism $\alpha : V \rightarrow W$ does this induce a vector space isomorphism on the dual ?
I am trying to understand a specific theorem in a book.
@TedShifrin first linear algebra was just calculation of matrices and second one just about change of basis and abstract vector spaces, but we didn't go in detail.
Try Friedberg, Insel, Spence ... or Artin's Algebra ... Didn't your graduate algebra course do decomposition of finitely generated modules over a PID or ED?
@AliCaglayan I am quite excited I am reading currently Allufi chapter 0. I finished first two chapters with all of the problems. I can't wait to go to linear algebra chapter.
The way it used to be done was like, second half of third quarter of calculus, about a week or two in analysis, second quarter of the algebra sequence (mixed in with ring theory, modules)
I remember I had an algebra book which started with matrices so that it could talk about matrix groups, and had a lot of linear. I ended up needing something more compact since I was in a rush, so I swapped to Herstein
@TedShifrin They define something called hyperbolic space as follows. Suppose V and $V^{}$ is vector space and its dual respectively. Define a bilinear form on $V \times V^{}$ as follows $((f,x),(g,y)) \mapsto f(y) + g(x)$.
At some point soon I do intend to go through at least groups more carefully, first time around I was rushing to group actions so there are still a lot of holes. I know Herstein is missing Jordan-Holder which Laci said people should know before starting his class (along with Sylow)
@Astyx: This one is sort of interesting. So given an $m\times n$ matrix $A$, it of course represents a linear transformation $T\colon\Bbb R^n\to\Bbb R^n$. You know how to identify its row space and column space with the image of $T^*$ (transpose) and image of $T$ respectively.
So you get isomorphisms $\lambda\colon\text{row space}(A)\to\text{column space}(A)$ and $\mu\colon\text{column space}(A)\to\text{row space}(A)$ in an obvious way. When are these inverse functions?
Let $f(x)=\sin(x), \text{for all x}$, if we want to write $f(x)=\tan(x) \cos(x)$, do we have to add the restriction that $x\not=\pi/2+n\pi \; (n\in \Bbb Z)$ ?
So I'll look for Artin as well. I've recently started looking at Aluffi and Lang, which seem good (though Lang is looking it's going to be a harsh read...)
@TedShifrin Looking at the first problem in your integration exams (exam #3). Why do you want to give your students a heart-attack? I almost had one before realizing a second later that it's easy.
I don't remember what you're talking about Balarka. But I think I had forewarned them numerous times (and they had numerous homeworks). This was a change the order problem, right?
@TedShifrin here is a categorical way to think about the question I asked you before. Suppose we have an isomorphism $\alpha^{\star} : V^{\star} \rightarrow W^{\star}$, then this induce $alpha^{\star} : V^{\star} \rightarrow W^{\star}$ given by $\phi \mapsto \phi \circ \alpha$. Then, $\alpha^{\star}$ is monic. In the category of finite dimensional vector spaces monic is the same as injective.
Also, $V^{\star}$ has same dimension as V and same with $W^{\star}$, so we are done.
Oh @Ted I think I might have an idea of how to solve the problem I told you about yesterday, about the signature of forms, can I run my (vague) idea by you?
Yeah, the signature of a 1-form $\omega$ in the punctured plane being the (unique) real number $k$ such that $\omega-k(\frac{-ydx + xdy}{x^2+y^2}$ is exact
Oh huh, "signature" how it was denoted in the problem. But yeah, my initial idea is to pullback to polar coordinates, so that $\omega - k\omega_0$ becomes $\Phi^*\omega - kd\theta$
So let $k=\frac{\int_{\mathbb{S}^1} \omega}{2\pi}$. Then given any closed loop around the origin, you can homotope it to the circle (since $\omega - k\omega_0$ is closed, the integral should stick), and the integral is 0. Then, if the loop doesn't contain the origin in the interior, it should follow from Stokes's theorem
Well maybe I need to prove smth like, if $c$ has winding number $n$ with respect to the origin, then $\int_c \omega = n\int_{\mathbb{S}^1} \omega$, for a closed 1-form $\omega$.
@Daminark Once you finish that problem, you might consider what happens in a doubly punctured plane. Specifically, compare what homotopy versus winding numbers say about the contour here.
Right. If you've justified that $|\sin x|$ doesn't exceed 1 for any real $x$---well, so long as $x\neq 0$, $1/x$ is certainly a real number as well. So it also applies for that.
@MATHASKER You're bold profile description, trust me the Math isn't very bad, if you look for understanding rather than ''Which formula will I be using in the exam ?''
Eh, I think it's arguable that the wording does suggest 'order matters'. A row of four, for instance, always consists of those students in some arrangement.
By contrast, if they just said "four students out of six go to see a movie/see a lecture/etc" then order definitely wouldn't matter.
@MATHASKER Since the boys must be adjoining, there are 4 places the boys can be clustered (123, 234, 345, 456) and 6 ways they can be arranged in each of those configurations. Then there are 6 remaining possible configurations for the girls in each case, so I believe the total number is 4*6*6 = 144.
@MATHASKER Let's say the boys, named A, B, and C, are grouped at the first three chairs. I'll use x's for the girls as they're not important for the moment.
They could be seated ABCxxx, or CABxxx, or BCAxxx, or ACBxxx, or CBAxxx, or BACxxx. Six ways when they're all at 123.
The same is true for when they're at 234, 345, or 456.
Then, to fill in the x's in each case, there are three girls and three slots, so 3!, or 6.
If you give me a few mins I can try and prove it, but I see what you're getting at (since it would be nonempty then it would have to intersect in an arbitrary point $y$
I understood your question, @Ali. Direct sum implies finite sums. I don't see how to use a product construction, where you can have infinitely many nonzero terms.
@Socrates: Yes, but it is usually less cumbersome to write the operation with the symbol. When you start doing associativity, you notation will get very cumbersome.
