I believe the answer to your question is yes, after fiddling with some "typical" counterexamples and finite spaces, but I have no rigorous proof @user193319
Is it easy to show that whenever $\langle v,w\rangle_1=0$, we have that $\langle v,w\rangle_2=0$? Where $\langle{.,.}\rangle_1$ is an inner product, and $\langle{.,.}\rangle_2$ also.
@AlessandroCodenotti Oh, very interesting! What were some of the "typical" counterexamples you checked? I'll have to continue thinking about this. If you do light upon a rigorous proof, I encourage you post it on the thread I started.
@user193319 infinite broom, the topologist sine curve, $[0,1]^2$ with the dictionary order topology, the comb space, those are the usual spaces used as sources of counterexamples about connectedness
Maybe I could use that since norms are equivalent, we also have that norms induced by inner products are equivalent, and therefore the second inner product must be zero iff the first inner product is zero
for example consider $\langle (a,b),(c,d)\rangle=(3a+b)c+d(a+2b)$, if I didn't mess up that should be an inner product wrt to which the standard basis of $\Bbb R^2$ is not orthogonal
Orthogonality is defined in terms of an inner product @sha
oh yea, another counter example could be the inner product for the space of functions. it makes sense that if you change the limits of integrations, some inner product will yield zero for two functions, while another won't
somehow I assumed that orthogonality is preserved under different inner products, but apparently that's false
In general if $V$ is a finite dimensional vector space and $A$ is a symmetric, positive definite bilinear form then $\langle x,y \rangle=x^tAy$ is an inner product
I've been tasked with showing that all strings composed of alphabet $\{a,b\}$ in which the number of $a$ = number of $b$ is a subset of a language defined as: Base $\{\epsilon\}$ and Step: for any string $xy \in \Sigma^{}$, that $axby$ and $bxay$ are $\in \Sigma^{}$. For the record, I really, really, despise discrete mathematics.
Suppose that I have a topological space whose Borel $\sigma$-algebra admits a minimal generating set, does this necessarily translate into a minimal basis for $\tau$? I don't think so
prove that after a recursive step, that property is preserved
and then prove the first base as following that property
then you can apply an induction kind of argument. the second recursive element would satisfy it because the first does, the third would because the second, and so on
@ShaVuklia Letting apart my mathematical problems which I also consider them poems (to some extent), yeah, I do write/read poems. Not that much lately since I'm pretty involved in stuff that requires a lot of time. How about you? Do you write/read poems? :-)
@MeowMix Given that $xy$ is in set #2 if $axby$ and $bxay$ are, I showed that for all elements of set #1 defined as: #a = #b, reducing both $axby$ and $bxay$ to $xy$ retained the property that #a = #b. I did this for both the arbitrary $xy$ string, and the null string (for the base case). While I thought this was okay, I am told that I am proving that #a = #b for any string in set #2.
Haha, that's really nice (math<=>poetry) - I'd almost forgotten! I also don't write/read poems that often, especially because I don't feel that much, but I managed to put this "nothingness" into words, giving it some recognition, I think.
But maybe I shouldn't bring up this subject in a math chat! @Waiting
So I've got one class (audit) which was going on until 1:20 (though someone was subbing in and I had no clue what he was saying so I probably should've sat it out), then I had to go quite some distance to get to the next class, so I tripped on the stairs
Hey guys - got a quick question: Consider a function f, where $f(x^*) = 0$ as the minimizer. The update steps is $x^{k+1} = x^k - \alpha f(x^k)$. An algorithm is globally monotone if $||x^{k+1} - x^*|| \leq ||x^k - x^*||$
If you were given that the function f is $f(x) = Ax + b$, and you're given the eigenvalues of $A$ and the vector $b$, how can you find an $\alpha$ such that it's globally monotone for all k in the sequence?
I think I'm a little stuck when I found the eigenvalues, found them to be positive definite. I wanted to say something along the lines of the $Ax$ portion could be decomposed into eigenvalue multiply by x, and since $f(x^{k+1}) \leq f(x^k)$ we have that it's globally monotone
but even then I can't say for sure that because $|f(x^{k+1})| \leq |f(x^k)|$, that $||x^{k+1} - x^*|| \leq ||x^k - x^*||$
@ShaVuklia Don't worry about it, almost everything is related to math, more or less, in a way or other, there is some math around even if sometimes this is not that obvious. Eventually I never ever heard poems hurt people ... :-)
It actually seems more interesting the more I think about it. Maybe there can be something in the verse of the poetry, and the word-game (but not the content!), which would be physically repulsing, disorienting and nauseating to me.
I like existential literature; Dostoyevsky has been and always will be one of my favorites. I guess I have switched to rather bizarre things over the course of days.
Hmm, now that I try to recall I think Murakami's novels are valid examples of things reading which I actually felt sick, without any apparent reason. The surrealistic setup is too disorienting.
actually I thought about 2 counterexamples to the other user's question but didn't recognize them as such... he posted on main and got an answer there though
@Ted Yeah I'm alive, the pain has mostly subsided. I went to Student Health Services, scheduled an appointment for Saturday, they said that if I do ever feel a recurrence of the ear buzzing, or other symptoms of head trauma, to go for emergency care
Beyond that, it doesn't seem like I have a concussion, it was probably just the immediate shock, since it lasted for under a minute, so that will be mostly for the fall pain.
@Eric I did not apply for summer school, I would have to supplement a lot of material on differential equations and all to be able to follow anything, which I don't anticipate doing
@Balarka Fell down, landed somewhat hard on my backside. At the time my hearing went funny for a second and I heard a beep/buzz
@Ted Thanks, I think I'm alright. Enough to make my great puns at least
"Now to apply the 3 tape recorder analogy to this simple operation. Tape recorder 1 is the Moka Bar itself it is pristine condition. Tape recorder 2 is my recordings of the Moka Bar vicinity. These recordings are access. Tape recorder 2 in the Garden of Eden was Eve made from Adam. So a recording made from the Moka Bar is a piece of the Moka Bar...
...The recording once made, this piece becomes autonomous and out of their control. Tape recorder 3 is playback. Adam experiences shame when his discraceful behavior is played back to him by tape recorder 3 which is God. By playing back my recordings to the Moka Bar when I want and with any changes I wish to make in the recordings, I become God for this local. I effect them. They cannot effect me." - William S. Burroughs.
If I have a matrix representation of an algebra, where matrices A, B, C generate a finite basis (AB, BA, ABC, CAB, etc), and you make a create some general element of that matrix representation, say M = 2AC+4CB+9ABC for example, then you've got this matrix M whose entries are linear combinations of that basis. What's the process called of pulling out those coefficients 2, 4, 9, 0, 0, 0, ... ?
I was wondering earlier if it wasn't decomposition, then I was trying to figure out if that's the case, what kind of decomposition is it? Maybe the kind of decomposition depends specifically on the matrix representation
see cause on wikipedia it says "decomposition, or matrix factorization is a factorization of a matrix into a product of matrices", but this is a linear combination of matrices. so it strikes me that this is just an issue of doing linear algebra except the vectors are linear combinations of matrices
