Alright, another question I should probably already know the answer to: Can you use any infinite set to index another infinite set, even if their cardinalities are mismatched?
@TedShifrin keep an eye out for it if you are ever near a middle eastern place. However, as I have learned now, some places don't do it with just garlic+oil+lemon juice.
Because there are only $\mathbb{N}$ element of a formal power series and the collection of all sequences of elements in an arbitrary field is going to be much larger.
Yeah I used my magic bullet one time, and my immersion blender the other.
Both times the emulsification broke part of the way.
I think I just need to try more garlic, or with a food processor that will allow me to blend while pouring oil. But I have this mini food processor I can try next time.
@TedShifrin See this math.stackexchange.com/a/335096/168764. It shows that the maximization of a function is equal to the minimization of -f. So, if a is the argument of q such that q attains its maximum, then a should be the argument of -q such that -q attains its minimum
So, let $K[[x]]$ be the vector space of formal power series over a field $K$. The set $\{x^i:i\in\mathbb{N}\}$ is not a basis for this vector space, as I am led to believe. To prove this, I am trying to adapt Cantor's diagonalization argument. The idea is to consider the the set of all power series with coefficients in $K$ and then to construct a power series from that set which is different from each element in that set in at least one place.
However, I think the issue I am encountering is that the length of a formal power series is not necessarily enough to account for all power series in the set.
That is far more arduous, @Semiclassic, because then you get a combination of the orthogonal basis vectors, whereas we're getting it as a combination of the original vectors quite painlessly. (Well, you still need to invert a symmetric matrix. So not painless.)
S is orientable so I get a non-vanishing section of NS. K is also orientable, in S specifically, so you get a non-vanishing section of NK inside of S. Then you get a trivialization of NK in M.
:51713118 Hi @TedShifrin... its me again... I've been thinking about this more... and I still am not convinced all the terms being different is the worst case...
Lets say $f(x) - f(y) = \frac{2}{3^2}$ Lets say $x_3 = y_3 = 0$, now let $x'$ be such that $x'_3 = \frac{2}{3^3}$, then $f(x') - f(y) = \frac{2}{3^2} + \frac{2}{3^3}$
Now let $y'$ be such that $y'_3 = \frac{2}{3^3}$, then $f(x) - f(y') = \frac{2}{3^2} - \frac{2}{3^3}$. So adding another term which is different brought the sum farther away from 0 in one case, but closer to 0 in another... It is clear that if all the terms are the…
@TedShifrin I am racking my brain for how this would relate to Whitney's theorem. I seem to recall it having to do with just refining an embedding to an orthogonal complement of a vector.
@TedShifrin according to WP (dangerous, I know) the factorization $A=QR$ has $Q$ orthogonal and square, and therefore $R$ rectangular if $A$ rectangular
Hence, they say that maximization of a function is equivalent to the minimization of its negation. When "they" say this, they are looking for argmax and not max
Right. Canonically it seems like the QR factorization is necessarily of that form, whereas a thin/reduced QR factorization has $R_1$ square (according to what WP says about the literature, anyways)
Some people I have met in the past don't like the idea of replication in the digital world. And no, I wouldn't make them publicly available unless you desired me to.
We are covering... let's see... we've done 2x2 matrices and eigenvectors/eigenvalues, we're going all the way up to spectral theorem and singular value decomposition
I'm in honours multivariable calculus/analysis and honours linear algebra (i.e. both proof based and geared for people wanting to go to grad school), so it doesn't sound entirely different from what I'm doing.
I'm not super fond of Axler, but Friedberg/Insel/Spence is good. Adams is not particularly theoretical and used to have a shocking mistake in it. I assume it got fixed.
What was the mistake in Adams? I agree - we're mostly using the professor's lecture notes/chalk board for anything to do with neighbourhoods, point-set topology, boundaries, etc.
and rely on assigned homework for practice with that
Consider $U=\{\textbf{x} \in \mathbf{R}^3; x_1+x_2+3x_3=0\}$ and $U=\{t(1,1,2)\in\mathbf{R}^3; t \in \mathbf{R}\}$. Show that $\mathbf{R}^3=U \bigoplus V$.
If one shows that $U\cap V=\{\textbf{0}\}$, which is easily shown, would that also imply $\mathbf{R}^3=U \bigoplus V$? Or does one need to show that $\mathbf{R}^3=U+V$? If yes, how? By defining say $x'=x_1+t,y'=x_2+t,z'=x_3+2t$ and hence any $\textbf{x}=(x',y',z') \in \mathbf{R}^3$ can be expressed as a sum of $\textbf{u}=(x_1,x_2,x_3) \in U$ and $\textbf{v}=(t,t,2t) \in U$ for $t \in \mathbf{R}$.
I thought I understood large parts of Axler's book but actually felt like I learned nothing. Probably it was meant for a different audience, but it was weird.
So, I have a question that says "Prove the following statements. Or, if a statement is incorrect, make a correct statement and prove it." and the vagueness of the "make a correct statement" part is making me laugh.
Can I ask a question? Let say I've a point $P = (2, 5)$ and another $C = (1, 1)$. $C$ is a child point of $P$, that means it use $P$ as reference, so its relative (local) coordinates are $C_l = (2, 5)$ while its world coordinates are $C_w = (3, 6)$. So, in order to get world coordinates, I must do $referencePoint + child$. And if I want to get relative coordinates $worldPoint - referencePoint$.
Is that right?
I just want to know if that is true or not. Because I am using it and it's not working in my program, so I want to know if I'm having a math error or a programming error.
This question is slightly confusing to me (the basis in (c.3) is a basis of the null space): "Using the basis in (c.3) to express each column of $A$ a combination of some independent columns of $A$, hence get a basis for the column space $\text{Col}(A)$."
(Sorry, the columns for which the corresponding row in the vector in the null space is non-zero) Though, can I necessarily say that removing one of those vectors gives me a linearly independent set?
brb, eating dinner. Maybe brain will move faster with food in me.
So, this is a vector in $\Bbb Z_2^{4\times 5}$: $$\begin{pmatrix}1 & 1 & 1 & 1 & 1\cr 1 & 0 & 0 & 0 & 1\cr 1 & 1 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 & 1 \cr \end{pmatrix}$$ with reduced row echelon form: $$\begin{pmatrix}1 & 0 & 0 & 0 & 1\cr 0 & 1 & 1 & 0 & 1\cr 0 & 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 0 & 0 \cr \end{pmatrix}$$ and the basis for the null space I found is $\{(1,1,0,1,1)^T,(0,1,1,0,0)^T\}$
My brain is missing the connection I should be making. Obviously, I shouldn't just look for vectors not in the null space, because $(1,1,1,1,0)^T$ isn't in the null space, but those columns contain a linearly dependent subset.
Is the linear transformation that takes the derivative of a polynomial from the set of all polynomials over $\mathbf{R}$ onto (surjective)? Since the polynomial “loses” a degree when differentiating, it can’t be onto, or?
Well, tell me where I'm going wrong: Since we're working in $\Bbb Z/2\Bbb Z$ we know that $-x=x$. So, $x_1=x_5, x_2=x_3+x_5,x_4=x_5$. Therefore any $x\in\text{Nul}(A)$ takes the form $x_5(1,1,0,1,1)^T+x_3(0,1,1,0,0)^T$
OK, maybe I didn't pay attention to the $\Bbb Z_2$ when I glance at it. With $z_3=1$ and $z_5=0$, I get $(0,1,1,0,0)$ and with $z_3=0$ and $z_5=1$ I get $(1,1,0,1,1)$.
Actually, since this is a $4\times 5$ matrix, removing two columns gives us a $4\times 3$ matrix, so it can't be invertible, but the logic you provided is much more stable.
@TedShifrin For the differentiation of a polynomial from the set of all polynomials over $\mathbf{R}$ ($P(\mathbf{R}$)) to be onto, the whole codomain has to be mapped, correct? Is this the case?
