Reputation
1,470
Next privilege 2,000 Rep.
Edit questions and answers
Badges
7 16
Newest
 Mortarboard
Impact
~22k people reached

  • 0 posts edited
  • 0 helpful flags
  • 212 votes cast
12h
comment Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?
Thanks for this. I will have to do my proof another way.
12h
comment Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?
Thank you for pointing out this easy counter-example.
2d
comment What is the Laurent series of $\frac{2}{z-1} - z$ in $1<|z|<2$?
@C.Dubussy I see it now. So is it true then that this Laurent series converges as long as $1 < |z| <\infty$ holds? Hence, it converges within the upper bound of the annulus in the question.
Apr
18
comment Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$?
Helpful. This could also be generalized to any finite dimensional space by taking any diagonal matrix with diagonal entries $2 \pi i$, I believe.
Apr
4
comment All solutions of $A^2+I=0$ in $M(n,\mathbb{C})$ are similar
How did you deduce the minimum polynomial from $A^2+I=0$? Is this a variant of Cayley-Hamilton? I calculated it as $p_A(x) = (i-x)^m (-i-x)^p$ which obviously does have repeated roots.
Apr
4
comment All solutions of $A^2+I=0$ in $M(n,\mathbb{C})$ are similar
In practice, we should be able to deduce the "appropriate" $m$ and $n$ from the characteristic polynomial of $A$, right?
Apr
4
comment All solutions of $A^2+I=0$ in $M(n,\mathbb{C})$ are similar
Shouldn't we mention that $A$ is diagonalizable over $\mathbb{C}$? I believe the diagonalizability of $A$ is why there are no $1$'s on the super-diagonal over the Jordan canonical form.
Mar
29
comment If $f(z)$ maps the unit disk onto itself $k$ times, prove that f(z) must be a rational function and show that the degree of its denominator $\leq k$.
@Chilango. Good point about the origin. Can you explain your suggestion and why that should work for $f(z)$? I've seen something similar called the Blaschke product, but not in relation to my question.
Mar
26
comment What does it mean for a function to tend uniformly to $\infty$ on every compact set?
Thank you. However, how/where did you get your definition of uniform convergence to infinity in your second paragraph?
Mar
7
comment For $T: V \to V$, suppose $A = A^*$ where $A = [T]_\mathcal{X}$. Find another basis/matrix where $B \neq B^*$ for $B = [T]_\mathcal{Y}$.
@FriedrichPhilipp. Thank you. How do you know to choose such a matrix $S$? As in, how do you know to choose a transition matrix that is not unitary? The intuition escapes me.
Feb
24
comment Let $f(z) = z^4 - 2z^3 + z^2$. Evaluate $\frac{1}{2\pi i} \int \frac{f'}{f} dz$ and $\int \frac{zf'}{f} dz$
Is there a theorem that states the result you are using for the sum of the zeros of a monic polynomial?
Feb
16
comment What is the form of the general mobius transformations that map the line $\Re(z)=2$ and the unit circle into concentric circles?
Also, how did you come up with the map $\varphi(\zeta)$? It seems to come out of nowhere for me.
Feb
16
comment What is the form of the general mobius transformations that map the line $\Re(z)=2$ and the unit circle into concentric circles?
Thanks. Can you explain your symmetry argument for finding $r$? Why should $\varphi(1/2)=r$ and $\varphi(0)=-r$?
Feb
9
comment What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
Thank you, that is helpful. Another question: how do you know to choose $D$ and $G$ as you have chosen? It seems a little arbitrary to me. For example, how do you know not to to choose the interior of the two circles in the domain? And why is $G$ chosen to be the space between the two circles in the range?
Feb
9
comment What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
Thanks for this. The mathematics makes sense to me but how do you motivate such an approach? For example, how did you know we needed to consider a transformation $h$ from $E$ to $F$?
Feb
8
comment If $T$ is an orthogonally diagonalizable linear operator in an inner product space, show that $T^*$ is also orthogonally diagonalizable.
@Omnomnomnom But aren't the $PDP^*$ meant to be understood as matrices in such a representation? Or as the composition of linear operators? In which case, it could also be written $P \circ D \circ P^*$?
Feb
7
comment Let $\{e_1,\ldots,e_n\}$ be an arbitrary basis in a finite dimensional inner product space. Prove $\exists \{f_1,\ldots,f_n\}: (e_i,f_j)=\delta_{ij}$
Does this really show they exist or just that $(A^T)^{-1}$ would have the $f_i$ as columns if they did exist?
Feb
1
comment Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint?
@Qidi, I have revised my question to clearly define $D$. I don't think this aligns with your suggestion, does it? Where can I find that definition of the differential operator, which you say is the usual convention?
Feb
1
comment Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint?
Integration by parts gives me $\langle Df, g \rangle = \sum\sum a_i \overline b_j - a_0 \overline b_0 - \int_0^1 f \frac{\partial \overline g}{\partial t} dt$. Does that give us something useful?
Jan
31
comment Suppose $T$ is diagonalizable in $\mathbb{C}$. Show $e^T = \sum_{\lambda \in sp(T)} e^\lambda P_\lambda$ is the matrix exponential series.
Thanks for the hint. Is there a reason to do this over the other suggested answers which use the series representation of the exponential function?