Let $\{\alpha_n\}_{n \in \mathbb{N}}$ be a bounded sequence of real numbers. Suppose $\{e_n\}_{n \in \mathbb{N}}$ and $\{f_n\}_{n \in \mathbb{N}}$ are orthonormal sequences in a Hilbert space, $\mathcal{H}$. Define $T : \mathcal{H} \rightarrow \mathcal{H}$ by $$Tx = \sum_{n \in \mathbb{N}} \alpha_n \langle x, e_n \rangle f_n.$$ 1. Show that $T$ is bounded. 2. Show that $T$ is compact if and only if $\alpha_n \rightarrow 0$.
I'm preparing for a test and I encountered this problem. I think in order for me to show that $T$ is bounded, I need to show that there exists $C \in \mathbb{R}$ such that $\left\Vert Tx \right\Vert \le C \left\Vert x\right\Vert$.
I think for people who don't really know what's happening - having a recipe (l'hopital - although often they don't know when they're allowed to apply) is "easier" than knowing taylor expansions..
Yes it must be for that. I was just surprised that it is used extensively on this site, although I've never heard about it at school (we only use Taylor expansions for that).
By induction on $n$: On $\mathbb C$, the characteristic polynomial of $A$ has one root. Hence $A$ has (at least) one eigenvalue $\lambda$. Take one eigenvector $x$ , consider a basis $(x,e_2,...e_n)$, then apply the induction hypothesis to the (n-1)*(n-1) bottom right hand block.
that proves that eigenvalues of A give eigenvalues of $A^k$ but i think the dimension of eigenspaces match up anyway (i.e. the $c$-eigenspace of $A$ is the $c^k$-eigenspace of $A^k$) so you get all of them
By induction on $n$: On $\mathbb C$, the characteristic polynomial of $A$ has one root. Hence $A$ has (at least) one eigenvalue $\lambda$. Take one eigenvector $x$ , consider a basis $(x,e_2,...e_n)$, then apply the induction hypothesis to the (n-1)*(n-1) bottom right hand block.
At $\sim$ 18 years old, you end high school with baccalauréat in France and then you go to study in prépa (2 or 3 years) to get into engineer schools or écoles normales
@Alessandro lol, my uni library doesn't have munkres so my lecturer suggested I look at Armstrong for proofs and Jänich for intuition (Jänich was suggested to me by @Mathein anyway)
I'm wondering what the courses in your uni look like since you're doing advanced algebraic number theory and intro to topology at the same time! @ÍgjøgnumMeg
@Alessandro I self-taught algebraic number theory for my undergrad dissertation, and I'm doing intro to topology over the summer so I have at least some topological background before I start a masters degree lol
@Alessandro I think it's common in most unis around the world, but my uni is very focussed on applied maths/stats so I essentially have to self-teach all of the pure stuff
@Nûr if I have all eigenvalues on the unit circle, I only need to pick a large enough n such that my eigenvalues are close to exp(2ipi(m/n)) for some m
@ÍgjøgnumMeg I'm going to focus on logic and set theory (they also have a cool course in descriptive set theory apparently) and I'll probably just do as much geometry/topology as possible with the remaining credits
That looked to me like one of those theorems where the proof is kinda ugly so you do it once and forget about it, but remember the result because it's really useful
Let $A$ and $B$ in $M_n(\mathbb C)$ such that the rank of $AB-BA$ is $1$. Prove that $A$ and $B$ are simultaneously triangularisable.
This generalizes the classical case $AB = BA$.
By induction on $n$, it suffices to show that $A$ and $B$ have a common eigenvector.
So, it would be sufficient t...
Does anyone know a symbolic language/package that provides the general term of a Taylor expansion? For example, for f(x) = 1-log(1+sqrt(1-x)) I can use Matlab's symbolic package to numerically obtain each Taylor coefficient. But I need an expression of the general term. (Actually I want to prove that they are all positive).
Or some table of Taylor series that goes beyond the usual exp, cos etc?
Does anyone see how to prove that if all eigenvalues of the matrices of a sequence tends to 0 then the matrices sequence tends to 0, without the Jordan form?
Hi, I'm trying to differentiate a quadratic form $Q(x)=x^T A x$. Can I use the product rule and write $Q(x) = (f \circ g)(x)$ for some $f,g$? I tried $g(x)=Ax$, but what is then $f$? I can't just set $f(x) = x^T x$, since then $(f \circ g)(x) = (Ax)^T Ax \neq Q(x)$. Can I write $Q$ as product of two functions anyhow?
I have another question regarding the definition of the directional derivative. For a function $f: U \subseteq \Bbb R^n \to \Bbb R^m$ the directional derivative along the vector $v \in \mathbb {R}^n$ at $x_0 \in U$ is defined by
Let V0 be the volume of a cuboid with the edges $a_0; b_0; c_0> 0$. A measurement of the edge lengths yields the values $a; b; c$ with the absolute errors $\Delta a=a-a_0; \Delta b=b-b_0 ;\Delta c=c-c_0$. The volume of the cuboid is thus determined with an absolute error $\Delta V = V-V_0$. How can we explain the following approximation formula for the relative error by means of differential calculus? $$\frac{\Delta V}{V_0}\sim \frac{\Delta a}{a_0}+\frac{\Delta b}{b_0}+\frac{\Delta c}{c_0}$$
Let $A$ and $B$ in $M_n(\mathbb C)$ such that the rank of $AB-BA$ is $1$. Prove that $A$ and $B$ are simultaneously triangularisable.
This generalizes the classical case $AB = BA$.
By induction on $n$, it suffices to show that $A$ and $B$ have a common eigenvector.
So, it would be sufficient t...
