If T is a linear transformation on $R^n$ with $||T - I||&lt;1$, prove that $T$ is invertible and that the series $\sum_{k=0}^\infty(I-T)^k$ converges absolutely to $T^{-1}.$ (Use the geometric series)
84212
Prove that if $\|A\| &lt; 1$, then $I-A$ is invertible. Here, $\|\cdot\|$ is a matrix norm induced by a vector norm.  This lemma is referred to as Neumann Lemma.  Any ideas on how to go ahead with this?  Thanks.
19712
Let $\sum a_{n}$ be an absolutely convergent series such that $$\sum a_{kn}=0$$ for all $k\geq 1$.   Help me prove that $a_{n}=0$ for all $n$.  Thank you!
111182
I have to solve this problem.      Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \infty$, defined by      $$T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\} $$   Prove that $T$ is compact.    My attempt is the following.  I have considered the operator $$ T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}. $$ $T_N$ is clearly linear and it is also compact. In fact given any bounded sequence $\{x^{(n)}\}_{n\geq 1}$ in $l^p$, the sequence $\{T_Nx^{(n)}\}_{n\geq 1}$ is bounded in $\mathbb{R}^N\subset l^p$. Then, since every bounded sequence in $\mathbb{R}^N$ has a convergent subsequence, it follows that $T_N$ is compact.   Now, I have to prove that $T$ is compact. For this I proved that $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$. In fact we have: $$ ||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}} $$ and, given $\epsilon&gt;0$, there exists $\bar n$ such that $|\lambda_n|&lt;\epsilon$, for every $n&gt;\bar n$. So if we take $N&gt;\bar n$, we get $$ \sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon $$ Therefore $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$ implies that $T$ is compact.   My questions are:  (1) is the proof of the compactness of the operator $T_N$ correct?  (2) is the conclusion correct?  Thanks
84836
Could any one just tell me a sketch of the proof of the followings?  $1$. If $n&gt;1$, does every linear transformation on $\mathbb{R}^n$ have an invariant subspaces of dimension $2$?  $2$ Is there a vector space $V$ and a linear transformation $T$ on $V$ such that it has exactly $3$ invariant subspace?  Thank you!
123580
Let $t\in \mathbb R$. Prove that $\displaystyle{\sum_{n=1}^{\infty}n^{3}\sin\left(\frac{1}{n}\right)-n^{2}+t}$ converges iff $t=\frac{1}{6}$.  While the $(\Rightarrow)$ direction was fairly easy (as $a_n$ should converge to $0$, forcing $t=\frac{1}{6}$), i'm facing some hard time trying to prove the second direction, that is $\displaystyle{\sum_{n=1}^{\infty}(n^{3}\sin\left(\frac{1}{n}\right)-n^{2}+\frac{1}{6})}$ converges. I did notice that $a_n=a_{-n}$ and tried to write this series as a Leibnitz series, but had no luck since the sequence doesn't decrease. I also tried to prove absolute convergence but then again, no luck. Thanks.
31680
Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.  Can we more generally conclude that $T$ is compact?  Assume $p&lt;\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology. On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$. EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true.
62497
I am a beginner of funcional analysis. I have a simple question when I study this subject.  Let $L(X)$ denote the Banach algebra of all bounded linear operators on Banach space X, $T\in X$ is invertible, then $||T^{-1}||=||T||^{-1}$? Is this result correct?
27650
It's a question from a test that I had, and I don't know how to prove this, so I am forwarding this to you.    $\sum \limits_{n=0}^{\infty }\:b_n$ is absolutely convergent series . How to prove that the series $\sum \limits_{n=0}^{\infty }\:(e^{b_n}\:-\:1)$ is also absolutely convergent?  We cannot assume here that $b_n\:\ge 0$ for every $n$.
13060
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.  In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.  Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.  How can I finish this proof?
90986
I've been given the following problem as homework:     Prove that there exists a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ such that $T(1,1) = (1,0,2)$ and $T(2,3) = (1,-1,4)$.   Since it just says prove that one exists, I'm guessing I'm not supposed to actually identify the transformation.    One thing I tried is showing that it holds under addition/multiplication in the sense of:  1) $T(x + y) = T(x) + T(y)$  2) $T(cx) = cT(x)$  3) $T((1,1) + (2,3)) = T(1,1) + T(2,3)$? But that's not necessarily true.  I don't know how I'd accomplish this given my limited knowledge.  What's the approach to solving a problem like this?
25809
I'm looking for the shortest and the clearest proof for this following theorem:  For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , I need to show $V= \ker T^n \oplus  $ Im $T^n$.  Any hints? I don't know where to start from.  Thank you.
96583
Let $B$ be a banach space and $T : B \to B$ be a bounded linear transformation.   If for identity transformation $I : B \to B$ , $||T-I||$&lt;1 , then $T$ is invertible.  || || is norm of transformation.
28920
Prove that if the positve term series $\sum^{\infty}_{n=1}a_n$ is convergent, also $\sum^{\infty}_{n=1}\sqrt{a_na_{n+1}}$ is convergent.  Prove that if the positive term series $\sum^{\infty}_{n=1}a_n$ and $\sum^{\infty}_{n=1}b_n$ are convergent, also $\sum^{\infty}_{n=1}a_nb_n$ is convergent.  I've tried to solve it using comparison test, but no results.
91794
Let  $\sum_{k=1}^{\infty}a_k$ be a series of real numbers. Suppose that $\sum_{k=1}^{\infty}a_kb_k$ converges for every bounded sequence {$b_k$} ? Prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely.  Here is my thought process:  Since {$b_k$} is bounded, every |$b_k$| $\leq$ M for all k.  So $\sum_{k=1}^{\infty}|a_kb_k|$ = $\sum_{k=1}^{\infty}|a_k||b_k|$ $\leq$ $\sum_{k=1}^{\infty}|a_k|M$ = $M$$\sum_{k=1}^{\infty}|a_k$| and since $\sum_{k=1}^{\infty}|a_kb_k|$ converges, $\sum_{k=1}^{\infty}|a_k|$ also converges.  However, I am having a hard time figuring out how to prove absolute convergence. I realize that they are similar questions on this website but they're asking to prove that $\sum_{k=1}^{\infty}a_kb_k$ converges, rather than $\sum_{k=1}^{\infty}a_k$ converging absolutely. Would really love any help, I'm quite stuck!
55965
Here is an exercise, on analysis which i am stuck.    How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?
53269
Let $(x_n)$ be a sequence of real numbers.  Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .              Thanks for any help.
64925
Let $\{a_n\}$ be a sequence of real number, let $\{n_k\}$ be an increasing sequence of positive integer and let $A_n=a_{n_k}+a_{n_{k+1}}+...+a_{n_{k+1}-1}$ for $k \in N$. Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges. Is the converse true?  I know that $\sum_{n=1}^\infty a_n$ converges implies that $lim _{n_&gt;\infty} a_n =0$, can I conclude that $A_n$ ->$0$ as well, even if I can conclude that, it still doesn't prove that $\sum_{k=1}^\infty A_k$ converges, am I correct?