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4 votes
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Is $\text{Id}-T$ always invertible when $\lim_{n\to\infty} T^n = 0$?

Injective: Let us first prove that $I-T$ is injective. For this we check that the kernel is trivial. Let $v$ be in the kernel of $I-T$, i.e. $(I-T)v=0$, then $v=Iv=Tv$. However, this cannot be as ...
Severin Schraven's user avatar
4 votes
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Prove that the sequence $(A_n(f))$ is convergent in the normed space $C[0,1]$. Prove that the sequence $(A_n)$ is not convergent in the operator norm.

To prove that $A_nf$ is convergent, just notice that $$ \|A_n(f)-f\|_{\infty}=\max_{[n/n+1,1]}|f(x)-f(n/n+1)|. $$ By continuity at $x=1$, we know that the latter maximum approaches zero as $n\to\infty$...
GReyes's user avatar
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2 votes
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How do I find subspaces and quotients of the vector spaces $V_i$ such that the maps become injective and surjective?

For the first problem, $W_1=V_1,W_2=g(W_1),W_3=fg(W_1)$, and similarly for the second problem $W_1=V_1/\text{ker}(fg), W_2=V_2/\text{ker}(f),W_3=V_3$ ADDED: Any such sequence for the first problem $$...
Ben's user avatar
  • 7,011
2 votes

Transformations of Function

I will give you a concrete example to see why does this make sense: $\textbf{Question}$:Let $D$ be the closed region surrounded by the $x$ axis, the $y$ axis, and $x+y=2$. Then compute $\iint_{D}e^{\...
Bowei Tang's user avatar
  • 1,300
1 vote
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Assume that $AT$ is a continuous operator. Show that $A$ is also a continuous operator.

You know that $y_n\to0$. Applying the Theorem to $T$ and $\{x_n\}$, you get $$ 0\leq \|x_n\|\leq m\|Tx_n\|=m\,\|y_n\|. $$ As $y_n\to0$, this shows that $x_n\to0$. Once you know this, $$ z=\lim_nAy_n=...
Martin Argerami's user avatar
1 vote

Completely non- normal matrix

The norm of the matrix $A$ is $$A=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\implies \| A \| = \frac 12 \left(|b| + \sqrt{|b|^2 + 4|a|^2} \right) $$ (for a proof see Corollary 2 from the ...
Exodd's user avatar
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1 vote

Is $\text{Id}-T$ always invertible when $\lim_{n\to\infty} T^n = 0$?

I assume that the space is complete. If $\|T^n\|\to 0$ then $\|T^k\|<1$ for some $k>0.$ Then the operator $I-T^k$ is invertible, by applying the Neumann series. We have $$I-T^k=(I-T)(I+T+\ldots +...
Ryszard Szwarc's user avatar
1 vote
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Why formula for operator in another basis is like this?

One way to think about this is as follows: Suppose that $T\colon V \to V$ is a linear map from a finite-dimensional vector space to itself. If we pick a basis $B= \{b_1,\ldots,b_n\}$ of $V$, then we ...
krm2233's user avatar
  • 4,851
1 vote

The left shift map and its image.

I claim that ${\rm Im} T$ is equal to the set of constant sequences $(x, x, x, \dots)$, where $x \in \mathbb{R}$ - we will call this set $A$. ($A \subseteq {\rm Im} T$) This is trivial; any sequence $(...
K. Jiang's user avatar
  • 7,992
1 vote
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Why is omitting a column that is a linear combination of other columns not considered an elementary operation in solving a system of linear equations?

You seem to be saying that $$ \begin{cases}2x+3y+z=1\\ x+3y+2z=3\end{cases} \qquad\text{ and } \begin{cases}2x+z=1\\ x+2x=3\end{cases} $$ are the same system? The second column is a linear combination ...
Martin Argerami's user avatar
1 vote

Prove that the canonical mapping from an infinite dimensional vector space to it's double dual is a one-to-one mapping.

Let $\psi: V \to V^{**}$ be the canonical injection $v\mapsto (g \mapsto g(v))$, we want to prove that $\psi$ is injective. If we have some non-zero $v\in V$ we can extend $\{v\}$ to a basis of $V$ ...
Kilian's user avatar
  • 111
1 vote
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understanding the dimension theorem proof and analysing it

Your proof is almost correct! The only thing is you need to argue why it is that $Tb_{k+1},\dots,Tb_n$ form a basis for $\text{im } T$. First we will argue that they are linearly independent. Suppose $...
moboDawn_φ's user avatar
1 vote
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Prove that $A$ is a bounded automorphism of the vector space $X$. Also prove that $\sigma(A) = \{ \lambda \in \mathbb{C} \mid |\lambda| = 1 \}$.

When $c=0$, $A$ is the identity operator, whose spectrum is just $\{1\}$, so $\sigma(A)=S^1$ doesn't hold. We assume $c\not=0$ in the next. If $\lambda\in\sigma(A)$, $|\lambda|\le\|A\|=1$, and we also ...
Just a user's user avatar
  • 16.8k

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