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If $f(x) = f(x − a) + f(x − b) − f(x − a − b)$, is $f$ $\Bbb Q$-linear?

A function as described is necessarely $\mathbb{Q}-$affine, not linear. In fact, the solution is hidden in your argument: you proved that $$f(x+y)=f(x)+f(y)-f(0),\qquad\forall x,y\in V.$$ In ...
ecrin's user avatar
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Generalized Rotational Matrix for n-dimensional Euclidean Vector Spaces

The right interpretation of that matrix is that it rotates in the $ij$-plane of $n$-space, rotating the positive $i$-axis towards the positive $j$-axis by some small amount (at least for $\theta$ ...
John Hughes's user avatar
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2 votes

Is it possible to efficiently create a matrix M in which the elements are the sum of all possible "path-products" of matrix A?

If I've understood you correctly, we have $$M = A + A^2 + A^3 + \dots = \frac{A}{I - A}.$$ (In general the notation $\frac{X}{Y}$ for two matrices is ambiguous since it could refer either to $XY^{-1}$ ...
Qiaochu Yuan's user avatar
1 vote
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Baby Rudin Theorem 9.7b: how to verify $\|A-B\|$ has the properties of a metric?

Linear transformations are functions. So yes, the equality $A=B$ means $Ax=Bx$ for all $x$.
Martin Argerami's user avatar
1 vote
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Understanding of Rudin Definitions 9.6c: $|A \mathbf{x}| \le \lambda|\mathbf{x}|$ for all $\mathbf{x} \in \Bbb{R}^{n}$ $\implies$ $\|A\| \le \lambda$

A little complement: Yes, and you have nearly established the following fact: The following three non-negative real numbers are equal, and are therefore all equivalent definitions of $\|A\|$: \begin{...
Asigan's user avatar
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0 votes

How do I find maximal quotients of subspaces of the vector spaces $V_i$?

(1) Clearly, $W_2\subset h(V_1)\cap g(V_3)$, so $W_1\subset h^{-1}(h(V_1)\cap g(V_3))=h^{-1}(g(V_3))$ and $W_3\subset g^{-1}(h(V_1)\cap g(V_3))=g^{-1}(h(V_1))$, and so $W_4\subset f(g^{-1}(h(V_1)))$. ...
Alex Ravsky's user avatar
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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
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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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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

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
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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

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
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0 votes

The left shift map and its image.

Let $\epsilon>0$ be arbitrary. For $\bar x\in X$ exists some $x\in \mathbb R$ and $N\in \mathbb N$ with $|x_n-x|<\frac\epsilon4$ for all $n\geq N$. You can now use this: \begin{align*}L^n(\bar x)...
Konstruktor'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
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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
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
0 votes

Reflection across a line?

Here is a perspective which requires only basic geometric intuition and knowledge of change of basis matrices. I give an alternative derivation using complex variables as well. What Should a ...
C Squared's user avatar
  • 3,612
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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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
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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
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0 votes
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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
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Let $X$ be a Banach space and let $A : X \to X$ be a linear operator such that $AT - TA \in \mathcal{B}(X)$ for every $T \in \mathcal{B}(X)$.

As you wrote, $(AT_f - T_f A)x = f(x)Au - f(Ax)u$, i.e. $f(Ax) u = f(x) Au - (AT_f - T_f A) x$. So $$ \eqalign{|f(Ax)| &= \|f(Ax) u\| = \|f(x) Au - (AT_f - T_f A) x\| \cr &\le \|f(x) Au\| + \|...
Robert Israel's user avatar
-1 votes

Action of matrices on right vector spaces

Definition: Let $R$ be a ring. A $\textbf{right $R$-module}$ is an additive abelian group $A$ together with a function $A\times R\to A$ such that for all $r,s\in R$ and $a,b\in A:$(i)(a+b)r=ar+br (ii)...
Bowei Tang's user avatar
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2 votes
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Let $(X, \|\cdot\|)$ be a normed space, $a_1,a_2,\ldots, a_n \in \mathbb{C}$ and $x_1,x_2,\ldots, x_n$ linearly independent vectors of the space $X$.

I will expand on the answer in the comments. For the first and third question, note that in order to describe a linear functional, it is sufficient to say what it does on a basis. Then the function on ...
Dean Miller's user avatar
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0 votes
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Let $X$ be a reflexive space and let $f \in X^*$, $ Px = x - \frac{f(x)}{\|f\|} y, \quad x \in X. $

We verify first that $P^2=P$. Observe that if $x\in X$, then $$P^2x=P(Px)=Px-\frac{f(Px)}{\lVert f\rVert}y=Px$$ where the second term vanished as $Px\in\ker f$ by part (a). This shows the claim. The ...
Lorago's user avatar
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1 vote
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Thus $MV - VM = V^2$. So the spectrum of $V^2$ is $\sigma(V^2) = (\sigma(V))^2=0$. Why??

It's just written awkwardly, here $\sigma(V)^2$ should mean $\{z^2 : z \in \sigma(V)\}$ and not $\sigma(V) \times \sigma(V) = \{(z, w) : z, w \in \sigma(V)\}$. That $\sigma(V^2) = \{z^2 : z \in \sigma(...
George C's user avatar
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4 votes
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We are interested in finding for which $\lambda$ the operator $A - \lambda I$ is not surjective.

You can solve for $f$ algebraically: $$f(x)=\frac{g(x)}{\ln x-\lambda}.$$ This gives a continuous result for any continuous $g$ iff $\lambda$ does not belong to the range of the logarithm on the ...
Lieven's user avatar
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1 vote
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How to know whether a set of points can be rotated to lie in positive orthant?

I think I have, many years later, stumbled on an answer to this question. Stack the set of $d$ points into a matrix $X \in \mathbb{R}^{N\times d}$. If the set of points can be rotated into the ...
Will Dorrell's user avatar
2 votes
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I want to prove the following: If $(X, \|\cdot\|)$ is a Banach space, then every Cauchy sequence in $(X, \|\cdot\|_1)$ converges in $(X, \|\cdot\|)$

You yourself have noted that $||x||\leq ||x||_{1}$ for all $x\in X$. Now suppose $x_{n}$ is a Cauchy sequence in $(X,||\cdot||_{1})$. Then, $||x_{n}-x_{m}||_{1}\xrightarrow{n,m\to\infty} 0$. So you ...
Mr.Gandalf Sauron's user avatar
2 votes
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$\|x\|_0 = \sup_{T \in G} \|Tx\|$ is equivalent to the norm $\|\cdot\|$ and in which all operators $T \in G$ are isometries.

If $T_0$ is the identity element of $G$ then $\sup_{T \in G}\|TSx\|\ge |T_0Sx\|=\|Sx\|$. Also, $\sup_{T \in G}\|TSx\|\le \|Sx\|$ because $\|Ty\|\le \|y\|$ for all $y$.
geetha290krm's user avatar
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3 votes

I want to prove the following: If $(X, \|\cdot\|)$ is a Banach space, then every Cauchy sequence in $(X, \|\cdot\|_1)$ converges in $(X, \|\cdot\|)$

If $\|x_n-x_m\|_1 \to 0$ then $\|(x_n-x_m)-A(x_n-x_m)\| \to 0$ and $\|A(x_n-x_m)\| \to 0$. By triangle inequality this gives $\|x_n-x_m\| \to 0$. So $(x_n)$ converges in $(X,\|.\|)$.
geetha290krm's user avatar
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Linear operator as linear combination of positive operators

Am I supposed to use 1. in order to prove 2. ? Yes. In fact, 1. is the hard part and 2. is the easy part. What you do, is write $$ T=\frac{T+T^*}2+i\,\frac{T-T^*}{2i} $$ and use part 1. on each of ...
Martin Argerami's user avatar
1 vote

Question about Euler angles and rotation about relative and fixed frames?

Observe that $R_{X,Y,Z}(\alpha,\beta,\gamma) =R_Z(\alpha)R_Y(\beta)R_X(\gamma)= R_{Z',Y',X'}(\alpha,\beta,\gamma)$. Explanation: In the Euler angle notation, once I have rotated about the $Z=Z'$ axis, ...
Wyatt Kuehster's user avatar
0 votes

Show that the given map is continuous in a Hausdorff space

$S$ is by definition the map that sends $\lambda = (\lambda_1,\ldots,\lambda_n)$ on $\sum_{k = 1}^n \lambda_kv_k$ where for all $k$, $v_k = T(e_k) \in F$. The way the $v_k$ are defined doesn't matter ...
Cactus's user avatar
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2 votes
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Determine the spectrum of the operator $A$. Is the spectrum composed only of eigenvalues of the operator $A$?

The two directions that you are missing: Now, put $x_n=(1,1,...,1,0,...)$, with the $1$s up to position $n$. Then $Ax_n=(1,1,...,1,1/2,0,...)$. So, $\|Ax_n\|/\|x_n\|=\frac{\sqrt{n-1+1/2^2}}{\sqrt{n}}$....
ameg's user avatar
  • 89
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Determine the spectrum of the operator $A$. Is the spectrum composed only of eigenvalues of the operator $A$?

You can solve the system of infinitely many simultaneous equations $\frac{x_i+x_{i+1}}2-\lambda x_i=0$ as $x_{i+1}=(2\lambda-1)x_i.$ This means that the kernel of $A-\lambda I$ consists of (multiples ...
Lieven's user avatar
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If the operator $A$ is self-adjoint and $\lambda$ and $\mu$ are distinct eigenvalues of $A$, then their corresponding eigenspaces are orthogonal.

(a) You have that $(A^*x,x)=(x,Ax)=\overline{(Ax,x)}=(Ax,x)$. Here the hypothesis that $(Ax,x)$ is real was used in the last equation. Therefore, $((A-A^*)x,x)=0$, for all $x\in H$. Let me call $(x,y)...
ameg's user avatar
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1 vote

Given two linear transformations T1 and T2, show that range T1 = range T2 if and only if there is an invertible operator S such that T1=T2S.

Since no one has answered this, I'll give it a shot, though please bear in mind that I'm just working through this stuff for the first time and could be missing something. (: As far as I can tell, the ...
Jake Khawaja's user avatar
1 vote
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Let $X$ be a Banach space and $M \subset X^*$. Prove: $M$ is bounded if and only if for every $x \in X$ {$\varphi(x) \mid \varphi \in M$} is bounded.

The direct implication is ok, if $𝑀$ is bounded. This means that there exists a constant $𝐶>0$ such that $\|f\|\leq 𝐶$ for all $ f \in 𝑀$. Using definition of $\|f\|$, it means that $\sup_{x\in ...
Ilovemath's user avatar
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1 vote
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Let $Y,Z$ be closed subspaces of a Hilbert space. Prove that $(Y^\bot + Z^\bot)^\bot = Y \cap Z$. Does the statement hold when $Y,Z$ are not closed?

For the first part, we observe that, $$\begin{split}h \in (Y^\perp + Z^\perp)^\perp &\Leftrightarrow h \perp k, \; \forall k \in Y^\perp + Z^\perp\\ &\Leftrightarrow h \perp k, \; \forall k \...
David Gao's user avatar
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2 votes
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Let H be a Hilbert space and $A \in B(H)$. Prove or disprove the following statements.

Hint for (a): Solution for (a): Hint for (b): Solution for (b):
Lorago's user avatar
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1 vote
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If the square of a matrix is negative definite, then the dimension is even.

If $A$ is a real symmetric matrix, then as you showed it is impossible for $A^2$ to be negative definite. However, it is possible to have $A^2$ negative definite if $A$ is not symmetric. For example,...
Robert Israel's user avatar
0 votes
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Defining $\mathcal{A}(S) = TS$ implies $\mathcal{A}$ and $T$ share the same eigenvalues. Need help with proof step.

As discussed in the comments, the backward direction has been shown. Now I show the forward direction. Let $\lambda \in E_T$. This implies there exists a nonzero $u \in V$ such that $$ Tu = \lambda u ...
Paul Ash's user avatar
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0 votes

$A^2 + I = 0$ show that the degree of $A$ is even

No no! You have to show it in such a way that it can be called proof and answer for any situation. I mean that the solution you have to provide for these questions should always answer. For example, ...
Overdeliver99's user avatar
5 votes

$A^2 + I = 0$ show that the degree of $A$ is even

Suppose $A^2+I_n=0$. Then $A^2=-I_n$. If $n$ is odd then $det(-I_{n})=-1$ so $-1=det(A^2)=det(A)^2$ but $det(A)\in\mathbb{R}$ so $det(A)^2\ge 0$. This is a contradiction. ($I_n$ is the $n\times n$ ...
Lucenaposition's user avatar
4 votes

Linear transformation definition redundancy?

Yes, from $T(v+w)=T(v)+T(w)$ for each two vectors $v$ and $w$, you can deduce that $T(2v)$ is always equal to $2T(v)$. However, it does not follow that, say, $T(\pi v)=\pi T(v)$.
Another User's user avatar
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