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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 ...
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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 $\theta$ ...
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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 $XY^{-1}$ ...
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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 $x$.
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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{...
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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 $W_4\subset f(g^{-1}(h(V_1)))$. ...
• 92.7k
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 ...
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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 $n\to\infty$...
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1 vote
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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)...
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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 ...
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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)...
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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 ...
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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 ...
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1 vote
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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 ...
1 vote
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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 $\sup_{x\in ... • 2,960 1 vote Accepted Let$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, $$\begin{split}h \in (Y^\perp + Z^\perp)^\perp &\Leftrightarrow h \perp k, \; \forall k \in Y^\perp + Z^\perp\\ &\Leftrightarrow h \perp k, \; \forall k \... • 7,665 2 votes Accepted Let H be a Hilbert space and A \in B(H). Prove or disprove the following statements. Hint for (a): Solution for (a): Hint for (b): Solution for (b): • 10.3k 1 vote Accepted If the square of a matrix is negative definite, then the dimension is even. If A is a real symmetric matrix, then as you showed it is impossible for A^2 to be negative definite. However, it is possible to have A^2 negative definite if A is not symmetric. For example,... • 453k 0 votes Accepted Defining \mathcal{A}(S) = TS implies \mathcal{A} and T share the same eigenvalues. Need help with proof step. As discussed in the comments, the backward direction has been shown. Now I show the forward direction. Let \lambda \in E_T. This implies there exists a nonzero u \in V such that$$ Tu = \lambda u ... • 1,376 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, ... 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$n\times n$... 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)\$.
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