# Tag Info

10

A matrix is nothing but a particular representation of a linear map (with respect to a choice of basis in source and target space). The formula is what results (naturally) if you look at the composition of such maps and write them down using a matrix.

7

One reason is that it gives you associativity with a vector: If $A$ and $B$ are matrices and $x$ is a vector, then $$(AB)x = A(Bx)$$ ETA: This doesn't say anything more than Thomas's answer, by the way; I thought it might help if it were presented in a more tangible way, though.

4

Let $V$ be a finite-dimensional vector space, $\mathcal{B}$ be a basis of $V$ and $f,g\in\textrm{End}(V)$. One has: $$\textrm{mat}_{\mathcal{B}}(f\circ g)=\textrm{mat}_{\mathcal{B}}(f)\times\textrm{mat}_{\mathcal{B}}(g).$$ I think the definition of matrix multiplication derives from there.

4

We say $\lim_{{x}\to{-\infty}}f(x)=L$ if $\forall\epsilon>0, \exists M \in \mathbb{R}$ such that $x<M \Rightarrow |f(x)-L|<\epsilon$ We say $\lim_{{x}\to{\infty}}f(x)=\infty$ if $\forall M>0, \exists N \in \mathbb{R}$ such that $x>N \Rightarrow f(x)>M$

4

It is just a matter of definitions. Let $z$ be a complex number. Then, $z$ may be written (uniquely) as $z = a + bi$. $a$ is said to be the real part of $z$. $b$ is said to be the imaginary part of $z$. $z$ is said to be a purely imaginary number if its real part is equal to $0$. Hence, $0$ is a purely imaginary number because its real part is $0$.

3

No, for any $(a,b,c)\in\Bbb N^3$ and $a+b+c\leq n$ then, $$\dbinom{n}{a}\dbinom{n-a}{b}\dbinom{n-a-b}{c}=\dbinom{n\qquad\qquad\qquad}{a,b,c, n-a-b-c}\;(n-a-b-c)!$$ Unlike the binomial coefficient, the usual convention is not to leave the last lower term of the multinomial coefficient implicit.   The sum of the lower terms is required to equal the ...

2

You can say that it converges in the extended real numbers, sometimes denoted $\overline{\mathbb{R}}$. However, usually in the context of the ordinary real numbers we do not say that it converges simply because $\infty$ is not a real number.

2

A finite sheeted covering is just a covering space of finite degree. In general a covering space of degree $n$ is a continuous surjective map $q : E\rightarrow X$ such that for every $x\in X$, there exists a neighborhood $V\subseteq X$ such that $q^{-1}(V)$ is a disjoint union of $n$ copies of $V$ each getting sent homeomorphically onto $V$ by $q$. Of course ...

2

A couple of relevant results can be found in chapter 4 of Steven Roman – Lattices and Ordered Sets. Using this, we can prove your claim under the additional assumption that there are no infinite chains. This also shows how your idea fits in with more conventional lattice theory vocabulary. Claim. Let $M$ be an upper semimodular lattice with no infinite ...

2

This is the correct definition. See, for example, page 7 of Isaac Goldbring's notes here. One thing to note is that this is just the usual $L^2$ space on $G$ with respect to $G$'s Haar measure $\mu$, normalized so that $\mu({1}) = 1$ (although the particular normalization doesn't matter). Since $G$ is discrete, $\langle f, g \rangle = \int f\bar{g} d\mu = ... 1 When you have a composition of function written as$f(g(x))$this is equivalent to$x \rightarrow g(x) \rightarrow f(g(x))$applying$g$and then$f$on the image of$g$. This operation makes sense only when the image of$g$is contained in the domain of$f$1 It is the same thing because$Kg$runs over$G/K$when$g$runs over$G$. However, the key point is that$\bar{\phi}(Kg) = \phi(g)$is well defined, that is, does not depend on the choice of the representative$g$for the coset$Kg$. This follows because$K = \ker \phi$, of course. 1 Formally, the statement is more correct. If you define a function$f:A\longrightarrow B$, you must specify the action of$f$over a generic element of$A$. In your case, a more pompous statement (but formally more precise) should be:$\bar{\phi}:G/K\longrightarrow \phi(G)$is defined in the following way: for every coset$\alpha \in G/K$, let be$g\in G$a ... 1 If one weakens the condition as said, that is considers sets$\Lambda \subseteq \def\P{\mathfrak P}\P(S)$that have (1)$S \in \Lambda$(2) If$A, B \in \Lambda$, then$A \setminus B \in \Lambda$(3) If$(A_n) \in {}^{\mathbf N}\Lambda$is increasing, then$\bigcup_n A_n \in \Lambda$. then every$\Lambda\$ satisfying (1)-(3) is - of course - a ...

1

Well, think about all these definitions as about rules written in blood and agreed by the community. This is slightly extreme and even somewhat wrong analogy, but the answer is in the "mood" of the question. With theorems it is slightly more complicated, they are result of many years of research and work of many scientists (well this is also true for some ...

Only top voted, non community-wiki answers of a minimum length are eligible