Hot answers tagged

11

The determinant is the volume of parallelepiped formed by the vectors. Wikipedia is more precise: Determinants express the signed $n$-dimensional volumes of $n$-dimensional parallelepipeds. The determinant is zero iff the vectors are linearly dependent. In this sense, it is a coarse measure of the independence of vectors. The finer measure of ...


6

Adding to lhf's answer, you can think of the 'geometric' meaning of this volume interpretation: think of how $A$ acts on the vector space $V$. The volume of the parallelepiped is connected to this interpretation in that the parallelepiped is the image under $A$ of the unit cube. In other words, the parallelepiped is a small-scale model of how $A$ acts on ...


3

By definition, this is the projective linear group, sometimes denoted $PGL(n, \Bbb R)$. There are only finitely many elements in $SL(n, \Bbb R)\cap Z(G)$, so the homomorphism $$SL(n, \Bbb R) \hookrightarrow GL(n, \Bbb R) \to PGL(n, \Bbb R)$$ is an immersion and (as Lie groups), $$\dim PGL(n, \Bbb R) = \dim GL(n, \Bbb R) - \dim Z(GL(n, \Bbb R)) = n^2 - 1 = ...


3

In 2D, the claim clearly holds for $\mathbf a=(1,0)$ and $\mathbf b=(\cos \theta,\sin\theta)$, and the relation is also invariant under scaling either of the two vectors or rotating both, hence it holds for arbitrary 2D vectors. For higher dimensions, just notice that the two vectors $\mathbf a$, $\mathbf b$ span a two-dimensional subspace, for which the ...


3

When you have $n$ objects to choose from, and you want to choose $r$, if the order doesn't matter, and repetition is not allowed, the answer is $\displaystyle{n\choose r}$. If the order does matter, then you'd get $n(n-1)\cdots(n-r+1)=P(n,r)$ by the multiplication rule. However, the order doesn't matter, so that means that every selection of $r$ objects ...


2

Short answer: The function $\mathbf{X},\mathbf{Y}\to \operatorname{trace}(\mathbf{X}^T \mathbf{Y})$ defines an inner product on the space $\mathbb{R}^{m\times n}$. Long answer: In the general setting of a finite-dimensional spaces $E$, the usual definition of Frechet derivative of function $f: E\to \mathbb R$ at point $x$ is the linear map $L\in \mathcal ...


2

The fraction $\frac{a}c$ is the number of apples per child, and $\frac{c}a$ is the number of children per apple. Suppose that you have $x$ children per apple, where $x$ can be any positive real number. No matter what $x$ is, each one of those $x$ children must be getting $\frac1x$ of an apple, so there are $\frac1x$ apples per child. Thus, in the ...


2

Here's the worst possible way to keep straight which direction the inequality goes: In a typical programming language strings are sorts alphabetically, so that for example "cat" < "dog", since "c" < "d". Observe that $$\text{IL < LI.}$$"Integral(limit) < Limit(integral)". This is in my opinion extremely awful, because it has nothing to do with ...


2

I highly recommend reading through Ravi Vakil's notes on spectral sequences. For convenience I'll outline the general points relating spectral sequences to diagram lemmas. Diagram lemmas usually begin with a double complex. That is a grid of maps composing to $0$. Here the blue bullet denotes the $(0,0)$ position (or the "origin") of the grid. This grid ...


2

One may recall that, as $M \to +\infty$, we have $$ \begin{align} \int_1^M \color{blue}{\frac1{x^2}}\:dx&=\left[ -\frac1x\right]_1^M=1-\color{blue}{\frac1M} \to \color{blue}{1}, \\\\ \int_1^M \color{red}{\frac1{x}}\:\:dx&=\left[ \:\ln x\:\right]_1^M=\color{red}{\ln M} \to \color{red}{+\infty}. \end{align} $$ Thus your question might be equivalent to ...


2

General comments Disclaimer Before I get into my best efforts at motivating these identities/drawing on some geometric intuition, I want to explain why I believe there won't be a very satisfying answer posted. Certainly, things like multiplication by $i$ and the unit circle and the unit hyperbola have geometric interpretations, but the trig functions ...


2

Let me be very frank here. For a beginner who is studying limits for the first time (meaning age around 16 years) the terms "sufficiently close" and "arbitrarily close" are very difficult to handle. This is primarily because such a student at his stage of learning is acquainted mainly with algebraic simplications/manipulations and the main focus in algebra ...


2

Topology: Why you can't turn your shirt inside out while wearing it.


1

You need to assume that $K<N$, not that $K>0$: you want to be sure that there is at least one $1$. Call a subset of $S$ odd if it contains an odd number of ones and even otherwise. First suppose that $N-K$ is odd. If $A$ is any subset of $S$, those $N-K$ ones are split between $A$ and $S\setminus A$, so exactly one of these complementary sets is odd. ...


1

(As others have remarked we have to assume $K<N$.) We are given a finite set $S$ and a function $f:\>S\to\{0,1\}$ with $f(a)=1$ for at least one $a\in S$. Then each subset $A\subset S$ has a parity $$p(A):=\sum_{x\in A}f(x)\quad{\rm mod}\ 2\ .$$ For any subset $A'\subset S':=S\setminus\{a\}$ the two sets $A_0:=A'$ and $A_1:=A'\cup\{a\}$ have ...


1

The formalization of Consider for example, the functors $(- \times B) \times C$ and $- \times ( B \times C): \textbf C \Rightarrow \textbf C$. For an object $\mathcal A$ we have the isomorphism $(A \times B) \times C \cong^{h_A} A \times ( B \times C)$ via $h_ A$. Now if we change $ A$ into $\mathcal A'$, we still have a ...


1

For any distribution that's roughly normal, you can get a good feel from the 68, 95, 99.7 rule mentioned in the comment. More generally, for arbitrary distributions, the standard deviation measures the variability of the data. More specifically, it measures the average distance of values from the mean. A small standard deviation means that most values are ...


1

The comment and @bubba 's answer offer useful technical information. For your use case Let's say for example you are software developer and you need to measure latency of the system over time and present this to the management I'd recommend some care in collecting measurements. The latency is probably highly dependent on system load, perhaps ...


1

For general distributions, Chebyshev's inequality is applicable https://en.wikipedia.org/wiki/Chebyshev%27s_inequality. It says that $1-\frac{1}{k^2}$ of the data falls within $k$ standard deviations of the mean. (E.g. $\frac34$ of the data falls within $2$ standards deviations of the mean.)


1

Let's see if we can make the idea of "arbitrarily close" precise and the best way to do that is by using open intervals on R. To really understand this, you have to understand the precise definition of a function on it's domain. Consider the precise definition of a function: Let A and B be nonempty sets.A function F from A into B is a nonempty subset of ...


1

Intro I'll present an incomplete interpretation framework. Let's being by considering two points in one dimension. We can easily generalize the results to more than one dimension. Our matrix is given by, $$M=\begin{bmatrix} 0 & |x_1-x_2| \\ |x_1-x_2| & 0 \end{bmatrix}$$ The eigenvalues are given by $$\lambda_1=-\lambda_2=|x_1-x_2|$$ The ...


1

Maybe it would help to consider that in cases like this if the integral of one function diverges and a similar function converges, then there is some cutoff in between where on one side it converges (gets smaller fast enough) and on one side it diverges (doesn't get smaller quite fast enough). For the functions you asked about consider 1/x^p as p decreases ...


1

I don't think of it as intuition. It is just taking $\cosh(x)=(e^x+e^{-x})/2$ with $e^{ix}=\cos x+i\sin x$ and allowing complex values to be used anywhere. This all, to me, gets justified by the power series converging everywhere in the complex plane.


1

You can visualize its connection using such math packages as (Maple, Mathematica, Matlab, etc). The answer for current task can be seen as central symmetry of graphics for absolute value(on the picture) and, analogously, for argument. $f(z)=|\cos(z)|,\quad z=x+iy$


1

At it's most basic and intuitive: The arithmetic mean isn't about the absolute difference, it's about the actual difference: $\sum_{k=1}^n ({x_k-\bar{x}}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + (x_3 - \bar{x}) + ... + (x_n - \bar{x})$ $= x_1 + x_2 + x_3 + ... + x_n - n*\bar{x}$ If $\sum_{k=1}^n ({x_k-\bar{x}}) = 0$, then $x_1 + x_2 + x_3 + ... + x_n = ...


1

I think @lhf 's answer is the best intuitive explanation. In response to your edit: a system of $n$ linear equations in $n$ unknowns will always have a unique solution if and only if the determinant is not $0$, because that is the condition that says the rank is $n$ so the corresponding linear map is one to one and onto. But in practice you don't use ...


1

I will try to explain this via an analogy to (linear) geometry into two dimensions, call the space $L$. You will have 4 "types" of flats: the empty set $\phi$, single points $\{\{p\}: p \in L\}$ and all lines $\{\{\ell\}: \ell \in L\}$, and $L$ itself. Single points cover the empty flat, a line that contains a point covers that point, and L covers each ...


1

One might get some if none intuition about a group by knowing where it acts. In this case this group acts naturally on projective space, and in fact transitively and faithfully.


1

Your picture of two horns attached at a point is correct - but from there you can stretch the attached section until you have two small horns joined by a line, and from there deform it until it's $T \vee S^1$. I've attached a hastily-drawn picture. In answer to your "general construction" question: hopefully that's more obvious now that you've seen how the ...


1

As you said in your post, the closest example I can think of where we take a derivative of a function wrt another function is the functional derivative: $$\frac{\delta J}{\delta f(x)} = \frac{\partial \Phi}{\partial f}-\frac{d}{dx}\frac{\partial \Phi}{\partial f'}$$ Where: $$J[f] = \int_a^b \Phi[f] dx$$ This derivative is defined in a vector space of ...



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