Abstract vector spaces | Chapter 16, Essence of linear algebra

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Recommends Bruno Reis · 2016-09 · Mathematics Stack Exchange · on “why a set of all $2 \times 2$ matrices can be a vector space?”

That's an example of abstract vector spaces. They call vector spaces, but since the "object" that you are looking into have the properties of subspaces: Closed under multiplication; Closed under addition; Contains the $ \left\{ 0 \right\} $ "vector", or in this case, the null matrix. You can affirm that it'll be a vector space. To "visualize" that, try to imagine any matrix 2x2 added to another 2x2 matrix. You'll still be in a 2x2 matrix world. Same happens when multipling them... And because the null matrix is in the 2x2 matrix world, you can assume that it'll be a subspace, hence, a vector space. Watch this [this video] to get a better understanding of that kind of abstraction. It's hard to vis

Recommends Arthur · 2019-06 · Mathematics Stack Exchange · on “Is there any relationship between the vector defined as a vector or line matrix and the vector defined in physics?”

You could take a look at this 3blue1brown video [this video] where he discusses this exact topic. It's the last video in a series, so there might be some things he mentions that you are unfamiliar with, but the overall discussion of what makes a vector should be possible to follow regardless.

Recommends Kyan Cheung · 2021-04 · Mathematics Stack Exchange · on “What is $e^{\frac d{dx}}$?”

(Note on notation: we will be counting rows and columns from $0$ , i.e. the first row will be row $0$ , and the cell on row $i$ and column $j$ of a matrix $A$ will be denoted $a_{ij}$ . Also, instead of writing $\frac d{dx}$ we will write $D$ .) In this answer we will be using the monomial basis [LINK] as it is the basis used by 3blue1brown in an earlier video [this video] and also the simplest basis. $$e^M=\sum^\infty_{n=0}\frac{M^n}{n!}$$ If we want to find $e^D$ , we need to find $D^n$ for natural $D$ . Suppose $$D^n=\left(\begin{matrix}\vec{v_0} & \vec{v_1} & \vec{v_2} & \vec{v_3} & \cdots\end{matrix}\right)$$ Then $D^n$ maps $x^k$ to the polynomial represented by $\vec{v_k}$ . Now, for natur

Cites templatetypedef · 2016-09 · Mathematics Stack Exchange · on “In Linear Algebra, what is a vector?”

In modern mathematics, there's a tendency to define things in terms of what they do rather than in terms of what they are . As an example, suppose that I claim that there are objects called "pizkwats" that obey the following laws: $\forall x. \forall y. \exists z. x + y = z$ $\exists x. x = 0$ $\forall x. x + 0 = 0 + x = x$ $\forall x. \forall y. \forall z. (x + y) + z = x + (y + z)$ $\forall x. x + x = 0$ These rules specify what pizkwats do by saying what rules they obey, but they don't say anything about what pizkwats are . We can find all sorts of things that we could call pizkwats. For example, we could imagine that pizkwats are the numbers 0 and 1, with addition being done modulo 2. Th

Cites Neil G · 2016-12 · Mathematics Stack Exchange · on “Does the polynomial derivative matrix have complex eigenvalues and what do they mean?”

I was watching this video [this video] , which describes a matrix that applies the derivative to the space of polynomials. It's pretty clear that this matrix has no real eigenvalues since $$\frac{d}{dx}kx^n \ne \lambda x^n$$ for any $k \ne 0$. What if one tries to solve for complex eigenvalues and eigenvectors? Do these exist, and if so, how can we reason about them intuitively?

Cites jnez71 · 2017-02 · Mathematics Stack Exchange · on “Physical interpretation of Laplace transforms”

There's really a lot that can be said, but I will only delve into one geometric idea: the laplace transform, like many integral transforms [LINK] , is a change of basis ("coordinate system"). I consider this a "physical" interpretation because it is geometric- you will be able to imagine the laplace transform's actions on a function much like you imagine how a matrix can geometrically transform a vector. This description is also mathematically precise. If you have studied linear algebra you have likely run into the concept of a change of basis [LINK] . Imagine you have a vector on $\mathbb{R}^2$ . You construct a basis $\{\hat{a}_x, \hat{a}_y\}$ to represent your vector as $$ v = a_1\hat{a}_

Cites MCCCS · 2017-08 · Mathematics Stack Exchange · on “What are some mathematically interesting computations involving matrices?”

Here's the polynomial derivative matrix (where $n$ is the degree of the polynomial): $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}\newcommand{\i}{\mathrm{i}}\newcommand{\text}[1]{\mathrm{#1}}\newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}\newcommand{\i}[0]{\mathrm{i}}\newcommand{\kume}[1]{\mathbb{#1}}\newcommand{\bold}[1]{\mathbf{#1}}\newcommand{\italic}[1]{\mathit{#1}}\newcommand{\kumedigerBETA}[1]{\rm #1\!#1}$ $$\large \mathbf {Derivative\s matrix\s

Cites erfink · 2017-12 · Mathematics Stack Exchange · on “Understanding Vector Spaces”

The beauty of mathematics is abstraction, but we often pay a price for this abstraction. For a concrete example, let's talk about what "$3$" means. When first teaching someone to count and add, we focus on three stones, three fingers, three dots on a page, three books, etc. All of these are fundamentally different objects, but we can abstract all of them with the notion of "$3$." Since we are now talking about a supremely abstract concept of "$3$", we can make statements like $$3 + 4 =7$$ without needing to separately show that this is true for three stones and four stones make seven stones three dots and four dots make seven dots three fingers and four fingers make seven fingers and repeati

Cites jnez71 · 2018-06 · Mathematics Stack Exchange · on “Can someone clearly explain the discrete Fourier transform (DFT)?”

The Fourier transform is a change of basis ("coordinate system") for the vector space of integrable functions. Specifically, it is a pure rotation onto the basis of complex exponentials (sinusoids). This description is both intuitively geometric, and mathematically precise. Let me explain: If you have studied linear algebra you have likely run into the concept of a change of basis [LINK] . Imagine you have a vector on $\mathbb{R}^2$ . You construct a basis $\{\hat{a}_x, \hat{a}_y\}$ to represent your vector $v$ as, $$ v = a_1\hat{a}_x+a_2\hat{a}_y $$ To express $v$ on an alternative basis, $\{\hat{b}_x, \hat{b}_y\}$ , we make use of the inner product (or "dot" product) that geometrically doe

Cites Arthur · 2019-08 · Mathematics Stack Exchange · on “The Derivative of a Linear Operator”

@Frasch I don't think that's the right video. Did you mean the last one in the series [this video] about abstract vector spaces? The polynomials are there regardless of whether we represent them as a column vector or not, and they have derivatives regardless of whether we represent differentiation as a matrix. The linearity of the derivative holds regardless of this representation. Yes, I believe he represented polynomials in column-vector form just to show the parallel to everything else covered in the series.