Dot products and duality | Chapter 9, Essence of linear algebra

What people said

Recommends Ben Grossmann · 2016-11 · Mathematics Stack Exchange · on “What does really mean by the picture below?”

It doesn't directly answer your question, but you might like this video [this video] which gives you one way to think about dot-products.

Recommends MCCCS · 2017-09 · Mathematics Stack Exchange · on “Proofs of the Cauchy-Schwarz Inequality?”

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}\newcommand{\i}{\mathrm{i}}\newcommand{\text}[1]{\mathrm{#1}}\newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}\newcommand{\i}[0]{\mathrm{i}}\newcommand{\kume}[1]{\mathbb{#1}}\newcommand{\bold}[1]{\textbf{#1}}\newcommand{\italic}[1]{\textit{#1}}\newcommand{\kumedigerBETA}[1]{\rm #1\!#1}$ Here's a simple proof: $|\vec{x}\cdot\vec{y}| \leq \|\vec{x}\|\|\vec{y}\| $ Substitute $|\vec{x}\cdot\vec{y}| = \|\vec{x}\|\

Recommends Charlie Parker · 2017-11 · Mathematics Stack Exchange · on “What is an intuitive way to understand the dot product in the context of matrix multiplication?”

@1110101001 I think I just gave up. However, I think there are some insights here: [this video] however I don't think I fully digested it in a way that 100% made sense to me. If you find a way to explain it better please feel free to share it .

Recommends Andres Mejia · 2019-11 · Mathematics Stack Exchange · on “Intuition behind the directional derivative being zero when gradient is perpendicular to the direction”

Actually, I think the "True" issue here is understanding why the dot product formula $(x_1,\dots,x_n) \cdot (y_1, \dots,y_n)=\sum x_iy_i$ is projection for unit vectors. Once you accept this, the gradient condition is quite natural. I recommend this [this video] video for understanding this bit :). Let $f: \mathbb R^n \to \mathbb R$ be a smooth function with gradient $\nabla f$ . Then, at a point $\nabla f(p)$ is actually a linear map $T_p(\mathbb R^n) \to T_{f(p)}(\mathbb R)$ , which is really just $\mathbb R^n \to \mathbb R$ . In other words, it is the first order linear approximation to your function. You can think of $T_p(M)$ as having basis $\partial/\partial x_i$ , which is just the tangent

Recommends Sean49 · 2020-04 · Mathematics Stack Exchange · on “Is there a derivation or geometric intuition for the distance formula?”

After lots of reading and thinking, I think I figured it out. For posterity, the following is my reasoning. I am not claiming this to be any kind of formal proof or derivation, but rather a way of looking at the problem that gives me physical intuition for why the distance formula and metric tensor give what I intuitively think of as "length" in (Euclidean) space. First, a physical intuition for why a dot product gives a projection is very well explained by 3Blue1Brown here [this video] . The idea of "projecting a vector onto itself" feels to me like a good intuitive procedure to measure its length, and this video shows why that procedure is produced mathematically by summing the products of two

Recommends Sebastian Nielsen · 2021-01 · Mathematics Stack Exchange · on “Dot product vs Matrix multiplication, is the later a special case of the first?”

What @Pawel said, additionally, though, I would like to add that there is a nice duality between $1\times 2$ matrices and 2d vectors. 3Blue1Brown covers this in the 9th episode of his series 'Essence of Linear Algebra' [this video]

Recommends Paul Sinclair · 2021-04 · Mathematics Stack Exchange · on “Distance from circle to line segment: What is the intuition or geometry behind this algorithm?”

There are two things needed here (after it has already been tested that the endpoints are outside the circle): The distance $d$ from the center $C$ of the circle to the line, and The location of the point $Q$ on the line which is closest to $C$ . If $d > R$ , then the line does not intersect the circle and we are done (if the line has no intersection, then neither does the line segment). If $d \le R$ , then the line does intersect the circle. But that doesn't mean the line segment intersects. It could be on a portion of the line away from the circle. For this we need $Q$ , the point on the line closest to $C$ . If $Q$ is between $P_1$ and $P_2$ , then the segment intersects the circle. If $P

Recommends Klevis Imeri · 2022-06 · Mathematics Stack Exchange · on “Why matrix dot product (in linear algebra) is equal to vector dot product (in geometry)? A link between linear transformation and vector projection?”

I found the answer myself. The thing a was looking for is called duality. But more precisely what I was missing is what you need to take a number line where the $\vec{v_1}$ lies. And then define $\hat{u}$ which is kind of the unit vector in that number line. So logicaly $\vec{v_1}$ = [some number] x $\hat{u}$ . Now the idea is to squish the 2D space into this number line with a linear transformations. You find where the $\hat{i}$ and $\hat{j}$ land, which I called $\hat{\hat{i}}$ and $\hat{\hat{j}}$ respectively. And construct a linear transformation matrice where $\hat{\hat{i}}$ and $\hat{\hat{j}}$ are represented with the $\hat{u}$ unit vector. Because they lie in the same line as $\hat{u}

Recommends smichr · 2024-05 · Mathematics Stack Exchange · on “Geometric proof of dot product and projection”

After reading the the short post here [LINK] I wonder if the curiousity about the nature of the equivalence of the geometric and algebraic forms of the dot product arises because the referant has been forgotten: the area of a parallelogram they both describe. Consider three vectors, $u$ , $u^*$ and $v$ where $u^*$ is oriented at $90^\circ$ to to $u$ , and $v$ cuts the corner at an angle of $\theta$ wrt $u$ . [LINK] We can compute the area of the upper parallelogram geometrically if we consider $u*$ to be the base (with length $|u^*| = |u|$ ) and note that the width/height is $|v|\cos\theta$ : the area is $|u||v|\cos\theta$ . We can also compute the area using the algebraic approach in Gauss'

Cites HERO · 2019-08 · Mathematics Stack Exchange · on “Dots Products and Duality”

The dot product of two vectors can be defined either as $$\vec{A}\cdot\vec{B}=\left|\vec{A}\right|\left|\vec{B}\right|\cos \theta$$ or as, $$\vec{A}\cdot\vec{B}=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$$ I want to know how these two definitions are geometrically connected. Mr. Sanderson made a video on this, dubbed “ Dot products and duality [this video] ”. However, that is too brief an explanation for me to get a grasp of what was explained. I'm just a high school student, and all of my understanding about linear algebra is merely based on Sal Khan and Grant Sandersons videos. I understand linear transformations (matrix-vector multiplications) and all the mathematical concepts used in the video, but I j