Cross products in the light of linear transformations | Chapter 11, Essence of linear algebra

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Strong rec Johnson · 2022-06 · Mathematics Stack Exchange · on “How to intuitively understand cross-product”

In the end, a friend showed me a great YouTube video by 3B1B that gives one of the best explanations of the cross product. For those that may wish to see it; [this video] Thank you to those that commented and helped me!

Recommends user832835 · 2020-10 · Mathematics Stack Exchange · on “Understanding 3blue1brown's derivation of the definition of the cross product”

I've been following through with 3blue1brown's linear algebra series, and I have a question regarding the definition [this video] of the cross product he gives. $$\begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \det\left(\begin{bmatrix} x & v_1 & w_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \end{bmatrix}\right)$$ where $p$ is the resultant vector from the cross product of $v$ and $w$ for any $x$ , $y$ and $z$ . 3blue1brown essentially says that since the determinant of a matrix is the area of the parallelepiped with side lengths of the column vectors, the determinant is also just the height of that parallelepiped times the base of it. And the dot product

Recommends Ben Grossmann · 2020-10 · Mathematics Stack Exchange · on “Explanation of the formula for determinant (and cross product).”

You might find my post here [LINK] to be helpful. The video that I refer to in my comment can be found here [this video] .

Recommends jng224 · 2021-02 · Mathematics Stack Exchange · on “Cross product and determinants”

There is a great 3b1b video [this video] on determinants. However, this questions seems to be about pure math to me.

Recommends StudentsTea · 2023-01 · Mathematics Stack Exchange · on “Hey, Poincaré! Do vectors have type?”

I think there are a few key points: Take note of how this video [this video] says to think of the dot product's left operand as a transformation function and the right operand as a tuple of column vectors, and that the transformation function takes a tuple of column vectors from one vector space and gives you their expression in the language of some other vector space . Which vector space? The one from which the column vectors of the right operand were taken. Build up the dot product equality expression one vector at a time. Or if you want to avoid some work, toggle row and column vectors in and out of the equality. When you toggle the number of column vectors in the left operand, you change the

Cites John Strong · 2017-04 · Mathematics Stack Exchange · on “Geometric interpretation of duality in optimization”

There are several beautifully written posts on stackexchange about duality. For example: A technical explanation of duality [LINK] that attempts to offer some intuitions including the insight that the primal and dual each sort of encode the other, so that you could reverse engineer one in order to reconstruct the other. In the same post, I found An interesting explanation of duality using epigraphs [LINK] This post on Quora [LINK] was quite good. It maps $f(x)$ (to be minimized) as a function of $g(x)$ the constraint (i.e. a function of a function), something no one else does, and garners some interesting graphic insights with this trick. This post on Science4All [LINK] illustrates the ying

Cites Adam · 2019-02 · Mathematics Stack Exchange · on “Cross product in $\mathbb R^n$ in Spivak's Calculus on Manifolds”

For $n=3$ , Spivak's definition is basically the way that 3Blue1Brown describes a cross product in [this video] , and I like it a lot more than the usual parallelogram-and-right-hand-rule approach. As for generalizing, there is no way to combine two vectors in $\mathbb{R}^n$ into a new vector that behaves the way $\times$ does in $\mathbb{R}^3$ but this linear functional approach is a perfectly good way to combine $n-1$ vectors in $\mathbb{R}^n$ in a way that behaves just like $\times$ does in $\mathbb{R}^3$ .

Cites Neptuneful · 2020-01 · Mathematics Stack Exchange · on “Geometric interpretation of the cofactor expansion theorem”

Already some good answers here but here's how I try to visualize it. It consists of steps that are redundant but nonetheless useful for visualization. This might be kinda long but I'm writing it in a way will be helpful to me if I need to revisit it, and therefore hopefully helpful to others. Let's begin with $$ A = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix} $$ and picture the paralellepiped formed by the 3 column vectors. $$ \left[ \left({\begin{matrix} a \\ b \\ c \end{matrix}}\right) \left({\begin{matrix} d \\ e \\ f \end{matrix}}\right) \left({\begin{matrix} g \\ h \\ i \end{matrix}}\right) \right] $$ Assume A is invertible (nonzero determinant). Without loss of ge

Cites Clemens Bartholdy · 2021-02 · Mathematics Stack Exchange · on “Why does function being linear imply it can be represented as matrix multiplication?”

In the video where 3blue1brown explains determinants [this video] , this point in question is brought up at 6:35 but what is the intuition/ justification for this statement?

Cites StudentsTea · 2023-01 · Mathematics Stack Exchange · on “Hey, Poincaré! Do vectors have type?”

As I ask more questions trying to sort out how the jargon of Linear Algebra fits together, I'm starting to think of both column and row vectors as sometimes having a type, if you'll allow me to borrow jargon from Computer Science. Here's what I mean: Vectors have an order. That is--where their components are written has a meaning, it isn't arbitrary (by that I mean that if $a \neq b$ then $\begin{bmatrix}a & b\end{bmatrix} \neq \begin{bmatrix}b & a\end{bmatrix}$ ), and that order-based meaning sometimes has a type in the Computer Science meaning of the word "type" . Sometimes. Maybe. Let there be a vector $\begin{bmatrix}a \\ b \\ c\end{bmatrix}$ in a space whose basis is $\hat{i}$ , $\hat{j