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I came across the notation $\det(v,w)$ where $v$ and $w$ are vectors. Specifically, it was about the curvature of a plane curve:

$$\kappa (t) = \frac{\det(\gamma'(t), \gamma''(t)) }{\|\gamma'(t)\|^3}$$

What is it supposed to mean?

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    $\begingroup$ Are they 2D vectors? You may want to see this section of the Wikipedia article on Curvature. As you should be able to see from there, $\color{blue}{\det(v,w)}$ means $\color{blue}{\det \begin{pmatrix} v_1 & w_1 \\ v_2 & w_2\end{pmatrix}}$. $\endgroup$ Mar 10, 2019 at 0:26
  • $\begingroup$ @MinusOne-Twelfth yes $\endgroup$
    – user
    Mar 10, 2019 at 0:26
  • $\begingroup$ That is the determinant of their components. $\endgroup$
    – Bernard
    Mar 10, 2019 at 0:27

3 Answers 3

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They formed a matrix by stacking $\gamma'(t)$ and $\gamma''(t)$ next to each other as column vectors. You can also regard it as the cross product of the two vectors if you extend both with a $z=0$ coordinate and take the z component of the resulting vector (that way you can relate it to the 3d formula in a way).

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Vectors in a plane $v,w$ can be written as column matrices: $$v=\begin{bmatrix}v_1\\v_2\end{bmatrix}, \ \ \ w=\begin{bmatrix}w_1\\w_2\end{bmatrix}.$$ Put several of such column matrices side by side, and you get a square matrix: $$(v,w)=\begin{bmatrix}v_1&w_1\\v_2&w_2\end{bmatrix}.$$ The determinant $\det(v,w)$ is simply the determinant of this square matrix: $$\det(v,w)=\det\begin{bmatrix}v_1&w_1\\v_2&w_2\end{bmatrix}.$$

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In general, the determinant of $n$ vectors $v_1$, $v_2$, $\dots$, $v_n$ in $\mathbb R^n$ is the determinant of the matrix whose columns are $v_1$, $\dots$, $v_n$ (in that order).

Seen as an application whose inputs are vectors, the determinant has nice properties:

  1. multilinear, that is linear in each variable: $$\det(v_1,\dots, a v_j+b w_j,\dots,v_n) = a \det(v_1,\dots, v_j,\dots,v_n) + b\det(v_1,\dots, w_j,\dots,v_n)$$

  2. alternating: switching two vectors transforms the determinant in its opposite

$$\det(v_1,\dots, v_i, \dots, v_j,\dots,v_n) = \det(v_1,\dots, v_j, \dots, v_i,\dots,v_n)$$

  1. The value on the canonical basis $(e_1,\dots,e_n)$ of $\mathbb R^n$ is $1$.

$$\det(e_1,\dots,e_n) = 1 $$

Actually, it can be proved that the determinant is the unique alternating multilinear form whose value on the canonical basis is $1$. Many (most?) Linear Algebra books use this as a definition of the determinant (before extending the definition to matrices and then linear applications). I think it is equivalent but more satisfying than introducing the determinant by the strange "expansion along a row" formula most PreCalculus textbook use.

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    $\begingroup$ I believe the issue here was not that he didn't know the definition of determinant, but rather the fact that he had a determinant of two vectors rather than a matrix. $\endgroup$
    – lightxbulb
    Mar 10, 2019 at 2:12

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