# What is the "determinant" of two vectors?

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?

• 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 $\color{blue}{\det \begin{pmatrix} v_1 & w_1 \\ v_2 & w_2\end{pmatrix}}$. Mar 10, 2019 at 0:26
• @MinusOne-Twelfth yes
– user
Mar 10, 2019 at 0:26
• That is the determinant of their components. Mar 10, 2019 at 0:27

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).

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: $$\det(v,w)=\det\begin{bmatrix}v_1&w_1\\v_2&w_2\end{bmatrix}.$$

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.

• 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. Mar 10, 2019 at 2:12