[Linear Algebra] 10. Cross products and duality

3-D cross product formula:

$$
\mathbf{\vec{v}} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}
;
\mathbf{\vec{w}} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}
$$
$$
\mathbf{\vec{v}} \times \mathbf{\vec{w}} =
\det
\begin{pmatrix}
\begin{bmatrix}
\mathbf{\hat{i}} & v_1 & w_1 \\
\mathbf{\hat{j}} & v_2 & w_2 \\
\mathbf{\hat{k}} & v_3 & w_3 \\
\end{bmatrix}
\end{pmatrix}
= \mathbf{\hat{i}}(v_2w_3 - v_3w_2) + \mathbf{\hat{j}}(v_3w_1 - v_1w_3) + \mathbf{\hat{k}}(v_1w_2 - v_2w_1)
$$

 

People are typically told to just believe that the resulting vector has the following geometric properties. 1). Its length equals the area of the parallelogram defined by $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. 2). It points in a direction perpendicular to both $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. And this direction obeys the right-hand rule.

---

"Linear transformation from space to number is associated with a unique vector in that space" is main idea of duality. In other words, performing the linear transformation is the same as taking a dot product with that vector. That vector is called the "dual vector".

What cross product is?

Before learning what 3-D cross product is, we can imagine that 3-D cross product is determinant of $3 \times 3$ matrix which columns are separate 3-D vectors ($\mathbf{\vec{u}}, \mathbf{\vec{v}}, \mathbf{\vec{w}}$). Geometrically, $3 \times 3$ determinant is volume of a parallelepiped spanned out by those three vectors. Of course, It's not 3-D cross product, because cross product use only two vectors, but this idea actually get close to what the real cross product is.

 

To understand the connection between the computation and the geometry of the cross product, let's define a certain linear transformation from space to line.

$$
\mathbf{\vec{v}} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}
\ 
\mathbf{\vec{w}} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}
\ 
\mathbf{\vec{u}} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}
\qquad
f\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) =
\det
\begin{pmatrix}
\begin{bmatrix}
x & v_1 & w_1 \\
y & v_2 & w_2 \\
z & v_3 & w_3 \\
\end{bmatrix}
\end{pmatrix}
$$

 

The "dual vector"(=$\mathbf{\vec{p}}$) associated with this 3-D to 1-D transformation is going to be the cross product of $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$.

 

Geometrically, the output is the volume of a parallelepiped spanned out by three vectors. And one important fact about this transformation is that it's linear which can bring the idea of "duality".

What we're looking is the special 3-D vector $\mathbf{\vec{p}}$:
$$
\mathbf{\vec{p}} =
\begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix}
\qquad
f\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) =
\begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix}
\cdot
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
=
\det
\begin{pmatrix}
\begin{bmatrix}
x & v_1 & w_1 \\
y & v_2 & w_2 \\
z & v_3 & w_3 \\
\end{bmatrix}
\end{pmatrix}
$$

Computationally

Taking the dot product between $\mathbf{\vec{p}}$ and $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ will give us $p_1 \cdot x + p_2 \cdot y + p_3 \cdot z$ where $\begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix}$ are the coordinates of $\mathbf{\vec{p}}$. On the right side, determinant can be organized by $x(v_2 \cdot w_3 - v_3 \cdot w_2) + y(v_3\cdot w_1 - v_1\cdot v_3) + z(v_1 \cdot w_2 - v_2 \cdot w_1) - (1)$ where those constants involve certain combinations of the components of $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. And, those constants which are the particular combinations of the coordinates of $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ are going to be the coordinates of the vector $\mathbf{\vec{p}}$ that we're looking for. Right side computation is very familiar with cross-product computation.

 

Collecting the constant terms in "Equation (1)" is no different from plugging in the symbols $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$ and $\mathbf{\hat{k}}$ to the first column and seeing which coefficients aggregate on each one of those terms.

 

Plugging in $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$ and $\mathbf{\hat{k}}$ is a way of signaling that we should interpret those coefficients as the coordinates of a vector.

 

Now we can answer the following question: If dot product between $\mathbf{\vec{p}}$ and $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ gives the same result as determinant which plugging in $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ to the first column of the matrix whose other columns have the coordinates of $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. What is vector $\mathbf{\vec{p}}$?

Geometrically

Remember, the geometric interpretation of a dot product between a vector $\mathbf{\vec{p}}$ and some other vector is to project that other vector onto $\mathbf{\vec{p}}$ and then multiply the length of that projection by the length of $\mathbf{\vec{p}}$.

 

To calculate volume of the parallelepiped, 1). take the area of the parallelogram defined by $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ 2). then multiply it, not by the length of $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ but by the component of $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ that's perpendicular to that parallelogram.

 

In other words, this linear function projects $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ onto a line that's perpendicular to both $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ then, 2). multiplies the length of that projection by the area of the parallelogram spanned by $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. But this is the same thing as taking a dot product between $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and a vector that's perpendicular to $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ with a length equal to the area of that parallelogram. Also taking a dot product between $\mathbf{\vec{p}}$ and $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is the same thing as computing that determinant of a $3 \times 3$ matrix whose columns are $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$, the coordinates of $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$.

 

Now we can answer the following question: If dot product between $\mathbf{\vec{p}}$ and $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ gives the same result as the signed volume of a parallelepiped defined by $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ along with $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. What is 3D vector $\mathbf{\vec{p}}$ 

Summary

Just to sum up what happened here, 1). Define 3-D to 1-D transformation by $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$, 2). then find dual vector from two perspective (numerically and geometrically).

 

In computational approach, explained about trick of plugging in the symbols $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$ and $\mathbf{\hat{k}}$ to the first column of the matrix and computing the determinant.

 

In geometrical approach, deduce that this duel vector must be perpendicular to $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ with a length equal to the area of the parallelogram spanned out by those two vectors.

 

Since both of these approaches give us a dual vector to the same transformation, they must be the same vector. So this computation has those geometrical property.