Lecture 13: Normal Matrix

Transforms Example

Consider a simple transform matrix $M$

$$\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

If we take the cross product between two vectors, we get one result:

$$(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$$

If we transform each by M, then take the cross product, we get something different (as expected):

$$M(1, 0, 0) \times M(0, 1, 0) =$$ $$(-1, 0, 0) \times (0, 1, 0) = (0, 0, -1)$$

What might not be expected is that if we apply the $M$ transform to the result of the first cross product, we get a different result:

$$M((1, 0, 0) \times (0, 1, 0)) = M(0, 0, 1) = (0, 0, 1)$$

This shows us that the result of a cross product transforms differently from a cross product. That’s because that result is not a vector - it’s a bivector.

Derivation

Let’s take a closer look. What do we get if we transform some vector $a$ by matrix $M$?

$$M \vec a = M (a_1 \hat x + a_2 \hat y + a_3 \hat z) = a_1 M_1 + a_2 M_2 + a_3 M_3$$

Here $M_i$ is the $i$th column of $M$.

So what is the cross product of two vectors $a$ and $b$ after they are transformed?

$$Ma \times Mb =$$ $$(a_1 M_1 + a_2 M_2 + a_3 M_3) \times (b_1 M_1 + b_2 M_2 + b_3 M_3) =$$

The cross product is distributive over addition, so we can write:

$$a_1 M_1 \times b_1 M_1 + a_1 M_1 \times b_2 M_2 + a_1 M_1 \times b_3 M_3 +$$ $$a_2 M_2 \times b_1 M_1 + a_2 M_2 \times b_2 M_2 + a_2 M_2 \times b_3 M_3 +$$ $$a_3 M_3 \times b_1 M_1 + a_3 M_3 \times b_2 M_2 + a_3 M_3 \times b_3 M_3 =$$

Any vector crossed with itself is 0:

$$a_1 M_1 \times b_2 M_2 + a_1 M_1 \times b_3 M_3 +$$ $$a_2 M_2 \times b_1 M_1 + a_2 M_2 \times b_3 M_3 +$$ $$a_3 M_3 \times b_1 M_1 + a_3 M_3 \times b_2 M_2 =$$

And cross products are anti-commutative ($a \times b = - b \times a$), so:

$$a_1 M_1 \times b_2 M_2 + - b_3 M_3 \times a_1 M_1 +$$ $$-b_1 M_1 \times a_2 M_2 + a_2 M_2 \times b_3 M_3 +$$ $$a_3 M_3 \times b_1 M_1 + - b_2 M_2 \times a_3 M_3 =$$ $$(a_2 b_3 - a_3 b_2) (M_2 \times M_3) +$$ $$(a_3 b_1 - a_1 b_3) (M_3 \times M_1) +$$ $$(a_1 b_2 - a_2 b_1) (M_1 \times M_2)$$

Okay, there’s the formula for the cross product components on the left, but what about those cross products on the right?

Well we can write this in vector form like so:

$$\begin{bmatrix} (M_2 \times M_3)_x * (a_2 b_3 - a_3 b_2) + (M_3 \times M_1)_x * (a_3 b_1 - a_1 b_3) + (M_2 \times M_3)_x * (a_1 b_2 - a_2 b_1) \\ (M_2 \times M_3)_y * (a_2 b_3 - a_3 b_2) + (M_3 \times M_1)_y * (a_3 b_1 - a_1 b_3) + (M_2 \times M_3)_y * (a_1 b_2 - a_2 b_1) \\ (M_2 \times M_3)_z * (a_2 b_3 - a_3 b_2) + (M_3 \times M_1)_z * (a_3 b_1 - a_1 b_3) + (M_2 \times M_3)_z * (a_1 b_2 - a_2 b_1) \\ \end{bmatrix}$$

Which shows us that we can rewrite this as matrix-vector multiplcation.

$$\begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix}^T \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \\ \end{bmatrix}$$

On the right is the result of the cross product between $a$ and $b$. On the left is a matrix we can multiply the result by to get what we have gotten by transforming both vectors before crossing.

So what is this matrix on the left?

$$\begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix}^T$$

Well, first let’s just point something out.

$$(M_2 \times M_3) \cdot M_1 = \text{det}M$$ $$(M_3 \times M_1) \cdot M_2 = \text{det}M$$ $$(M_1 \times M_2) \cdot M_3 = \text{det}M$$

This follows from the formula for determinant - if you multiply everything out, it’s the same quantity.

However, all other dot products are $0$, e.g.:

$$(M_2 \times M_3) \cdot M_2 = 0$$ $$(M_2 \times M_3) \cdot M_3 = 0$$

This also follows from the multiplication:

$$(M_2 \times M_3) \cdot M_2 =$$ $$\begin{bmatrix} M_2y M_3z - M_3y M_2z \\ M_2z M_3x - M_3z M_2x \\ M_2x M_3y - M_3x M_2y \\ \end{bmatrix} \cdot \begin{bmatrix} M_2x \\ M_2y \\ M_2z \\ \end{bmatrix} =$$ $$M_2y M_3z M_2x - M_3y M_2z M_2x + M_2z M_3x M_2y - M_3z M_2x M_2y + M_2x M_3y M_2z - M_3x M_2y M_2z = 0$$

This is interesting because it gives us:

$$\begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix} M = \begin{bmatrix} \text{det}M & 0 & 0 \\ 0 & \text{det}M & 0 \\ 0 & 0 & \text{det}M \\ \end{bmatrix}$$ $$\frac{1}{\text{det}M} \begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix} M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

Which, by the definition of matrix inverse, means that:

$$\frac{1}{\text{det}M} \begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix} = M^{-1}$$ $$\begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix} = (\text{det}M) M^{-1}$$ $$\begin{bmatrix} M_2 \times M_3 \\ M_3 \times M_1 \\ M_1 \times M_2 \\ \end{bmatrix}^T = (\text{det}M) (M^{-1})^T$$

There we go - inverse transpose is how to transform the result of a cross product. We need to divide by the determinant as well, but since we always normalize, this only really matters for flipping normals.

$$(\text{det}M) (M^{-1})^T \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \\ \end{bmatrix}$$

Next, we will cover why exactly it is that the result of a cross product is this different entity that obeys different rules.