# Cross products in the light of linear transformations | Essence of linear algebra chapter 11

Hey folks! Where we left off,
I was talking about how to compute a three-dimensional cross product
between two vectors, v x w.
It's this funny thing where you write a matrix, whose second column has the coordinates of
v,
whose third column has the coordinates of w,
but the entries of that first column, weirdly, are the symbols i-hat, j-hat and k-hat
where you just pretend like those guys are numbers for the sake of computations.
Then with that funky matrix in hand,
you compute its determinant.
If you just chug along with those computations, ignoring the weirdness,
you get some constant times i-hat + some constant times j-hat + some constant times k-hat.
How specifically you think about computing that determinant
is kind of beside the point.
All that really matters here is that you'll end up with three different numbers
that are interpreted as the coordinates of some resulting vector.
From here, students are typically told to just believe that
the resulting vector has the following geometric properties.
Its length equals the area of the parallelogram defined by v and w.
It points in a direction perpendicular to both of v and w.
And this direction obeys the right hand rule
in the sense that if you point your forefinger along v
and your middle finger along w
then when you stick up your thumb
it'll point in the direction of the new vector.
There are some brute force computations
that you could do to confirm these facts.
But I want to share with you a really elegant line of reasoning.
It leverages a bit of background, though.
So for this video I'm assuming that everybody has watched chapter 5 on the determinant
and chapter 7 where I introduce the idea of duality.
As a quick reminder, the idea of duality is that
anytime you have a linear transformation from some space to the number line,
it’s associated with a unique vector in that space
in the sense that performing the linear transformation
is the same as taking a dot product with that vector.
Numerically, this is because one of those transformations
is described by a matrix with just one row
where each column tells you the number that each basis vector lands on.
And multiplying this matrix by some vector v is computationally identical to
taking the dot product between v and the vector you get by turning that matrix on its side.
The takeaway is that whenever you're out in the mathematical wild
and you find a linear transformation to the number line
you will be able to match it to some vector
which is called the “dual vector” of that transformation
so that performing the linear transformation
is the same as taking a dot product with that vector.
The cross product gives us a really slick example of this process in action.
It takes some effort, but it's definitely worth it.
What I'm going to do is to define a certain linear transformation from three dimensions
to the number line.
And it will be defined in terms of the two vectors v and w.
Then, when we associate that transformation with its “dual vector” in 3D space
that “dual vector” is going to be the cross product of v and w.
The reason for doing this will be that understanding that transformation
is going to make clear the connection between the computation and the geometry of the cross
product.
So to back up a bit,
remember in two dimensions what it meant to compute the 2D version of the cross product?
When you have two vectors v and w,
you put the coordinates of v as the first column of the matrix
and the coordinates of w is the second column of matrix
then you just compute the determinant.
There's no nonsense with basis vectors stuck in a matrix or anything like that.
Just an ordinary determinant returning a number.
Geometrically, this gives us the area of a parallelogram
spanned out by those two vectors
with the possibility of being negative, depending on the orientation of the vectors.
Now, if you didn't already know the 3D cross product
and you're trying to extrapolate
you might imagine that it involves taking three separate 3D vectors u, v and w.
And making their coordinates the columns of a 3x3 matrix
then computing the determinant of that matrix.
And, as you know from chapter 5
geometrically, this would give you the volume of a parallelepiped
spanned out by those three vectors
with the plus or minus sign
depending on the right-hand rule orientation of those three vectors.
Of course, you all know that this is not the 3D cross product.
The actual 3D cross product takes in two vectors and spits out a vector.
It doesn't take in three vectors and spit out a number.
But this idea actually gets us really close to what the real cross product is.
Consider that first vector u to be a variable
say, with variable entries x, y and z
while v and w remain fixed.
What we have then is a function from three dimensions to the number line.
You input some vector x, y, z and you get out a number
by taking the determinant of a matrix whose first column is x, y, z
and whose other two columns are the coordinates of the constant vectors v and w.
Geometrically, the meaning of this function is that
for any input vector x, y, z, you consider the parallelepiped defined by this vector
v and w
then you return its volume with the plus or minus sign depending on orientations.
Now, this might feel like kind of a random thing to do.
I mean, where does this function come from?
Why are we defining it this way?
And I'll admit at this stage of my kind of feel like it's coming out of the blue.
But if you're willing to go along with it
and play around with the properties that this guy has
it's the key to understanding the cross product.
I'll actually leave it to you to work through the details of why this is true
based on properties of the determinant.
But once you know that it's linear
we can start bringing in the idea of “duality”.
Once you know that it's linear
you know that there's some way to describe this function as matrix multiplication.
Specifically, since it's a function that goes from three dimensions to one dimension
there will be a 1x3 matrix that encodes this transformation.
And the whole idea of duality
is that the special thing about transformations from several dimensions to one dimension
is that you can turn that matrix on its side
and, instead, interpret the entire transformation as the dot product with a certain vector.
What we're looking for is the special 3D vector that I'll call p
such that taking the dot product between p and any other vector [x, y, z]
gives the same result as plugging in [x, y, z] as the first column of a 3x3 matrix
whose other two columns have the coordinates of v and w
then computing the determinant.
I'll get to the geometry of this in just a moment.
But right now, let's dig in and think about what this means computationally.
Taking the dot product between p and [x, y, z]
will give us something times x + something times y + something times z
where those somethings are the coordinates of p.
But on the right side here, when you compute the determinant
you can organize it to look like some constant times x + some constant times y + some constant
times z
where those constants involve certain combinations of the components of v and w.
So, those constants, those particular combinations of the coordinates of v and w
are going to be the coordinates of the vector p that we're looking for.
But what's going on the right here
should feel very familiar to anyone
who's actually worked through a cross-product computation.
Collecting the constant terms that are multiplied by x, y and z like this
is no different from plugging in the symbols i-hat, j-hat and k-hat to that first column
and seeing which coefficients aggregate on each one of those terms.
It's just that plugging in i-hat, j-hat and k-hat
is a way of signaling that we should interpret those coefficients as the coordinates of a
vector.
So, what all of this is saying
is that this funky computation can be thought of as a way to answer the following question:
What vector p has the special property
that when you take a dot product between p and some vector [x, y, z]
it gives the same result as plugging in [x, y, z] to the first column of the matrix
whose other two columns have the coordinates of v and w
then computing the determinant?
That's a bit of a mouthful.
But it's an important question to digest for this video.
Now for the cool part which ties all this together
with the geometric understanding of the cross product that I introduced last video.
I'm going to ask the same question again.
But this time, we're going to try to answer it geometrically
What 3D vector p has the special property
that when you take a dot product between p and some other vector [x, y, z]
it gives the same result as if you took the signed volume of a parallelepiped
defined by this vector [x, y, z] along with v and w?
Remember, the geometric interpretation of a dot product
between a vector p and some other vector
is to project that other vector onto p
then to multiply the length of that projection by the length of p.
With that in mind, let me show a certain way to think about
the volume of the parallelepiped that we care about.
Start by taking the area of the parallelogram defined by v and w
then multiply it, not by the length of [x, y, z]
but by the component of [x, y, z] that's perpendicular to that parallelogram.
In other words, the way our linear function works on a given vector
is to project that vector onto a line that's perpendicular to both v and w
then, to multiply the length of that projection by the area of the parallelogram spanned by
v and w.
But this is the same thing as taking a dot product
between [x, y, z] and a vector that's perpendicular to v and w
with a length equal to the area of that parallelogram.
What's more, if you choose the appropriate direction for that vector
the cases where the dot product is negative
will line up with the cases where the right hand rule for the orientation of [x, y, z],
v and w is negative.
This means that we just found a vector p
so that taking a dot product between p and some vector [x, y, z]
is the same thing as computing that determinant of a 3x3 matrix
whose columns are [x, y, z], the coordinates of v and w.
So, the answer that we found earlier, computationally
using that special notational trick
must correspond geometrically to this vector.
This is the fundamental reason
why the computation and the geometric interpretation of the cross product are related.
Just to sum up what happened here
I started by defining a linear transformation from 3D space to the number line
and it was defined in terms of the vectors v and w
then I went through two separate ways
to think about the “dual vector” of this transformation
the vector such that applying the transformation
is the same thing as taking a dot product with that vector.
On the one hand, a computational approach
will lead you to the trick of plugging in the symbols i-hat, j-hat and k-hat
to the first column of the matrix and computing the determinant.
But, thinking geometrically
we can deduce that this duel vector must be perpendicular to v and 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 that wraps up dot products and cross products.
And the next video will be a really important concept for linear algebra
“change of basis”