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• 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 thedual vectorof 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 itsdual vectorin 3D space

• thatdual vectoris 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 ofduality”.

• 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 thedual vectorof 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

Hey folks! Where we left off,

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

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tai posted on 2021/02/16
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