Hello, hello again. So, moving forward I will be assuming you have a visual understanding
of linear transformations and how they're represented with matrices the way I have been talking about in the last
few videos. If you think about a couple of these linear
transformations you might notice how some of them seem to
stretch space out while others squish it on in. One thing that turns out to be pretty useful
to understanding one of these transformations is to measure exactly how much it stretches
or squishes things. More specifically to measure the factor by which the given region
increases or decreases. For example look at the matrix with the columns 3, 0 and
0, 2 It scales i-hat by a factor of 3 and scales j-hat by a factor of 2 Now, if we focus our attention on the one
by one square whose bottom sits on i-hat and whose left
side sits on j-hat. After the transformation, this turns into
a 2 by 3 rectangle. Since this region started out with area 1,
and ended up with area 6 we can say the linear transformation has scaled
it's area by a factor of 6. Compare that to a shear whose matrix has columns 1, 0 and 1, 1. Meaning, i-hat stays in place and j-hat moves
over to 1, 1. That same unit square determined by i-hat
and j-hat gets slanted and turned into a parallelogram. But, the area of that parallelogram is still
1 since it's base and height each continue to
each have length 1. So, even though this transformation smushes
things about it seems to leave areas unchanged. At least, in the case of that one unit square. Actually though if you know how much the area of that one
single unit square changes it can tell you how any possible region in
space changes. For starters notice that whatever happens to one square
in the grid has to happen in any other square in the grid no matter the size. This follows from the fact that grid lines
remain parallel and evenly spaced. Then, any shape that is not a grid square can be approximated by grid squares really
well. With arbitrarily good approximations if you
use small enough grid squares. So, since the areas of all those tiny grid
squares are being scaled by some single amount the area of the blob as a whole will also be scaled also by that same single
amount. This very special scaling factor the factor by which a linear transformation
changes any area is called the determinant of that transformation. I'll show how to compute the determinate of
a transformation using it's matrix later on in the video but understanding what it is, trust me, much
more important than understanding the computation. For example the determinant of a transformation
would be 3 if that transformation increases the area
of the region by a factor of 3. The determinant of a transformation would
be 1/2 if it squishes down all areas by a factor
of 1/2. And, the determinant of a 2-D transformation
is 0 if it squishes all of space onto a line. Or, even onto a single point. Since then, the area of any region would become
0. That last example proved to be pretty important it means checking if the determinant of a
given matrix is 0 will give away if computing weather or not
the transformation associated with that matrix squishes everything into a smaller dimension. You will see in the next few videos why this is even a useful thing to think about. But for now, I just want to lay down all of
the visual intuition which, in and of itself, is a beautiful thing
to think about. Ok, I need to confess that what I've said
so far is not quite right. The full concept of the determinant allows
for negative values. But, what would scaling an area by a negative
amount even mean? This has to do with the idea of orientation. For example notice how this transformation gives the sensation of flipping space over. If you were thinking of 2-D space as a sheet
of paper a transformation like that one seems to turn
over that sheet onto the other side. Any transformations that do this are said
to "invert the orientation of space." Another way to think about it is in terms
of i-hat and j-hat. Notice that in their starting positions, j-hat
is to the left of i-hat. If, after a transformation, j-hat is now on
the right of i-hat the orientation of space has been inverted. Whenever this happens whenever the orientation of space is inverted the determinant will be negative. The absolute value of the determinant though still tells you the factor by which areas
have been scaled. For example the matrix with columns 1, 1 and 2, -1 encodes a transformation that has determinant Ill just tell you -3. And what this means is that, space gets flipped over and areas are scaled by a factor of 3. So why would this idea of a negative area
scaling factor be a natural way to describe orientation flipping? Think about the seres of transformations you
get by slowly letting i-hat get closer and closer
to j-hat. As i-hat gets closer all the areas in space are getting squished
more and more meaning the determinant approaches 0. once i-hat lines up perfectly with j-hat, the determinant is 0. Then, if i-hat continues the way it was going doesn't it kinda feel natural for the determinant
to keep decreasing into the negative numbers? So, that is the understanding of determinants
in 2 dimensions what do you think it should mean for 3 dimensions? It [determinant of 3x3 matrix] also tells
you how much a transformation scales things but this time it tells you how much volumes get scaled. Just as in 2 dimensions where this is easiest to think about by focusing
on one particular square with an area 1 and watching only what happens to it in 3 dimensions it helps to focus your attention on the specific 1 by 1 by 1 cube whose edges are resting on the basis vectors i-hat, j-hat, and k-hat. After the transformation that cube might get warped into some kind
of slanty slanty cube this shape by the way has the best name ever parallelepiped. A name made even more delightful when your
professor has a nice thick Russian accent. Since this cube starts out with a volume of
1 and the determinant gives the factor by which
any volume is scaled you can think of the determinant as simply being the volume of that parallelepiped that the cube turns into. A determinate of 0 would mean that, all of space is squished
onto something with 0 volume meaning ether a flat plane, a line, or in
the most extreme case onto a single point. Those of you who watched chapter 2 will recognize this as meaning that the columns of the matrix are linearly
dependent. Can you see why? What about negative determinants? What should that mean for 3 dimensions? One way to describe orientation in 3-D is with the right hand rule. Point the forefinger of your right hand in the direction of i-hat stick out your middle finger in the direction
of j-hat and notice how when you point your thumb up it is in the direction of k-hat. If you can still do that after the transformation orientation has not changed and the determinant is positive. Otherwise if after the transformation it only makes
since to do that with your left hand orientation has been flipped and the determinant is negative. So if you haven't seen it before you are probably wondering by now "How do you actually compute the determinant?" For a 2 by 2 matrix with entries a, b, c,
d the formula is (a * d) - (b * c). Here's part of an intuition for where this
formula comes from lets say the terms b and c both happed to
be 0. Then the term a tells you how much i-hat is
stretched in the x direction and the term d tells you how much j-hat is stretched in the
y direction. So, since those other terms are 0 it should make sense that a * d gives the area of the rectangle that our favorite
unit square turns into. Kinda like the 3, 0, 0, 2 example from earlier. even if only one of b or c are 0 you'll have a parallelogram with a base a and a height d. So, the area should still be a times d. Loosely speaking if both b and c are non-0 then that b * c term tells you how much this parallelogram is stretched or squished in the diagonal direction. For those of you hungry for a more precipice
description of this b * c term here's a helpful diagram if you would like
to pause and ponder. Now if you feel like computing determinants
by hand is something that you need to know the only way to get it down is to just practice it with a few. There's not really that much I can say or
animate that is going to drill in the computation. This is all tripply true for 3-rd dimensional
determinants. There is a formula [for that] and if you feel like that is something you
need to know you should practice with a few matrices or you know, go watch Sal Kahn work through
a few. Honestly though I don't think those computations fall within
the essence of linear algebra but I definitely think that knowing what the
determinate represents falls within that essence. Here's kind of a fun question to think about
before the next video if you multiply 2 matrices together the determinant of the resulting matrix is the same as the product of the determinants
of the original two matrices if you tried to justify this with numbers it would take a really long time but see if you can explain why this makes
sense in just one sentence. Next up I'll be relating the idea of linear transformations
covered so far to one of the areas where linear algebra is
most useful linear systems of equations see ya then!
I'm going to take linear algebra this fall, and /u/3blue1brown is changing my perspective on this topic day-by-day. When I had dealt with matrices before, mindlessly finding the discriminant and whatnot, they seemed clunky and boring to me - now I actually understand the meaning behind what I was doing. Props to you.
From now on, parallelepipeds will be referred to as "slanty-slanty cubes".
Could anyone provide some insight into how one takes this understanding of the determinant and generalises to the summation formula with all the permutations?
I've learned much much much more from his videos then in an entire linear algebra course. I always proved things "from the numbers" as he says. I had heard this view of determinet before but it didn't really click and wasn't covered in class. Super super helpful.
Using the geometric reasoning presented in the video, I was able to conclude that if a system of linear equations has infinite solutions, then the determinant must be zero. However the converse is not true.
Furthermore, if a system of linear equations has no solution, then the determinant must be zero. Again the converse is not true.
Finally, if the system has a unique solution, then the determinant must be non-zero. The converse is true in this case.
I don't have a proof for any of them, so that's why I'm asking: are they true statements?
PS. This was just some visualizing, without putting pen to paper, so I'm sorry if I made any mistakes.
I have difficulty retaining abstract math concepts without a illustrative analogy. The fact that the determinant is the volume scaling of a unit (hyper)cube is perfect.
This explains intuitively why a Det=0 matrix is not invertable: if your transformation reduces your volume to zero, it's impossible to work backwards uniquely to restore the original structure.
In higher dimensions, does the sign of the determinant still indicate a unique orientation? This holds true in two and three dimensions because the two/three transformations that invert a single unit vector all result in the same new orientation. Does inverting the second unit vector in four dimensions result in a different orientation than inverting the fourth unit vector? My instinct tells me that the number of possible orientations seems to have to do with number of ways to order a closed loop of objects, which exceeds two as the number of objects surpasses three. However, I also feel that rotation in higher dimensions seems to be more powerful. Hopefully what I'm asking makes sense.
Haha I remember for the determinant I would do top diagonal minus opposite diagonal for 2x2 matrices and 3x3 there was a formula to multiply the top row with the remaining 2x2 matrices. And alternate the signs. So formulaic and so drab. This is so much more interesting!
Question: How did people visualize something like this in the 1800s? Also what software is the author of the video using?
I just taught a linear class this summer, and I wish I could have taught it like these videos. There is no way I could have drawn all the visuals on the board though. And even if I had tried, it wouldn't have made the one day I took on introducing determinants, and made it probably 3 to 4.