Solve Linear Systems with Geometric Algebra — A New Way to See Equations

Transcript

So let's start by looking at a simple system of two linear equations and let's try to solve it uh to interpret it first before delving into trying to solve it. Okay. So, we are going to check we're going to check this system of equations. 2x + y = 3 and 2x + 4 y = 6. Let's not pay attention to the fact that they are linearly no nothing.

Forget about that. The traditional interpretation is that this system is an intersection of two lines. So this is how is traditionally taught at least the way it was for me and it means that we can write the system in the following way as y = - 2x + 3 and y = 3 - no sorry - 12 of x + 3 as well. And sorry, this is not a three. This is uh three halves, which is one half.

And now we are going to paint the first one in yellow. Let's make it thick. to be clear and the second one in green in here. So let's take the first one we set yellow and a little bit bigger. So we know it and we know that it has it cuts the axis in the plus three.

go up here and it has a slope minus2 meaning that it goes in the following way. So it will be this way here. This will be our first equation probably speaking. And the second one which is green [snorts] intersects in 1/2. So intersect here and now has a slope minus one2.

So it goes down h let me check minus one/2 will be exactly this way here. Okay. And by observing these two lines we know that they should intersect in this point which is the point one one. And that's the solution to our equations. If the system will be if the equations will be parallel then it will have no solution.

But let me tell you there is a different way of seeing this system a different geometry which is interpreting this system as a combination of vectors. And no way is better than other. But this second way will be particularly efficient in terms of the geometrical interpretation. And the way we do this is by picking up this here and this here and this one here. and interpret this as vectors.

This is the same as if we will write 2 x + 14 y equals 3 6. And what this means again we paint this one green this one red this one blue is that we have to find two scalars x and y that multiplying these two vectors the green and the red one gives us the blue. So if we come back down here and we paint them for example the green one which is 22. We found that this is this is the 22. Okay.

And now we have the red one. H red one which is 14. So it goes one and four up will be this one. And now we grab it and move it here. And we want to see if this produces the blue vector which is 3 six.

So it's three to the right. And then we have one, two, three, four, five, six will be here. And this will be our system of equations. If we give them names, for example, we can call this one A, B, and C. We can write our equation as a x + b y equals c.

And it's a single equation that deco composes into one for each of the components of the vectors. [snorts] And here will be a b c. And we can see that precisely a + b equals c. And therefore from here we can obtain directly that x = 1 and y = y one also. This will be a second interpretation that is not unknown.

It's uh very well taught in matrix when we write the the vector as matrices but it will be very useful in a second. The reason why it will be very useful is the following. If we start by this equation here that is our general equation. Notice that we can isolate xy by exterior multiplication. Oh, I wrote over myself by exterior multiplication.

How we do that? Well, very easy. Let's move also this thing here. Let's get rid of this one. And then we can paint the arrow here. How do we do this? Well, notice that if we multiply by exterior by a Then the first term is a zero because this is a w a x plus a wb b y equals a w c.

So this term is zero and we have that y well a wb y equals a w c. And what this is telling us is that we are comparing the area of this B vector and the direction with the one from this B vector which we can draw here in the following way. So a watch b will be let's go here. So we need a and we need another a that I will paste here for later. And we need B which is here.

And we need another beam that we place here in the origin. And I almost nailed it. That would have been very nice. And this area here, this is a wedge b, right? And now we do a similar procedure with a and c. So we have another a copy.

I will not over We can get rid of this. I will not. Okay, this is A. This is C. And this second area this is a wedge c.

And what we are doing in this equation that we were having here is comparing them. How does the ve vector a compares with the AC? And the result is that they have the same area and y is one. That is what all of this is telling us. But we can because in geometric algebra we have the geometric product and the geometric product allows for the inverse of blades. We can multiply by a wb minus one and write y equ= a b minus one a wedge c.

And this expression can only be obtained via geometric algebra because otherwise this object is not well defined. And similarly instead if instead of doing an exterior product by A you do it by b. Let's say a we do b wedge a x plus b wedge b y equals b wedge c. This has a zero. And we can define we can find x as minus a wedge.

No sorry as yeah let's write b a minus one b w c which is the same as minus a w b minus one b wed c which equals a wedg b minus one c wedged b. So I absorb the minus to the other side and you will see later why I wrote why I decided to write it in this particular order. So our solution is directly derived by saying y or let's start with x equals a wb minus one cb and y equals a wb minus one a w c. And this will be our geometrical way of solving this equation. And the only remaining problem that one might have is that you have to calculate a wb.

And for that and without h further proof, I will let you know here and put the references in the notes that the inverse of any blade. So we have a blade that is A1 wedge, A2 wedge, A N or no let's say A K is this equal to minus one elevated to K K minus one half divided by A1 A2 D a K squar and then here the blade itself A1 A2 D which A K and this is the expression for the inverse of a blade. In our case, in our case, a h which b minus one will be k equals 2. So the numerator is one. One divided by a b² absolute value.

Important [snorts] a wedged b. And this will be the inverse of of it. Sorry, I think here I am mistaken without the absolute value. This should be a b square. So you can see that if we now have a HB a b minus one this is equal to a b² and a hb² and this equals to + one which is what we want from the inverse element.

So with this what we have found is how to solve the equation geometrically. Let's go back to our original system. This one here to solve it in this particular case. So in this particular case I'm going to save this for later because I may need it. A was 2 2 which equals to 2 E1 + 2 E2 and B equals to 14 which was E1 + 4 E2 and C was 3 E1 + 6 E2.

Okay. And now to solve this to solve the system of equations we need to do a couple of products. First of all we need a wedge b which we can write as the grade of a b which equals to the great selection of 10 + 6 e12. Therefore this is 6 e12. Or alternatively you can do the multiplication directly 2 E1 + 2 E2 which E1 + 4 E2 and the key element here is not forgetting any element of the distributive of the distributivity.

So E1 E1 E is zero. E1 E2 this is 8 E12 and then we have E2 E1 + 2 E21 E2 E2 0 and this is equal this is equals to minus 2 E12 and therefore this is 6 E12 whatever way you prefer this is just to show you that it's the same. Now we calculate C wedge B because this is what we had up here for a solution of X that we want first. C H WP and we can do similar procedure and C WB is 3 E1 + 6 E2. Remember that the the order is very important.

And we cannot say B w C is C w which now it comes to 4 E2 and again we have that this is equal to 12 E12 and then we have min -6 E12 which equals 6 E12 and from the expression before we have that X equals a HB minus one which we have to calculate. Let's go here on the right to calculate AHB minus one. We say that this is AB² which is minus 36 in the denominator and in the numerator we have 6 E12. Right? So [snorts and clears throat] this is equal to CHP in terms of number six E12 / minus 36 and here we have again six E12 for CB and this indeed is + one as we wanted to. And now for calculating y.

Y was a a w b minus one a w c and a b minus one. We have it. We only have to calculate now a wedge C which is 2 E1 + 2 E2 wedge 3 E1 + 6 E2 and this is 12 E12. Let's we know how this is going to go. 12 minus 6 E12 which indeed is 6 E12.

And now we can say that y equals 1 / - 36 upp minus 6 uh e12 if we simplify here that I did not do before but can be done or you know you can always also turn around this one but let's say also here six E12 and thus this is another one and this is how you solve a system of equations via geometric algebra [snorts] which might look very complicated in comparison with you know if in a very simple system like this I would for sure not use all the power of geometric algebra to calculate the solution. I will use probably substitution to obtain the solution because it's very straightforward. But when you have a system of four, five, six equations, then it becomes a problem and then is when the geometric algebra approach becomes very powerful because we can make the solution systematic. We can obtain very very very easily and construct immediately the solution that we have that we want and in a manner that preserves the quality or the grammar rules and adjints etc that are present in other methods. And we will see now how we can use the exact same method to so to invert a matrix without adjints without grammar rules etc.

which are definitely powerful but can be very confusing tools in linear algebra.