Birne Binegar: Okay, I guess, we get underway so today's lecture we're gonna finish up hopefully our discussion on.
Birne Binegar: matrices and there in verses and then.
Birne Binegar: that's the end of the material that's going to be covered, on the first exam However I don't want to have the first exam immediately because there's a problem set actually two problems that problem such four and five that are due on Monday, and so I want to postpone.
Birne Binegar: The first exam.
Birne Binegar: The earliest possible date would be Wednesday of next week, so we can go to the problem sets but I actually think it'd be better be Fridays, so that.
Birne Binegar: There any difficulties on the problems that we can address those on Monday, and then we could do a more general review a full review on Wednesday and then do the exam on Friday, so let me pull a class if they're not.
Birne Binegar: Because of Maybe other exams or other things actually actually we can I can postpone that decision until Friday.
Birne Binegar: So you can look at your schedules first but I I would think that the likely time for the exam is going to be a week from this Friday.
Birne Binegar: But this Friday I may ask you if you would be willing to have the exam as early as Wednesday, which you might do if you have two other exams next Friday.
Birne Binegar: I don't want to push the exam off any further to the following week because we'll have essentially a bunch of material that will start to cover and you'll forget about this material.
Birne Binegar: On well the calculation will material we developed so far, so, in fact, the the course kind of makes a shift after the first exam as we start to do.
Birne Binegar: less emphasis on calculations, but more emphasis on tying things together in a theoretical fashion until I don't want to get too far ahead of ahead of myself with the new material before we've actually had a test on on.
Birne Binegar: Material up to this point so.
Birne Binegar: Let me just see a show of hands that there any strenuous objections to Nick a week from Friday or having the exam.
Birne Binegar: Nothing OK so i'm going to pull my other classes as well because i'm going to give the.
Birne Binegar: Two Sections I want to give the exam on the same day so i'll pull them as well, but I suspect that it's going to be a week from Friday that we do a first exam.
Birne Binegar: Not a week from today we'll do a full review session to prepare you for the exam and then Monday next week kind of working backwards Monday next week homework sets will be do if there are any issues on.
Birne Binegar: On the problem sets, particularly the road reduction we can try to address those in class as well okay so.
Birne Binegar: Questions so far.
Birne Binegar: yeah people generally don't have questions at the beginning of class but don't interrupt me if you have questions during class.
Birne Binegar: So what we're going to do today is i'm going to quickly review the material that we covered in monday's lecture on on matrix industry matrix in verses then we're going to do a.
Birne Binegar: Relatively hard calculation of matrix inverse that that calculation is going to involve row reduction and i'm going to i'm doing a strenuous real reduction problem just to show you that they can be done, even if things don't work out nicely.
Birne Binegar: And then, what i'm going to end today's lecture is a fun it's called the fundamental theorem because what it's going to do it's going to tie together lots of concepts that we've.
Birne Binegar: developed and you see that, yes, even though we have one calculation or another calculation that somehow the things are all related to one another.
Birne Binegar: And that's actually going to be at least semantically the sort of discussion that we're going to have in in detail, following the first exam as we now begin to see how these linear algebra constructs work together.
Birne Binegar: Okay, so let me get started.
Birne Binegar: Work yeah good, so I am matrix inverse it's the analog of a multiplicative inverse for numbers for numbers, you say.
Birne Binegar: To with them one half that's a multiplicative inverse of two because one half times two equals, one which is the identity element, with respect to multiplication well for matrices we have a similar idea we actually have two conditions, because for matrix multiplication.
Birne Binegar: Be times a it's not necessarily equal to a time be order multiplication makes a difference, however.
Birne Binegar: For this question of whether or not we can multiply a matrix times another and produce the identity matrix it turns out that these two conditions are equivalent to one another.
Birne Binegar: If the times a is going to equal the identity matrix it's also going to be true that a times B as equally identity matrix but that's kind of hard to prove so I won't even attempt it in this course.
Birne Binegar: Effectively, though, if you simply have B times as equal to the identity matrix you're going to know that be is actually the inverse of a.
Birne Binegar: don't mess, the first year and we showed that if you do have a square matrix and it has an inverse there's only one possibility for and so on the rest of monday's lecture was devoted to figuring out well how do we actually find the inverse of a given matrix.
Birne Binegar: So.
Birne Binegar: Then we're using this notation a super script minus one to indicate the inverse the unique inverse of the matrix.
Birne Binegar: Okay, so what we developed last time was an algorithm for calculating that inverse, but let me remind you of the sort of theoretical material that went into the calculation at minus first.
Birne Binegar: First of all, we have the fact that the inverse if it does exist it's a unique matrix.
Birne Binegar: And so, basically, the problem finding a to minus first is finding some matrix times a to produce it, the identity matrix if you have this multiplicative identity, then you know automatically that see is your guy that's the guy you want.
Birne Binegar: Alright, so.
Birne Binegar: Then the problem saying well how do we find such a matrix see, and for that I you introduce a theoretical tool, not a practical tool for doing calculations, but it's it's on this slide you'll see the utility of these elementary matrix.
Birne Binegar: The idea of an elementary matrix well, you have these real operations that you can perform on any matrix.
Birne Binegar: And, in particular, you kind of perform them on the identity matrix when you take an identity matrix can you do an elementary row operation on it, you create a new matrix we call that the elementary matrix corresponding to the given role operation.
Birne Binegar: Now what's nice about these is the elementary makes these give us an alternative way of implementing reduction so here.
Birne Binegar: On the left hand side, I mean you have some elementary rotation interchanging rose scale are multiplying a row.
Birne Binegar: you're placing a row by it's some of a multiple of another row some elementary roll operation Now you can either do that directly to the matrix.
Birne Binegar: Or you can construct this elementary matrix by doing that same operation to the identity matrix and then doing matrix multiplication rather than operating directly on the matrix itself.
Birne Binegar: doing it this way it's going to take longer right because you have to do, yet the real operation to one matrix and then you have to do matrix multiplication.
Birne Binegar: But the point is is that here we can implement that real operation using a matrix multiplication and, in fact, what we're looking for is something that when you multiply.
Birne Binegar: A by a matrix you get the identity we're making we're looking for a matrix Nice that will convert a to the identity matrix and we know that using row operations, we can convert matrices to other major cities, so the thing we might try to do is.
Birne Binegar: Reduce a matrix to the identity matrix.
Birne Binegar: That would look something like this, I have a sequence of elementary row operations at systematically transform a given matrix to reduce role Echelon porn which were supposing is going to be the identity matrix well because of the existence of these elementary matrix these.
Birne Binegar: These the sequence of elementary real operations could be done by instead of operating directly on the matrix.
Birne Binegar: operating on the identity matrix to get these elementary matrix ease and then just carrying out this matrix multiplication.
Birne Binegar: But look.
Birne Binegar: At this equation here is equal to this product of matrix these times a this product of matrices has to be the inverse of a because.
Birne Binegar: This product considered as a single matrix times a produces the identity.
Birne Binegar: matrix we're satisfying.
Birne Binegar: That equation and that's going to make that left hand factor will be able to identify is a the miners first.
Birne Binegar: Until that's that's the idea that underlying the following calculations what we're going to try to do this we're going to try to construct figure out what this matrix it like calculating it directly.
Birne Binegar: And it turned out.
Birne Binegar: So what you'll see on the next slide is you do not have to individually calculate calculate these individual factors what we'll be able to do is do a single row reduction and then.
Birne Binegar: This product is going to be automatically formed at the end.
Birne Binegar: So the way that's done.
Birne Binegar: Is we don't operate on just a itself rather.
Birne Binegar: We off we operate elementary row operations on a a joined with the identity matrix now what's going to happen is that we're going to try to put this matrix.
Birne Binegar: The original matrix to start it off with and the identity matrix of the same size would put them right next to another make a matrix with double the columns the same number of rows so the original matrix.
Birne Binegar: Then will calculate it reduce your national unformed Well, we know how to do that, you.
Birne Binegar: worked at get rid of pivots below pivots until you reach row Echelon form and then you reach scale pits and poor out entries about pivots to get it in reducing isilon porn that can be done to any matrix including this augmented matrix and.
Birne Binegar: What will happen is that.
Birne Binegar: When you are able to.
Birne Binegar: Reduce a matrix a to the identity matrix this double matrix ends up getting reduced to the identity matrix and then something else is going to happen to this left this right hand factor, and in fact that right hand factors, going to be exactly the inverse of a.
Birne Binegar: So what i'm saying here is if.
Birne Binegar: You start off with the coefficient matrix a and joined it with Ai and then you start applying row operations to it.
Birne Binegar: And you don't have to convert these guys to elementary makes your season at all you just have to reduce until you reach reduce row Echelon form and then one of two things is going to happen.
Birne Binegar: If the reduced lunch on form factor on the and whatnot factor, but they love left hand side of this augmented matrix if that gets reduced to the identity matrix.
Birne Binegar: Then.
Birne Binegar: What happens to the right hand side under the same sequence of row operations is going to be exactly E rk.
Birne Binegar: are one thing sequence of elementary row operations, and in fact this is just a the minus first or you get something else.
Birne Binegar: On the left, that is either this left hand side real reduces to the identity matrix or doesn't if it does.
Birne Binegar: Then, whatever happened to the right hand side that's going to be identifiable as the inverse of your matrix and on the other hand, if a does not it's not reducible to the identity matrix then you're going into this situation and that means there's no inverse.
Birne Binegar: Okay, so that's.
Birne Binegar: The calculation of the inverse of the matrix what it boils down to is the real reduction of a single matrix you have to double.
Birne Binegar: The number of columns you get a an A but you just row reduce this matrix see what POPs out if you end up in this situation.
Birne Binegar: Where the reduced rational on form of this matrix has the identity matrix on the left hand side, then you know that whatever happened to the right hand side that's exactly what your interest is otherwise a is not going to have an inverse.
Birne Binegar: Okay, so, in the end it just boils down to this row reduction calculations.
Birne Binegar: Alright, so let's do.
Birne Binegar: A real reduction calculation will use real reaction to solve linear systems.
Birne Binegar: And now we're going to do is use rubbish option to figure out the inverse of the matrix though here's a three by three matrix.
Birne Binegar: it's a square matrix because matrix in verses only exist when you have the same number of roses, you have calls three by three that's good first thing we'll do is we'll take the given matrix and will join to it, the identity matrix at the same size.
Birne Binegar: All right, and now what we're gonna do is we're going to take this matrix this line is kind of imaginary just helping us separate.
Birne Binegar: This double matrix into its constituent parts what we'll do is we'll start reducing this matrix So what do we got to do first well, we got the pivot up in this upper left hand column.
Birne Binegar: We have this pivot sitting below it as well, so we got to get rid of it now.
Birne Binegar: This examples.
Birne Binegar: going to be a little bit more strenuous because, unlike most of the examples i've done before this pivot is to rather than one, so if i'm going to clear out this entry.
Birne Binegar: I know I have to multiply the first row by negative three but i'm also going to have to divide it by two so that you actually get that cancellation, so, in fact, what you have to do, and sometimes unavoidable sometimes you do have to introduce fractions into this role reduction calculation.
Birne Binegar: Right so let's do that if we replace the second row by it, some with minus three half times the first row that to in the denominator is going to cancel that and the three and the numerator coming in with a negative, so I just gonna what allows us to cancel that guy.
Birne Binegar: And so what we get.
Birne Binegar: up a little bit what we get.
Birne Binegar: Is a new matrix now look all the way across this matrix to figure out where the pivots are they're all going to occur on this side until we've got it in real Echelon form, at least up to the second row but.
Birne Binegar: The third row, we have a pivot below a pivot so we're gonna have to clear out this guy.
Birne Binegar: So.
Birne Binegar: Maybe the next step.
Birne Binegar: display this maybe the next step and, of course, you have choices.
Birne Binegar: The way you actually reduce to reduce your national and form there's many, many different path to that final result.
Birne Binegar: So I can do a couple things at this point, I could maybe multiply the second row by to make it simpler to add to this rowan anyway i'm going to want that pivot to equal one eventually anyway.
Birne Binegar: I could also multiply the first row by one half to get that pity for the one but actually my concern i'm going to delay working with this guy until the very end, like.
Birne Binegar: primary concern initially is to get it into row Echelon form not reduce cholesterol and.
Birne Binegar: So i'm multiply the second row by like to simply to get this pivot equal to one so it's a little bit easier to work with.
Birne Binegar: Like corrected the second row, and now you see it's very easy to cancel this guy with them effectively two times the original second row.
Birne Binegar: And so what we'll do is we'll simply replace the third row by some with the second row, and that will produce a matrix now, at least in grow Echelon form because the pivots.
Birne Binegar: are occurring in the proper downward staircase.
Birne Binegar: tickets are sort of start in the upper left hand corner and they systematically if you drop down the pivots move off to the right okay now that's halfway through.
Birne Binegar: yeah.
Birne Binegar: No that's reduce your actual on form so yeah there were no more conditions we're going to impose but, once you reach this stage your strategy changes that's why i've separated the two.
Birne Binegar: yeah eventually we're going to have to make that guy one, and this guy also have to be made one, but we also going to have to clear out the entries above the pits but the reason why I stopped here is because.
Birne Binegar: What happens when you reach rational and form is the position of the pit it's not their values, but the position of the pivots inside the matrix it's not going to change any further.
Birne Binegar: you're always going to have a pivot in the first row first call always will have a pivot, no matter what you do.
Birne Binegar: working towards reduce cholesterol and form the position of the pivots inside that matrix it's not going to change anymore what you do is you fix them up i'm making sure they're all equal one and you clear out entries above them, but the position of the pivots doesn't change.
Birne Binegar: That, by the way, allows you to see already at this early stage if we're solving linear equations how many solutions here would be.
Birne Binegar: Because, if you look forward to the end of the calculation, where you write down the solution as a hyper plane, the columns that do not have pivots correspond to free parameters.
Birne Binegar: But you can see where the pivots are going to be already at this stage, and the for this matrix are going to be here here and here.
Birne Binegar: There are no columns without pivots and so there's going to be unique solution if I were solving a linear system, but in fact what we're trying to do here is find the inverse of the matrix so.
Birne Binegar: i'm just pointing out some nuances of row reduction that really aren't.
Birne Binegar: The kind of meaningless in the current context of.
Birne Binegar: Finding the inverse of the matrix, but if we are solving a linear system.
Birne Binegar: which looks more like this, then we would be able to count the number of free parameters in the solution as soon as we reach relational on form we don't have to go all the way to reduce cholesterol unformed to see the nature of the solution.
Birne Binegar: Okay, so let's continue we're not it reduce cholesterol on form yet we're essentially halfway through that process.
Birne Binegar: So the next step is to maybe fix this pivot i'm going to multiply it by minus one and now at least the first to last two rows of pivots equal to one I still have this guy here i'm gonna worry about him at the very end.
Birne Binegar: till the first thing i'm going to do next, and we're going to do is i'm going to get rid of the century that's sitting above a pivot in the lower row.
Birne Binegar: So I add two times the third row, the second row use that as a replacement for the second bro you get this good starting to look a little cleaner now, but we have an entry here as well, again i'll use this bottom row it's pivot it clear out the entry above it.
Birne Binegar: Until for get rid of that guy I subtract four times the third row from the first row to clear out that entry.
Birne Binegar: And then finally.
Birne Binegar: Get the entries above and below pivots all equal to zero, I need to clear out this entry.
Birne Binegar: And so that just going to involve subtracting.
Birne Binegar: From row one, the second row.
Birne Binegar: And you end up with this matrix and now finally i'll do something about this too.
Birne Binegar: So what I delete it because I didn't want to introduce fractions and it by delay fixing this to the very end, I can now the by the whole row by two without introducing any fractions.
Birne Binegar: And so what you get in the end is.
Birne Binegar: This matrix on the right hand side or the left hand side, which is the identity matrix according to this rule here.
Birne Binegar: The reduction of our original matrix produced the identity matrix on the left and some other matrix on the right.
Birne Binegar: Well, this guy the 7533 minus two minus two three minus two minus one that matrix is going to be exactly the inverse of thing and the calculation that's done so.
Birne Binegar: This role reduction calculation that was useful for solving linear systems, you now see it has another application, it helps us find the inverse of the matrix.
Birne Binegar: Okay yeah.
Birne Binegar: Ah.
Birne Binegar: Oh, you you're probably right I probably yeah I got a typo there.
Birne Binegar: yeah so y'all try to remember that my.
Birne Binegar: y'all i'll try to correct an online don't try to get back to my office yeah that should have been negative, seven.
Birne Binegar: Okay yeah yeah it was pretty clear alright so.
Birne Binegar: That there's a calculation of them have an inverse of the matrix if you know rubbery option it's not hard at all.
Birne Binegar: it's straightforward maybe strenuous but it's straightforward actually find its inverse and that's that's going to be what's going on in homework set five but, but you should already be pretty familiar with row reduction once you've done problem set for.
Birne Binegar: OK so.
Birne Binegar: Now we come to.
Birne Binegar: sort of a strange.
Birne Binegar: Departure from what i've been emphasizing up to this point what i've been emphasizing up to this point in the course is how to do calculations.
Birne Binegar: How to solve linear systems, how to row reduce matrices how to do scale or multiplication how the multiplying matrices together what's going to happen next, is we're going to transition over into.
Birne Binegar: A broader understanding of what's going on, because these various calculations are linked to one another.
Birne Binegar: So i'm going to give you a theorem that applies to any square matrix.
Birne Binegar: And what it's going to do it's going to relate, a problem of finding solutions to a linear system or a it's a coefficient matrix to the problem of finding its inverse to the problem of reducing it to the identity matrix cetera.
Birne Binegar: So here's here's a statement i'm going to list five statements that I you're going to have the property that if you can verify any one of them, then you'll know that the other four statements are true.
Birne Binegar: So the first one is that this square matrix actually does have an inverse so calculation wise that's going to mean that if I.
Birne Binegar: Take out my matrix and I joined it with the identity matrix i'm getting i'm guaranteed that it's going to.
Birne Binegar: The left hand side is going to rub reduced to the identity matrix.
Birne Binegar: The existence of this inverse, though, is is athlete necessary and sufficient condition could be able to solve any linear system where a appears as the coefficient matrix and, in fact.
Birne Binegar: it's going to be a solution that's actually unique it's not going to be.
Birne Binegar: more elaborate high for playing the other than a single point.
Birne Binegar: here's a third condition which looks like a special case of the second one so here be as some some vector that I put on the right hand side of a linear system, however, if I choose D equal to zero, I get a stronger statement.
Birne Binegar: Here I only.
Birne Binegar: want to know about the existence of a unique solution so it's kind of a week requirement see a stronger if you actually replace the by the zero vector then in fact there's only the, the only solution is X is equal to the zero vector.
Birne Binegar: And then.
Birne Binegar: A fourth condition is that when I wrote reduce a to reduce row Echelon form that that has to be the identity matrix.
Birne Binegar: And then finally.
Birne Binegar: A fifth property is is that a itself can be factorization and it's going to be a product of elementary matrices.
Birne Binegar: Though this is yeah this is sort of like analogous to be able to take an arbitrary integer and write it as a product of PRIMES.
Birne Binegar: The statement here is is similar if you have a matrix a that over to the inverse exists, for example, then you will know that it has to be factorization is a product of.
Birne Binegar: Elementary matrix needs that we create by applying row operations to the identity matrix until we actually able to factor is a matrix.
Birne Binegar: Alright, so what i'm going to do to prove that theorem is i'm going to show that the first statement applies the second second and.
Birne Binegar: Third, third apply for fourth, fifth and then the fifth takes me back to the start, so, in fact, if any one of these statements true you.
Birne Binegar: Just by following this chain of implications if see is true, then D is true, that is true, and as true, then as to be is also true, so you just it's it's a closed circle of implications and so that's generally how you prove.
Birne Binegar: The equivalence of state and you show that it's a closed chain.
Birne Binegar: Right so let's do that I didn't write these statements down, but as I go through the proofs.
Birne Binegar: You i'll restate this these two conditions now it's all remarked very on in the course, this is not approved course so i'm not going through these proofs just so that you.
Birne Binegar: Can memorize them and then apply them on the exam you will not see this proof on the exam However, I do want to go through this step by step, to show you the logic of what's going on, because at.
Birne Binegar: Its most fundamental level this theorem is about connecting the logic that we've already developed.
Birne Binegar: So let's do this first one i'm going to show that statement, a was that not a has an inverse and statement D was that a linear system where I use a as a coefficient matrix always has a unique solution.
Birne Binegar: Alright, so i'm going to show this first i'm going to suppose that a isn't vertebral, and that means it has an inverse it'll be note by a the minus first.
Birne Binegar: What I claim is first of all.
Birne Binegar: Then I have a solution to such a system, so what i'm gonna do is i'm going to substitute into this equation at minus first D and verify that the equation holds.
Birne Binegar: So here's a X substitute for.
Birne Binegar: This value of X and i'll try to show that the right hand side eventually evaluates to just be so here, it goes you just regroup those multiplicative factors.
Birne Binegar: eight times eight minus first is equal, the identity matrix the identity matrix times any other matrix is equal to that second matrix, and so there you go.
Birne Binegar: Here i've explicitly displayed at minus first be as the solution of this equation.
Birne Binegar: Alright, so I haven't quite.
Birne Binegar: I haven't quite prove the statement here all I shown is that I, I do have a solution, and here it is what remains do complete this statement is to show you that it's unique.
Birne Binegar: Okay, so what we'll do is we'll suppose that we have another solution of this equation.
Birne Binegar: Alright, so if y is the solution it satisfies the same equation what i'm going to do now is i'm going to multiply both sides of this equation by the inverse of a which we are supposing exists, then eight minus first time.
Birne Binegar: Here i'm just multiplying offside from the left by a the minus first, then the right hand left hand side just kind of simplify to the identity matrix and find out why, and then I times, why just why, and now you see that, why is the solution.
Birne Binegar: And I simply multiply the equation that satisfies by eight of minors first it tells me that y is equal to eight a minor first be.
Birne Binegar: Well that's not a new solution that's the same solution that we already discovered, and so, if we have any other solution, it had to be the one that we originally found and so we're done we found a unique solution.
Birne Binegar: Okay, so.
Birne Binegar: that's how the logic works and we're done with the first step of this five.
Birne Binegar: Five statement theorem which showed that the first statement implies the second.
Birne Binegar: Okay, so now let's next thing in the chain would be that the second statement implies the Third, the second statement was that.
Birne Binegar: A linear system eight X equals B has a unique solution so we'll assume that and then, what I want to conclude is that the only solution of this linear system where the right hand side is zero that the only solution of that is X is equal to zero.
Birne Binegar: Well, this is really easy, first of all we're supposing that we only have one solution, no matter how we choose be so if I choose be equal to zero.
Birne Binegar: i'm still only going to have one solution it's a special case of condition be but.
Birne Binegar: On the other hand, if I simply substitute X the zero vector for X in this equation it's not going to be right, because.
Birne Binegar: Any matrix time to zero vector is going to produce the zero vector So what do I know I know I know I have unique solution of this equation, and I found a solution, so the unique solution is that zero vector.
Birne Binegar: Alright, so that's.
Birne Binegar: That handles that chain that link in the chain.
Birne Binegar: Alright, so now.
Birne Binegar: we'll start with a third statement that X equals zero vector is the only solution of this equation.
Birne Binegar: And what i'm going to want a nap conclude is that the coefficient matrix a has to be row reducible to the identity matrix, how do we do that well remember the way we find solutions of linear systems is we use row reduction.
Birne Binegar: And what i'm saying here is that X equals zero is the solution well what's that stage, just before we write down the solution of a linear system.
Birne Binegar: Well, we have this augmented matrix it's somehow we did real elementary row operations till you reach reducible Ashkelon form and we get to this augmented matrix of the solution.
Birne Binegar: And so, here it is, if you look at the equations corresponding to this augmented matrix it's exactly the statement that X is equal to zero vector here we have.
Birne Binegar: One times X one plus zero times X, to the zero times X three etc is equal to zero second row is saying X two X equals zero last row is saying the N.
Birne Binegar: Variable X n is equal to zero so i'm replicating this solution with this augmented matrix.
Birne Binegar: Well okay so stepping back from the solution to an augmented matrix Well, we know this augmented matrix had to come from reduction, the original matrix, and so what we know.
Birne Binegar: Since this is the form of the augmented matrix of the solution, I can only reset by applying elementary row operations and so that, like coefficient matrix was converted into the identity matrix and so.
Birne Binegar: If X equals zero is the only solution to this equation, then the coefficient matrix has to reduce the row reducible to this left hand side, which is exactly the identity matrix.
Birne Binegar: Okay, good so.
Birne Binegar: I think we're done with that.
Birne Binegar: yeah i'm just repeating that argument that.
Birne Binegar: Because that augmented matrix was obtained by.
Birne Binegar: Reducing this matrix to it's reduced relational and form which could include that in fact a itself had to be reproducible to that identity matrix.
Birne Binegar: All right now we're up to the force statement.
Birne Binegar: There i'm now going to assume that there's a way of reducing a to the identity matrix.
Birne Binegar: And what i'm to show is that a itself is a product of elementary matrices.
Birne Binegar: Well, this is kind of replicate much of the argument, I gave at the start a class, if I have a sequence of row operations that systematically converts a to the identity matrix.
Birne Binegar: I can implement those role operations via the corresponding elementary matrices instead of acting directly on a matrix I could.
Birne Binegar: Do that role operation to the identity matrix compute this elementary matrix factor and then do matrix multiplication to have the same effect is our one times are one, acting on a, and this is the elementary matrix corresponding r1 multiplying a from the Left.
Birne Binegar: But the sequence of elementary real operations can be implemented instead buy a product of elementary matrices.
Birne Binegar: All right now What this shows you is that, in fact.
Birne Binegar: what's written here it's actually.
Birne Binegar: A the miners first right because this product of matrices times a producer to identity matrix that product has to be a terminus first what i've.
Birne Binegar: argued, thus far is that not a itself, but rather the inverse of as a product of elementary matrices the way I get the stronger statement that itself is also have a product with elementary matrices well.
Birne Binegar: That argument is has two steps to it, first of all.
Birne Binegar: The first thing you do is you show that every elementary matrix has an inverse and.
Birne Binegar: they're easy to figure out what those in verses are because we understand those elementary operations it's been tied to particular row operations and so.
Birne Binegar: Let me do an example here.
Birne Binegar: Suppose.
Birne Binegar: suppose I have an elementary operation that switches to rose.
Birne Binegar: well.
Birne Binegar: So that's going to be now.
Birne Binegar: going to be a corresponding let's see row I interchanging.
Birne Binegar: roget OK, so the corresponding elementary matrix you would switch the I enroll now I and J throw up the identity matrix you get this elementary matrix.
Birne Binegar: But.
Birne Binegar: multiplying is elementary matrix see all it doesn't switch rose.
Birne Binegar: And what I was trying to say here is that the inverse of this guy.
Birne Binegar: is exactly going to be itself so elementary row operation, corresponding to interchanging these two rows because.
Birne Binegar: If I interchange two rows and I do that operation, again I get back to where I started with and so effectively.
Birne Binegar: The inverse of this elementary matrix it's exactly itself, and if you had an operation that say central I to get some.
Birne Binegar: multiples another row what's going to happen in this case.
Birne Binegar: This if I create the corresponding elementary matrix let me write it this way are on ghosts are plus Lambda rj take that.
Birne Binegar: That particular elementary row operation I calculated elementary matrix and then I compute its inverse well what this is going to be.
Birne Binegar: it's going to be another elementary matrix that undoes this operation and effectively what it's going to be is the ice row is going to be replaced by.
Birne Binegar: Our I minus Lambda rj so here when I did if I change the ice row by adding something to it, you undo that what I do is I subtract it off again.
Birne Binegar: But this is just another elementary row operation right it's just where the parameter Lambda has switched signs but it's still the some of our I with a multiple with another row, so, in fact, what happens here is that each of these elementary matrices.
Birne Binegar: There in verses are going to be other elementary matrices so what you do, then, if a the minus first is equal to.
Birne Binegar: E rk.
Birne Binegar: er one, so this product of elementary matrices so remember to use a.
Birne Binegar: Non renewable Echelon for me to get the identity matrix That means that this product development matrix This is exactly the inverse hey I take.
Birne Binegar: The inverse of both sides of that equation.
Birne Binegar: At the minus first the minus first it's going to equal a and what's going to happen on the right well when I introduce.
Birne Binegar: In verses of major cities on Monday, I said that you can calculate the.
Birne Binegar: If you know, a product of maker sees you can calculate.
Birne Binegar: The inverse of that matrix by first of all.
Birne Binegar: Writing it as a product and then get the inverse you take the inverse is that the original factors, but you reverse the order, so this is going to be our one.
Birne Binegar: To the minus first second factor would be to to the minus first.
Birne Binegar: first factor is going to be the bar okay to the minus first.
Birne Binegar: What i'm using here.
Birne Binegar: Is the rule that a B to the minus first is equal to be the minus first aid minus first, but if I have more factors or.
Birne Binegar: The digitalization is simple, when you take the inverse of a product of matrices you get a product of the inverse is but the order of the factors is reversed.
Birne Binegar: And so what I did.
Birne Binegar: From this equation, I see how I can write a the minus first as a product of elementary nature sees.
Birne Binegar: Then I take the.
Birne Binegar: multiplicative inverse of both sides of this equation Ada minus first to the minus first is exactly equal to a and then it's now expressed by.
Birne Binegar: Converting each of these factors into their in verses and reversing the order of the product like at the inverse and then, finally, what i'm using is the observation that when you undo an elementary operation.
Birne Binegar: You get another elementary operation, and so, even though these guys are stated as the inverse is of elementary matrices they're also elementary matrix season selves and we're done we've.
Birne Binegar: argued that.
Birne Binegar: here's my.
Birne Binegar: yeah this is this assessment we showed on the on Mondays lecture and then here's here we're applying it.
Birne Binegar: Once we write a is a two minute first at first, we take the inverse of this product, and then we get a product of the inverse is written in the reverse order, but each of these in verses of an elementary matrix is another elementary matrix we've got.
Birne Binegar: Our matrix a displayed as a product of elementary matrices.
Birne Binegar: Okay, so.
Birne Binegar: Now we have to close the loop.
Birne Binegar: So the loop is closed, when we can show that if a is a product of elementary matrices that it's necessarily convertible.
Birne Binegar: And so, here it is, but each of these elementary he sees his convertible and so a is written as a product of elementary matrices and now we do that same calculation, we take the multiplicative inverse of both sides of that equation, and that displays.
Birne Binegar: That.
Birne Binegar: here's a formula for the inverse of a if it's reducible to the identity matrix if it's a product with elementary matrices, then we can calculate this inverse by simply using that product rule for calculating in verses.
Birne Binegar: Okay, and so now like gone full circle.
Birne Binegar: All right now what's going to happen.
Birne Binegar: After well I a little bit before the midterm because i'm what my intention on Friday is to move forward with material, but that will not be covered, on the first exam.
Birne Binegar: But what we're going to start to do on Friday is start to utilize these chains of implications to attract more information and see for a higher level connections from the things we've been calculating all along.
Birne Binegar: Alright, so i'll stop here asked questions at the ready yeah.
Birne Binegar: Well, so um.
Birne Binegar: I think it's very, very likely that the exam is going to be a week from this Friday, however.
Birne Binegar: i'm giving everyone an opportunity to look at their own schedule and we'll discuss it again on Friday if that's a reasonable date.
Birne Binegar: Okay, other questions.
Birne Binegar: Okay, good i'll see you on Friday, please have a look at the at the homework problems, at least in homework set for so we can identify any problems, you might run into in reduction.
Birne Binegar: what's.
Birne Binegar: what's the elimination.
Birne Binegar: reduction process, it was it was invented by gals.
Birne Binegar: Sometimes it's called gals Jordan.
Birne Binegar: yeah.
Birne Binegar: I think I think, maybe gals is just getting row I salon form and gals Jordan is going all the way to reduce shorts on what.
Birne Binegar: They should be done by next Monday that'd be like you'll have that until the end of the Semester.