Birne Binegar: Okay, so um.
Birne Binegar: let's see here.
Birne Binegar: Okay, so, first of all I brought all the exams from midterm one mentor that I have in my possession, I have them up front here today, so if you haven't collected your.
Birne Binegar: Your first two exams, you can pick them up after lecture today what we're when I plan to do today is this i'm just go through the last three problems on the practice exam these problems, deal with material that we haven't.
Birne Binegar: That we're not touched upon in the first two exam so you should pay attention today.
Birne Binegar: there'll be problems very much like this on the final exams.
Birne Binegar: And what we're dealing with here is actually i'm not I don't have any explicit problems on not planning any specific problems on the final exam involving determinants because calculation of the permanent anyway, is.
Birne Binegar: The first step in actually finding eigenvectors now I can values, and so what the last three problems on the exam are going to deal with, for the most part is finding I give vectors and eigenvalues and also this business of.
Birne Binegar: orthogonal decomposition to respect with subspace.
Birne Binegar: Okay, so.
Birne Binegar: Let me start asking if there are any questions on what we've covered so far.
Birne Binegar: Okay, so.
Birne Binegar: Let me step through these problems i'll try to work them out directly on on the practice exam So you see, you have plenty of room to answer these questions and.
Birne Binegar: Maybe design your responses on the actual exam following.
Birne Binegar: My examples here so.
Birne Binegar: we'll begin with his problem eight were were given a three by three matrix.
Birne Binegar: And we're here it is 110001001 and we're asked to find the eigenvectors in eigenvalues of that matrix.
Birne Binegar: Okay, so.
Birne Binegar: What do you do, that is, first of all, you need to find the eigenvalues and that's done by calculating the characteristic polynomial.
Birne Binegar: Which is determinant of the matrix you get like subtracting from your given matrix.
Birne Binegar: A parameter Lambda along that main diagonal.
Birne Binegar: What will get here.
Birne Binegar: Yes, one minus Lambda 1000 minus Lambda or simply negative Lambda wow.
Birne Binegar: Sorry, you can't hear me.
Birne Binegar: yeah.
Birne Binegar: Alright, so the last row here is going to be 001 minus Lambda.
Birne Binegar: Okay, so first thing we need to do is calculate that determinant it's going to give us a polynomial and Lambda whose route that is solutions up PA of Lambda equal to zero, those are going to give us exactly or eigenvalues.
Birne Binegar: don't.
Birne Binegar: We have to take a determinant of a three by three matrix notice that, in this last row, we only have one non zero entry.
Birne Binegar: So an ideal strategy for calculating that determine inches just do a Co factor expansion along that last row so that would be equal to well zero from that first term zero from the second term, and then we have a third entry.
Birne Binegar: One minus Lambda and we have a sign minus one that power minus one is raised to the summit, the row index and in column index at this.
Birne Binegar: Item we're looking at here so that's going to be three plus three and then we have the determinant of the corresponding minor, which is the.
Birne Binegar: Smaller matrix I get by deleting the row and column that contain that three three entry that's going to be one minus Lambda one zero minus Lambda.
Birne Binegar: And then let's see we'll have a factor of one minus Lambda this is ends up being minus one to an even power so that will just be a factor of one and then this.
Birne Binegar: Two by two matrix is determinative simply going to be one minus Lambda Lambda because it's a upper triangular matrix and for those matrices to determine and it's just the product along the main diagonal Okay, and so right away, we see.
Birne Binegar: We have to eigenvalues.
Birne Binegar: One of these factors has to equal zero if the characteristic polynomial is going to equal zero so either Lambda half that equals zero or Lambda equals one.
Birne Binegar: Those are two.
Birne Binegar: eigenvalues.
Birne Binegar: And now, what we'll do is for each Eigen space will find a corresponding basis factor so for the audience face is zero this is going to be.
Birne Binegar: Zero Eigen space.
Birne Binegar: And when you calculate it and you find the solution set of.
Birne Binegar: A linear equation, where you use a minus zero times I so that's the eigenvalue looking at and that's just going to be no space of a.
Birne Binegar: record for subtracting nothing from the main diagonal and that's going to be the north face.
Birne Binegar: let's see our matrix is 110001 and I may as well, you notice the second and third row or exactly the same when you calculate the no space what you're looking for.
Birne Binegar: is to reduce row Echelon form of your coefficient matrix well I might as well just observed that.
Birne Binegar: The first thing I would do to reduce this guy to row Echelon four minutes I get rid of this pivot below the pivot in the second row well, I would just subtract the second row from the third, so what i'm going to write down here is i'm.
Birne Binegar: Assuming i've done that real reaction, and this is the.
Birne Binegar: Reduce row Echelon form of this matrix.
Birne Binegar: and
Birne Binegar: Now, what we do is we try to.
Birne Binegar: find a basis for list solution set well the equations are X one plus two, plus zero X equals zero and zero times X one zero times X two one time X three is going to have to equals zero.
Birne Binegar: Until the equations we get are going to be X one plus x is equal to zero and X three is equal to zero that third row doesn't tell us much it just tells us at zero is equal to zero.
Birne Binegar: Next, what we do is we.
Birne Binegar: look again at that matrix in reducing Echelon form and we observed that the.
Birne Binegar: Second column here or a paper up that second column does not contain a pivot that's going to be X two is a free parameter so i'll take the free parameter in this equation and i'll move it to the opposite side.
Birne Binegar: We get X one is equal to minus X two, and this is free.
Birne Binegar: And X three is a fixed variable, it has to equal zero and then from these you see that your solution vector.
Birne Binegar: Well, the first component you got to be minus X two second component will be X, to the third component has to be zero so that's simply some constant times the vector minus 110 and that constant vector to be it's being multiplied by the three parameter that's your eigenvector.
Birne Binegar: corresponding to the eigenvalue.
Birne Binegar: Lambda equals zero we're going to have the Eigen vector minus 110.
Birne Binegar: That take care of.
Birne Binegar: The first eigenvalue we have a second eigenvalue or which we have to do a similar come computation.
Birne Binegar: So now we'll look at the E one I guess based move this up it's going to be the no space.
Birne Binegar: Of the matrix I get when I subtract the eigenvalue one from a so let's look at a when I subtract one from the diagonal a a minus I that's going to be the north base.
Birne Binegar: Of the matrix sexy it's going to be 0100 minus one one and 000.
Birne Binegar: Which is the same, at least, that no spaces, the same as the role Echelon form of this matrix you can see, pretty good pretty clearly what that's going to be is.
Birne Binegar: yeah I can eliminate that it's just going to be the null space 010001000.
Birne Binegar: Upon row reduction this matrix becomes this, and from that you can see that what you require of your solutions is the X to has to equal zero X equals zero X one there's no pivot in that column so X one is free.
Birne Binegar: And so, your solution variable since the last two solution vector since the last two components have two equals zero it just going to be some multiple of the first standard basis factor.
Birne Binegar: And so, your eigenvalue your eigenvector for the eigenvalue Lambda he was one it just going to be that 100.
Birne Binegar: Okay, any questions on that.
Birne Binegar: part when we do software, I can better 00 yeah.
Birne Binegar: So X rays are free, the reward is X three.
Birne Binegar: Well, so X X three is there's a pivot in that call.
Birne Binegar: So where is it here, you see here, this is the reducer Echelon form that matrix in here net third column, you have a pivot.
Birne Binegar: And so that you get you get an x Ray is going to be set equal to zero so it's no longer free it's got there's.
Birne Binegar: yeah what happens is when you have a pivot you have a first non zero entry, and that means you can.
Birne Binegar: Take the corresponding variable and you'll be able to solve for that, but the other.
Birne Binegar: variables that might occur in the equation there necessarily going to be free until you move those to the other side of the equation don't for this this row it simply says X Ray is equal to zero it's not a free variable.
Birne Binegar: Okay.
Birne Binegar: If I were free variable corresponding X Ray they would have to be another another non zero entry net row so you'd have but.
Birne Binegar: You only have one pivot in a row, because you've now reduced everything to row isilon form, so you only if you have a.
Birne Binegar: You have an anon Tyrol pivot, then you have other non journal entries they can't live in columns that have pivots.
Birne Binegar: And so the first entry is always going to be the thing you can solve for and the variables, corresponding to the other entries in a.
Birne Binegar: In a row, one in the matrix is reduced social on form they're going to correspond automatically the free variables and you just push them to the other side.
Birne Binegar: That way, you always have one variable you can solve for for each row you'll be the first the first variable that's the one you can solve for and the other ones are going to be free.
Birne Binegar: Well, you eat you work it out and you use the number of columns without pivots it's going to give you the number of free parameters.
Birne Binegar: And that's going to give you the number of free parameters is going to give you the dimension of this space, so in this in this example we only had one call without a pivot.
Birne Binegar: So we only have one three parameter and that's why our Eigen spaces one dimensional is it's basically the solution set of linear system, you have using out of the coefficient matrix okay.
Birne Binegar: Okay, so um.
Birne Binegar: Oh there's a second part to this question, but we can answer it fairly easily in Part B.
Birne Binegar: The question is, is this matrix dying likable and the next problem lastly diagram is a matrix, but we can answer this question already.
Birne Binegar: What we need to diagonal is a three by three matrix are three linearly independent eigenvectors how many we found well, we found one linearly independent eigenvector for the eigenvalue.
Birne Binegar: zero and we found so that's this guy and then we found one linearly independent eigenvector for the eigenvalue one but we don't have any more eigenvalues and so.
Birne Binegar: The maximum number of linearly independent eigenvector shoot down as to.
Birne Binegar: We need 343 by three matrix in order to diagonal is it so the answer to be is no we don't have enough linearly independent eigenvectors we don't have three linearly independent eigenvectors, so we cannot die and license three by three matrix.
Birne Binegar: Okay.
Birne Binegar: Do another one, so in problem nine.
Birne Binegar: i'm getting i'm giving you a matrix that will be diagnosable and I just want you to die realize it fine sorry they get it off screen here once you define major cities see in a diagonal matrix D, such that see the miners first time a time see is a diagonal matrix.
Birne Binegar: Now the way you've solved that issue solved the eigenvector I first of all, the eigenvector I can value from for the given matrix a.
Birne Binegar: Then what you'll do is that so long as you have to linearly independent eigenvectors you can construct an inverted matrix si.
Birne Binegar: si si has to be in verbal you can construct this matrix using to linearly independent eigenvectors for the original matrix ED and then the diagonal elements of this diagonal matrix D i'm just going to be the eigenvalues corresponding the eigenvectors used to construct see.
Birne Binegar: So let's uh let's begin I solve.
Birne Binegar: The eigenvector.
Birne Binegar: I can bought value problem.
Birne Binegar: For the given matrix so that means we look at P, a Lambda equals, to determine.
Birne Binegar: minus one minus Lambda 513 minus Lambda that determine is going to be the characteristic polynomial for the given matrix and what it's going to be, is it looks like.
Birne Binegar: minus one minus Lambda times three minus Lambda minus five so that's going to be like the negative three cross terms or minus three plus one look like minus two Lambda.
Birne Binegar: And then, plus Lambda squared minus five which I guess is.
Birne Binegar: gonna be.
Birne Binegar: Lambda squared minus two Lambda minus eight which now I can factor ISIS Lambda minus four times Lambda plus two.
Birne Binegar: Alright, so.
Birne Binegar: After I computed is, I had a polynomial I had to actually simplify a little bit, but then once I get it in.
Birne Binegar: standard form it's pretty easy to factor eyes, you have two factors to import eight and you can actually arrange for their cross turn to produce this minus two pretty easily, and so what we have, or two eigenvalues.
Birne Binegar: Lambda equal to four and Lambda equal to negative two, by the way.
Birne Binegar: Knowing the eigenvalues I already know, but that diagonal matrix has to be you'll be 400 negative two I just use the eigenvalues like down and that that will give me the matrix.
Birne Binegar: We have to figure out is that matrix see and that's going to be determined once we figure out the eigenvectors for each of these eigenvalues.
Birne Binegar: So for Lambda equal to for that corresponding Eigen space is going to be the no space.
Birne Binegar: Of the matrix you get I subtracting for from the diagonal elements of the original matrix a so minus one minus four is going to be negative five, then we have five, and then we have one.
Birne Binegar: Three minus four is going to be negative one it's pretty clear that both these rows.
Birne Binegar: become minus one.
Birne Binegar: minus 100 alright so i've.
Birne Binegar: Looked at this matrix and I see that both rows are proportional to one minus one, this one is just minus five times a second row, so this one's going to simplify.
Birne Binegar: To this reduce show Echelon form, and from that oh one thing to point out.
Birne Binegar: I mentioned this before.
Birne Binegar: And this example of a computation of the north of an Eigen space, we had a zero with the bottom of the matrix of our real Echelon for.
Birne Binegar: notice that also occurred.
Birne Binegar: Previously, when we were calculating know the the Eigen space for the zero again space in the preceding problem and the one I get stations proceed problem we always had a zero row that's.
Birne Binegar: that's actually a consequence of this, of setting this determinant equal to zero you're guaranteeing that the matrix is not going to be.
Birne Binegar: Real reducible to the identity matrix when you subtract and eigenvalue from those diagonal elements so think of this zero row as a confirmation that your calculation is proceeding as as expected.
Birne Binegar: All right now.
Birne Binegar: This is the coefficient matrix of the linear system in reduce relational on form we convert back to equations to get the equations of the solution.
Birne Binegar: That first row says X one minus X two has to equal zero second row says zero has to equal zero and from this, we see that a solution vector.
Birne Binegar: Well X two is going to be a free parameter and X, one can be expressed in terms of X to what you're going to get it's simply this vector one one, so let me.
Birne Binegar: Show you quickly here, if you have X one minus X is equal to zero and zero is equal to zero, this is free, because it's the second variable that occurs in the equation, corresponding to a row, the reduce show excellent form, so that tells me X one.
Birne Binegar: X equal X two second equation doesn't tell me anything my solution variable solution vector.
Birne Binegar: it's going to look like.
Birne Binegar: X two X two or simply X two times one one.
Birne Binegar: Alright, so I hope this calculates the sort of going from the equations coming from the matrix and reduce your actual on form to the corresponding basis factors.
Birne Binegar: I hope we were done this enough times that this is sort of straightforward, but if not here's the short calculation that allows me to go from.
Birne Binegar: This coefficient matrix in reduce your actual on form who the basis factor for the solution set actually X to any constant times this vector is going to be a solution of this matrix.
Birne Binegar: Well linear system with that guy as the coefficient matrix so the upshot of this is for the eigenvector corresponding the Lambda equal, for we have this vector one one.
Birne Binegar: All right, we have a second eigenvalue wham the equal negative two until will calculate also it's again space that's going to be the north face up well, so now what i'll do is i'll subs to.
Birne Binegar: Negative two for Lambda and basically calculate the north face of the corresponding matrix so when Lambda is equal to negative to what we're going to get is.
Birne Binegar: look like minus three when I oh sorry there'll be plus one.
Birne Binegar: When I subtract negative two from negative one i'm going to get plus one by one, and then three minus a negative two is going to be five or so and that's going to really do real quickly 21500 good.
Birne Binegar: Again, you got that disrobe, which is a good check that our calculation is proceeding as it should now we write down the equations.
Birne Binegar: corresponding to that coefficient matrix and reduce your actual form beta intelligent X one is equal to minus X.
Birne Binegar: minus five times X two and I take this the opposite side that's right take up the minus sign, and so the solution vector in this case it's going to be.
Birne Binegar: minus five one times a free parameter X two, and so the Eigen vector corresponding to the eigenvalue minus two is going to be minus five one all right, and now we're in a position to write down.
Birne Binegar: The matrix see that was going to diagonal is a.
Birne Binegar: i've already written down the diagonal matrix D that he gets converted into to write down the matrix see when I have to do is write down the corresponding eigenvectors in the same order that I wrote these eigenvalues So the first Eigen value was for and it's eigenvector.
Birne Binegar: was one one.
Birne Binegar: And then the second I can value is negative, to its force learning that guy so the second like injectable uses minus five one all right, and then I put a box around D and see.
Birne Binegar: If i've answered the problem I found the two by two matrix see in a diagonal matrix D such this identity is automatic.
Birne Binegar: OK so.
Birne Binegar: The soul is diarization problems you have to find both the eigenvalues in the eigenvectors the eigenvalues go into this matrix D and all they are is going to be a diagonal elements and you just list your your eigenvalues in order along that main diagonal.
Birne Binegar: While you choose some order could have been minus two could have been minus 244 minus two, but if you chose if you chose a different order here you'd have to switch these columns So the important thing is to correlate the eigenvalues with their eigenvectors.
Birne Binegar: If it was three by three.
Birne Binegar: Longer calculation, but it'd be exactly the same.
Birne Binegar: yeah you would pick you would pick some ordering you would pick some ordering that the eigenvalues and then write it down down a diagonal matrix and then, when you try to construct the matrix see just make.
Birne Binegar: The first guy go the first column it's going to be the vector corresponding the first diagonal entry.
Birne Binegar: Okay let's see we have one last problem to do.
Birne Binegar: And this is going to deal with.
Birne Binegar: orthogonal D compositions.
Birne Binegar: and
Birne Binegar: I like these problems because they really.
Birne Binegar: there's sort of a geometric intuition to them, but a few metric intuition isn't that very helpful in actually getting to the solution to it rather an order to solve problem 10 you have to understand.
Birne Binegar: Both the notions of Sub spaces and bases and what they're what they're good for.
Birne Binegar: So the way on one solve these problems where you need it decompose a given vector with respect to a subspace issue you define.
Birne Binegar: bases for the sub base and you also need to find and in doing so you'll get this piece automatically this piece of basically going to come from knowledge of the basis for this space this.
Birne Binegar: This piece, here comes from knowledge of a different subspace but it's the set of vectors that are perpendicular to every vector and w.
Birne Binegar: So.
Birne Binegar: The first thing to note is that.
Birne Binegar: that these two vectors are obviously linearly independent so so note that dumb out the vectors 110 and.
011.
Birne Binegar: Our basis.
Birne Binegar: For w a reason, I can tell that simply by inspection is that this factor has a component that can't be canceled with the other vector and similarly.
Birne Binegar: This factor has a component that can't be cancelled, with a contribution from this, so if I had a linear combination of these guys and that was the equals zero i'd have to take the coefficient of fee to is equal to zero, but then I wouldn't be able to.
Birne Binegar: Well, I I couldn't cancel this guy without setting whatever is multiplying this factor equal to zero, and similarly I couldn't cancel the third.
Birne Binegar: component, without taking multiple at the second factor equal to zero, so these guys are pretty much linearly independent, let me say, like Dan explicitly what i'm saying here now if.
Birne Binegar: V1 and V2 are linearly independent, then it means the only solution.
Birne Binegar: X One the one plus x two the two equals zero is X one equals zero.
Birne Binegar: And X is equal to zero now, if you look at these two vectors X one times 110 plus x two times 011 that can only equal to zero vector.
Birne Binegar: Well, upon calculation, you need X one plus two times X zero then we'll get X one plus x two in the second component and then we'll just get.
Birne Binegar: X, to the last component, you see, that the only way i'm going to get zero is for that first component.
Birne Binegar: I have to take X one is equal to zero and get the last component that equals zero, I have to take X is equal to zero, so the only linear combination.
Birne Binegar: Of these two vectors that produces the zero vector is the one where X one X two or equal to zero for that reason the vectors are linearly independent.
Birne Binegar: That sort of phenomena and you can spot whenever you have a list of vectors for which one are the vectors has a component that can't be canceled with the other vectors.
Birne Binegar: Okay, so I know that.
Birne Binegar: The vectors given actually constitutes.
Birne Binegar: A basis for the subspace.
Birne Binegar: What we look at next is a second subspace w this is going to consist of all the vectors.
Birne Binegar: inside our three such that X dotted with an element of w equals zero.
Birne Binegar: The scalability for all w inside are given subspace see what we're trying to do is, we have a.
Birne Binegar: set of two independent directions that's capable of reaching every vector and w but this vector doesn't live in w we need some additional directions, so that we can actually get coordinates for this sector, so what we do is we.
Birne Binegar: Create Nigel reset substations we're looking for additional directions and the best natural way to find additional directions, is to find perpendicular directions.
Birne Binegar: directions that are perpendicular to everything you started off with, for example, if I wanted to build the three dimensional space of this room.
Birne Binegar: And I knew I had a basis for the blackboard all you need is an additional direction that is perpendicular to like given to direction, then I get a basis for the entire.
Birne Binegar: Entire entire room and so that's actually the purpose of this guy it allows us to extend this basis for w to a basis for the entire vector station which things live.
Birne Binegar: Alright So what is this This turned out to be the north face.
Birne Binegar: Of the matrix we get my taking basis vectors for w as wrote 110011.
Birne Binegar: Because.
Birne Binegar: If you had a.
Birne Binegar: bunch of row backers.
Birne Binegar: And you had a column vector X one X n.
Birne Binegar: This matrix product which is produce be one dotted X be to God with X get down to be okay with act so.
Birne Binegar: This matrix product, how do you compute it you take God products of rows with the one and only call them at the second factor feel the one one component it's going to be the first row times that one and only call second component can be zero so this thing evaluate this.
Birne Binegar: When you write it this way, we have a linear system and what is it says it says that X is perpendicular to each of these factors, and so, for looking for.
Birne Binegar: A subspace of vectors that are perpendicular to every vector inside w what we do is we write the basis vectors of w is rose, and then we think of that is defining a linear system that we can solve.
Birne Binegar: don't calculate this no space and when it will be also the north face of 110 let's see reduce relational and form is going to be one sorry it's going to be zero minus 111.
Birne Binegar: Right, so I quickly did the role reaction.
Birne Binegar: To the way I got the this matrix here is by taking.
Birne Binegar: I basis vectors for the subspace w in writing image rose that gives me a different subspace and I calculate the null space and we can see what it's going to be X, one has to equal X three x two is going to have to equal minus X three.
Birne Binegar: that's the equations i'll get.
Birne Binegar: From these two rows up a coefficient matrix and reduce your rational and form and i've already identified X three is the pre parameter and put it on the right hand side so.
Birne Binegar: Our solution vector is just going to be X three times one minus one one.
Birne Binegar: Alright, so this is the basis vector for this no space and so it's a beta spectre w perpendicular.
Birne Binegar: All right now, the next step, I should be writing this on paper so let's see what we did is first we first fine.
Birne Binegar: A basis for another seven days, which is defined as a set of vectors.
Birne Binegar: X inside or three such that X dotted with w always equals zero for all w in tied our original space and.
Birne Binegar: What happened in is that.
Birne Binegar: That w perp it's going to be the no space.
Birne Binegar: Of the matrix we get by writing our basis vectors is rose.
Birne Binegar: And that worked out to be when all was said and done.
Birne Binegar: little space 110011.
Birne Binegar: equals to no space.
Birne Binegar: One zero minus 1011.
Birne Binegar: And that led us to a vector.
Birne Binegar: Everything proportional X three and it will be one minus one one and that one basis.
Birne Binegar: vector for w PR all right now, the reason for.
Birne Binegar: constructing this like jewelry subspace w perp is actually define this additional basis factor, because we're working in a three dimensional space 123 dimensions.
Birne Binegar: We only have a two dimensional subspace, so we need a an additional direction to actually coordinate eyes are three dimensional space what this calculation finding a basis specter for this perpendicular suspect does it gives us the additional direction we need, so now we have.
Birne Binegar: Three doctors.
Birne Binegar: First, two were given to us.
Birne Binegar: 011 and now we have a third vector one minus one one.
Birne Binegar: Are these are going to be three linearly independent vectors.
Birne Binegar: In a three dimensional vector space until they provide a basis.
Birne Binegar: For our three that means every three dimensional vector can be written as a linear combination of these guys so in particular.
Birne Binegar: The are given vector was.
Birne Binegar: or sorry are given back to your head component 111 so that's a three dimensional vector, and so it can be represented as a linear combination of these three actors i'm going to write them vertically So you see that linear combination 110.
Birne Binegar: Plus C two times 011.
Birne Binegar: Plus C three times one minus one one.
Birne Binegar: Now what i'm gonna do.
Birne Binegar: Is i'm going to figure out exactly sorry.
Birne Binegar: i'm going to figure out exactly what those numbers see one and see to.
Birne Binegar: Our that will see once you to MC three once I figure out what those numbers are.
Birne Binegar: Some of the first two vectors since i'm using basis vectors are w this part it's going to live in.
Birne Binegar: it's going to live in the perpendicular in the subspace w and this piece it's leftover it's going to be.
Birne Binegar: Like outside of who enable compile it can be identified with this deep perk.
Birne Binegar: So how do we find.
Birne Binegar: Those numbers see one cpu and see three again effectively, we have a linear system to solve this time it's a non homogeneous we have.
This.
Birne Binegar: This multiplication here, you can think of that as well take the each of these vectors and write them as columns.
Birne Binegar: 0111 minus one one.
Birne Binegar: And then.
Birne Binegar: time to call them vector C one C to C three what I have is a homogeneous or non non homogeneous linear system, the constant factor is equal to a certain coefficient matrix times.
Birne Binegar: an unknown vector.
Birne Binegar: But what we want our those components, so that we can.
Birne Binegar: make this expansion explicit, so what we'll do is we'll solve that linear system by taking the coefficient matrix 111 minus 1011.
Birne Binegar: Take the constant vector on the right hand side is in homogeneous term, this is the.
Birne Binegar: augmented matrix for that system will reduce to row Echelon form 1011 let's see.
Birne Binegar: One minus one is 01 minus zero is one negative one minus one is negative two and one dies wanted zero and then.
Birne Binegar: doesn't look right.
Birne Binegar: Actually it's going OK, and then we have 011 and on the last call we're going to have one and that's going to reduce further 2100010001 and last call we end up with 232 30.
Birne Binegar: And one third.
Birne Binegar: All right, so.
Birne Binegar: Why was I looking at this.
Birne Binegar: augmented matrix it's got i'm trying to find these coefficients he wanted see two and see three such that are given vector be can be expressed in terms of this compound basis basis for w combined with the basis for it.
Birne Binegar: orthogonal compliment.
Birne Binegar: So these first two columns correspond to a basis for w last column correspond to a basis for w PR.
Birne Binegar: I write those right the given vector is the last column, and I will reduce all the way to reduce row excellent form.
Birne Binegar: And now the last step is to convert back to equations in this case, I have the identity matrix so that first row tells me that X one is equal to two thirds second row tells me that.
Birne Binegar: solution must have to equal to two third, and the third row says that the third component has to equal 130 so notice in this case, we have a unique solution.
Birne Binegar: Well that's appropriate because these three factors are basis factors.
Birne Binegar: That means that anytime you expand a vector in terms of your basis factors there's only one choice of coefficients that will work, but we calculated or exactly those choices let's call them C one C two C three so we can interpret them in this fashion, so that means the let me write.
Birne Binegar: That down here so.
Birne Binegar: What we found here is that this matrix.
Birne Binegar: 110011 and one minus one one and then.
Birne Binegar: I use this matrix in the last call or that vector and last call that that reduces to.
Birne Binegar: matrix 1110000002 thirds two thirds one third.
Birne Binegar: All right, that was on the blackboard and now, that means that be.
Birne Binegar: it's going to equal.
Birne Binegar: To 30.
Birne Binegar: times.
Birne Binegar: 110 see two is two thirds two thirds times 011 and then the last factor.
Birne Binegar: Is as coefficient one third one minus one one and now, some of these two guys.
Birne Binegar: Sorry.
Birne Binegar: Here we go so.
Birne Binegar: So what I did.
Birne Binegar: Having found a basis for the entire vector space, I take the given vector and I expand it in terms of that basis well defined the numbers, I have to solve another linear system that's done using row reduction.
Birne Binegar: Once I get the solution see one has to equal to 32 has equal two thirds and see three has to equal one third I didn't write out the expansion and then I know that the first two terms are constructed using basis vectors or w.
Birne Binegar: So let's just write that guy down it's going to be.
Birne Binegar: Two thirds, the first component then i'll have four thirds, and the second component and last component is going to be two thirds again so sorry going off off the page here.
Birne Binegar: Because.
Birne Binegar: Let me show you because i'm i'm just taking, I have a be itself is this entire.
Birne Binegar: combination of three vectors but the first two vectors live inside w So the first component is going to be two thirds times one plus two.
Birne Binegar: Times zero i'm just going to get two thirds, and the second component, I have two thirds times one plus two thirds times one and in the last component that I kind of falling off the page there that two thirds times zero plus two thirds times one just going to be two thirds.
Birne Binegar: And then, then the perpendicular piece.
Birne Binegar: it's just going to be this last one so that's going to be.
Birne Binegar: One third minus one third one third.
Birne Binegar: And then, those are the that's the orthogonal decomposition he is equal to this vector that lives inside the original subspace w and then there's a leftover piece perpendicular piece and that's exactly that last piece.
Birne Binegar: Okay, so.
Birne Binegar: yeah so i'm sorry.
Birne Binegar: So look carefully, there was no homework explicitly assigned on this orthogonal decomposition because it didn't really make sense to sign homework this last week, but a problem, like this is going to be on the final exam so.
Birne Binegar: The printed solutions that are actually posted I tried to be explicit.
Birne Binegar: With my verbiage of exactly how you do this study, though solution to be able to repeat it, the basic idea is going to be to first of all find a basis for your subspace w.
Birne Binegar: Then you need to find a basis for an additional subspace that you get by taking the original set of basis vectors and writing them inch rows.
Birne Binegar: and finding a basis for the solution set that matrix times X equals zero, then what you'll have our as many basis vectors coming from w in the perpendicular subspace that you get a basis for the entire vector space and you then have to.
Birne Binegar: write your given vector in terms of that new basis for our three and but you do another row reduction to get those coefficient and then you just split your answer yeah in half to get the dw in deeper.
Birne Binegar: Okay, oh what's that.
Birne Binegar: i've worked out the final jam any any questions.
Birne Binegar: Okay, so I sent out an email this morning with a correction, the final exam is going to be in this room, on Wednesday at 10 o'clock and so, if I stated that wrong now, whatever the university schedule is that's what we're going to help have the final exam but.
Birne Binegar: it's going to be, Wednesday, at 10 in this room.
Birne Binegar: Sorry.
Birne Binegar: Oh, you may be right, it may be at eight o'clock.
Birne Binegar: you're right you're right, it is, it is it eight o'clock Wednesday may 8 at eight o'clock in 312 your classroom okay.
Birne Binegar: Alright, so i'll see everyone then good luck studying, not only for my final but for your others as well.
Birne Binegar: So I know a way.
Birne Binegar: That I can get this thing.
Birne Binegar: Can I do that or like you did you have to write each.
Birne Binegar: be kept it right these guys just columns so what happened is that first thing you calculated you use that as the first call.
Birne Binegar: Second, one, the second column third you have to you have to make sure they have the right orientation, or else you don't get the right matrix yeah.
Birne Binegar: Like I can do it in like I don't know how they do it to make them, but I can kill switch to my brain and and just get you this answer what will you do this point because I didn't do this part.
Birne Binegar: If I only gave you this rights without this ah.
Birne Binegar: Probably all understand what you're doing but it's always say this to to explain as much as possible, because that way, if you do something wrong, I can at least give you partial credit.
Birne Binegar: card I don't understand the English part of it's like.
Birne Binegar: But like what I understand is that we have to put a certain value and if it's not equal to, we have to test, something we have to test to make sure that if we take a vector that starts inside this set and we multiply it by any number that we're still living inside the set.
Birne Binegar: To let's take this vector it lives inside the set because one plus zero Plus zero does equal one, so the components of this vector satisfy this condition so w lives inside capital yeah let's multiply it by zero.
Birne Binegar: And what happens, you get zero, but now zero Plus zero Plus zero equals zero, it does not equal one so so this vector doesn't live inside here.
Birne Binegar: But if you look back sorry you look back at the definition of.
Birne Binegar: What it means to be a subspace you have to if you start off with a backdoor and a number and you multiply together, you have to stay in that set.
Birne Binegar: But what you found is that if you multiply this factor, even though it starts in w after you multiply by zero it leaves w so this condition is not satisfied, and so it's not a subspace okay.
Birne Binegar: yeah.
Birne Binegar: What else like these we just memorize them right just memorize them but but, as you can see, knowing these things you can, if you have a question about nothing you working now go back and look at the definition and and see what and see what kind of object you're looking for.
Birne Binegar: you please it's like you don't want like I can see my homework assignment, I cannot see the campus well I haven't I haven't completed it yet, but.
Birne Binegar: What you can do is take your your score on web assign cigarette and get get the percentage score okay take the percentage and just multiply that by 50 so that's still 50 points is going to be the maximum you can get for homework in this course.
Birne Binegar: Thank you so much, oh yeah yeah what's your name.
Birne Binegar: lesson again.