Birne Binegar: Okay, so.
Birne Binegar: First of all, the final exam is Monday at eight o'clock that right what Monday at 10 o'clock I think in this room yeah.
Birne Binegar: And so what i'm going to do today is I started going over a practice exam on Wednesday, what I want to do is.
Birne Binegar: Do the last three problems of the.
Birne Binegar: Practice final exam first because that's material that has actually appeared on a preceding exam and there was one problem that I didn't.
Birne Binegar: Quite finish on Wednesday that dealt with material that was already covered on the on the second exam that's just a promo what you have a linear transformation i'll do that at the very end, but I want to make sure that I covered the material that.
Birne Binegar: dealing with problems that come from that aspect of the course that occurred after the second exam so let me start with the.
Birne Binegar: The eighth problem on the practice exam where the idea is to find the eigenvalues in eigenvectors of a given matrix.
Birne Binegar: Alright, so.
Birne Binegar: The way you perceive the solvent eigenvector eigenvalue problem is first you find the eigenvalues that's done by solving a certain polynomial equation.
Birne Binegar: Then you take the route to that all normal equation those will be your eigenvalues and for each of those you find a corresponding eigenvector once you've correlated eigenvalues with their eigenvectors, then the problem is done.
Birne Binegar: So in this problem eight we've given a three by three matrix So the first thing we do.
Birne Binegar: Is we look for to eigenvalues by setting a certain polynomial equals zero what that polynomial is its determinant of the matrix we get from the written given matrix.
Birne Binegar: By subtracting a parameter Lambda from its main diagonal elements so here we'll have one minus Lambda one zero for the first row 010 minus Lambda.
Birne Binegar: One second row 001 minus Lambda alright, so we have a determinant to compute now on the practice final exam and so likely, also on the actual final exam you take on Monday, there are no.
Birne Binegar: Problems where you explicitly have to be computer determinant but right away in this problem eight, you have to computer determinant so let's look at the matrix which we need computers determinant and you see that along this last row.
Birne Binegar: Most of the entries are zero you only have a non zero entry in this three three flop, so what we use as a co factor expansion.
Birne Binegar: Along the third row.
Birne Binegar: use that to compute this determinant so from the first entry, we have zero they don't have a second contribution from the second entry, but both of those entries, because the entries of the matrix themselves are zero.
Birne Binegar: This last one that's where we have the act like do a computation so we have the entry itself.
Birne Binegar: Then we have a sign factor that comes about by figuring out the position of this entry it's in the third row and third column so we'll take minus one to the three plus three and then we have the determinant.
Birne Binegar: Of the matrix we get might.
Birne Binegar: The corresponding minor, and that would get by taking row and calm removing those from the matrix we started off with and then taking determining the two by two matrix that remains well that's pretty.
Birne Binegar: Much just going to be one minus Lambda one zero and then zero minus land or negative Lambda that's a two by two matrix.
Birne Binegar: And it's determinant cheek we have a one minus Lambda the sign factor just becomes a factor of one and determine, we need to compute this get that by the different types of dive into products.
Birne Binegar: minus Lambda time minus Lambda.
Birne Binegar: And so that's going to work out to be one minus Lambda squared times Lambda in an overall minus sign okay so.
Birne Binegar: Actually, what i'm looking for are the numbers Lambda such that this polynomial equals zero.
Birne Binegar: And so you see either got to have the first factor equal to zero, that would mean Lambda is equal to one or you need the second factory little zero meaning Lando would equal zero in that case alright So these are eigenvalues.
Birne Binegar: All right, we also need to find the eigenvectors and for to do that.
Birne Binegar: yeah we can look at, for example, he one.
Birne Binegar: Like in space so that's the.
Birne Binegar: eigenvalue.
Birne Binegar: one.
Birne Binegar: Again space.
Birne Binegar: And that's obtained by.
Birne Binegar: Looking at the north face of the matrix we get by removing that eigenvalue from the dive and our lunch at the original matrix so JESSICA, the.
Birne Binegar: lead a minus one times the identity matrix and that's going to be the normal space.
Birne Binegar: Well, what happens when we subtract one from there, we get 010 when we subtract one from the second diagram we get minus one one, and then the third row we get 000 all right.
Birne Binegar: notice that zero with the bottom i'm sorry, this is not coming up clear diode that's a little better.
Birne Binegar: notice the zero with the bottom here when you calculate these no spaces, you should always.
Birne Binegar: At least one you row reduce the matrix euro Echelon form, you should always have a zero at the bottom if you're finding an Eigen space because.
Birne Binegar: The condition here that this Paul o'neill banishes that's guaranteeing that this matrix cannot be reduced the identity matrix and that's where this that consequence is showing up here, because you have zero row here this matrix which comes about by substituting a root.
Birne Binegar: Of this equation into this matrix once.
Birne Binegar: This matrix cannot be diagonal I if I such to.
Birne Binegar: The eigenvalue one for Lambda there because that's what your athlete imposing here by setting this polynomial equals zero you're assuring that matrix of this form cannot be it's not in vertical cannot be reduced to the identity matrix.
Birne Binegar: So that zero row with the bottom it's a good check make sure that things are going okay.
Birne Binegar: Now, with now that we've got no it's not quite in row Echelon form, yet, though, reduce relational on form so it's the same thing it's an all state.
Birne Binegar: wealth and pretty clear this is going to reduce 21011001 sorry and that's going to be 000.
Birne Binegar: What you can do is you can add a multiple the second row to the first road to get rid of this entry there and then you end up with a matrix in reduce row Echelon form and so for calculate you know space we're looking for the solution set of this matrix times and.
Birne Binegar: A vector of unknowns is equal to the zero vector well here's somebody do show national on form but that could coefficient matrix.
Birne Binegar: And we can see that the equations implied or zero times X one one times X two plus zero times X three has two equals zero and so from this matrix in reduce your national on on the equations.
Birne Binegar: relevant to that are X two is equal to zero, the second row telling us X three is equal to zero third row is telling us zero is equal to zero.
Birne Binegar: And then, what we need solution vector notice here that X one just because there's no non zero entry net calm doesn't mean that the corresponding variable is immaterial rather, this is the actually the most important variable for this example because.
Birne Binegar: These two equations are equations for X two X three those are columns corresponding columns with pivots.
Birne Binegar: And so, these are the variables, you can solve for and what we find is that X two X three have to separately equals zero X one corresponds to a call about a pivot so it's a free parameter so that means when we write down a solution vector.
Birne Binegar: The second and third variables have to equal zero, but the first one is free.
Birne Binegar: Until we get X one times 100 and then here's your eigenvector it's a it's a basis vector for the solution set of the linear system with this as a coefficient matrix until what we get is the fact eigenvector corresponding to the eigenvalue one consists of a single vector.
Birne Binegar: it's.
Birne Binegar: Just the basis for the solution set.
Birne Binegar: Then you have.
Birne Binegar: A second eigenvalue zero so we'll do the same thing we'll look at an old space.
Birne Binegar: Of a minus zero times the identity matrix which is just no space.
Birne Binegar: Of the original matrix.
Birne Binegar: Which is just going to be the null space.
Birne Binegar: Of let's see.
Birne Binegar: If you look at the original matrix i'm going to do a real reduction immediately.
Birne Binegar: The third row was just a multiple exactly identical to the second row So if I take this matrix and I subtract a second row from the third i'll get zero at the bottom, I have is 110001 and then after removing this row using real reduction don't get zero zooms deal with the bottom.
Birne Binegar: Again zero row with the bottom of this no space means things are going as expected.
Birne Binegar: This is a coefficient matrix that's going to allow us to determine exactly not the eigenvector corresponding the eigenvalues zero now we have to do is write down the eigenvector.
Birne Binegar: So if we convert back to equations here we have X one plus x two plus zero has to equal zero and the first row second row is telling me that zero Plus zero plus one times X three has to equal zero.
Birne Binegar: And then last row just says zero is equal to zero, which I see that X, one has to equal minus X two and that the.
Birne Binegar: second row saying X three is equal to zero, so, in fact, for this matrix here columns one in column three contain pivots that's why I can solve for X one X three.
Birne Binegar: X two is going to be left free.
Birne Binegar: Until when I write down and solution vector it's going to look like minus X two X two zero.
Birne Binegar: equals X two times minus 110 all right, and then this is this vector is your eigenvector corresponding to the eigenvalues zero and the vector above here is the eigenvector corresponding to the eigenvalue one.
Birne Binegar: And with that with the eigenvalues determine one and zero in their corresponding eigenvector vectors determine I fall at answer Part A, not the eigenvalues or one and zero and the corresponding eigenvectors on 100 N minus 110.
Birne Binegar: All right, so that answers Part A.
Birne Binegar: Part B.
Birne Binegar: Is the question is that matrix dying alive.
Birne Binegar: When you try to die eliza matrix what you need or for enough linearly independent eigenvectors for a three by three matrix you need three linearly independent eigenvectors.
Birne Binegar: In part A, we found literally independent eigenvectors but how many did we find for the eigenvalue one.
Birne Binegar: You only have one linearly independent eigenvector and for the eigenvalues zero, we only have one linearly independent eigenvector so we found.
Birne Binegar: Two linearly independent eigenvectors, but we need three linearly independent eigenvectors through diagonal is a three by three matrix So the answer is the matrix is not dying alive.
Birne Binegar: So the answer to be no not dying and likeable.
Birne Binegar: Because we have only.
Birne Binegar: Two linearly independent.
eigenvectors.
Birne Binegar: And we need three linearly independent eigenvectors if we're going to dive in lies a three by three matrix.
Birne Binegar: Okay, any questions on that.
Birne Binegar: And um the next problem on their practice exam you're actually to diagonal is a matrix but try to keep things simple i'm just gonna have you dying like a two by two matrix.
Birne Binegar: Alright, so here, it is the first thing you need to do well, let me, let me outline this.
Birne Binegar: Guy go.
Birne Binegar: and buy anything matrix.
Birne Binegar: The main thing you need is you need.
Birne Binegar: And literally.
Birne Binegar: Independent eigenvectors.
Birne Binegar: But once you have those.
Birne Binegar: What you do is you can form a matrix see by writing the factors that's called.
Birne Binegar: And then, corresponding to see there will also be a diagonal matrix.
Birne Binegar: Where the non zero entry Tucker along this main diagonal.
Birne Binegar: So you have zero everywhere except along that main diagonal.
Birne Binegar: And the numbers that you put here if you've actually intent on diagnosing a matrix a are going to be the eigenvalues of the matrix a.
Birne Binegar: matrix see consistent the eigenvectors of the matrix a the way they're correlated is that that first diagonal entry has to be the eigenvalue corresponding to this eigenvector the second diagonal entry has to be the eigenvalue correspond to the second eigenvalue you fail.
Birne Binegar: And so yeah if you set this up correctly, then automatically.
Birne Binegar: you'll have see the minus first a C is equal to D and the diagonal ization of your matrix a is complete.
Birne Binegar: Alright, so let's do this procedure for this particular matrix.
Birne Binegar: So we start with finding the eigenvalues zero equal to the characteristic polynomial of the matrix, which is the determinant of minus one minus Lambda 513 minus Lambda.
Birne Binegar: Which is going to be minus one minus Lambda times three minus Lambda is a two by two matrix all computers determinant by taking the difference of the diagonal product minus five which is.
Birne Binegar: Look look like minus three.
Birne Binegar: crush terms or negative three plus one minus two Lambda plus Lambda squared minus five which is Lambda squared minus two Lambda minus eight which is Lambda minus four.
Birne Binegar: Lambda plus two Okay, so I work that characteristic polynomial out until I reached its factorization because once I factor is this polynomial I can resonate spot the roots of PA of Lambda is equal zero I even the Lambda equal to four or Lambda equals negative to.
Birne Binegar: The eigenvalues.
Birne Binegar: Our Lambda equal four and negative to.
Birne Binegar: Next, so we found the eigenvalues The next thing we need to do is find the eigenvectors one for each eigenvalue so we'll look first at the E4 again space that will be the no space of the matrix.
Birne Binegar: I get by subtracting the number four from the diagonal elements of a.
Birne Binegar: No space of let's see that will be negative 551.
Birne Binegar: minus one shot yeah so i'm truck subtracting for from three elegant not negative one, and then that makes reduces to.
Birne Binegar: A matrix one minus 100 okay good again I got zero with the bottom once I computed the row Echelon form of this coefficient matrix that's a good sign that things are going as planned.
Birne Binegar: Now, what I do is I convert back to equations X one minus X is equal to 00 equals zero coming from this coefficient matrix and from that I see the X, one has to equal X two, and so my solution vector is going to be the form X 211 and that's my eigenvector the.
Birne Binegar: Lambda equal four equals one one.
Birne Binegar: Right, I found for the first eigenvalue, for I found each corresponding eigenvector.
Birne Binegar: let's do the same thing for the second eigenvalue.
Birne Binegar: This will be end up being a plus two times I, which will be no space.
Birne Binegar: See.
Birne Binegar: 1515.
Birne Binegar: Which is yeah it's gotta be no space.
Birne Binegar: At the matrix one five after after i've done reduce this coefficient matrix to it reduce show Echelon form and from this, you can see that the solution vector basically has to have the first component equals minus five times the second component that's going to end up being X two times.
Birne Binegar: minus five one.
Birne Binegar: And that then tells me that the eigenvector corresponding the eigenvalue minus two is the vector minus five one.
Birne Binegar: Now to complete the problem what am I got i've got two linearly independent eigenvectors.
Birne Binegar: And i've just given a two by two matrix so I have enough linearly independent eigenvectors to construct and convertible matrix out of the eigenvectors and a diagonal matrix out of the eigenvalues what it's going to be is let's see.
Birne Binegar: For see what will use are these two eigenvectors first one was one one second one was minus five one.
Birne Binegar: And then the diagonal is the.
Birne Binegar: diagonal matrix deed, that is produced when are you see two diagonal is a is going to be well corresponding to this first eigenvector i'll use the first eigenvalue which I for.
Birne Binegar: i'll put for there and then in the second column, I have another eigenvalue minus two I put that is the eigenvalue in the second call and now i've got the answer to.
Birne Binegar: The problem nine i've got.
Birne Binegar: A diagonal matrix and.
Birne Binegar: an invisible matrix see that will convert your given matrix into a diagonal matrix this identity, what is automatically satisfied.
Birne Binegar: Alright, but just put a box around this.
Birne Binegar: And that would be the answer to problem nine.
Birne Binegar: Okay let's do.
Birne Binegar: The last problem on the exam this one.
Birne Binegar: Is a problem where we need to decompose a vector with respect to a subspace and while I have not assigned any homework on this, I did a couple examples on on on an example on Monday, another example.
Birne Binegar: On last Friday about doing these D compositions, let me remind you of the strategy that one is to follow to.
Birne Binegar: decompose a vector of respect to a subspace.
Birne Binegar: With respect to.
Birne Binegar: subspace what's nice about this problem and why I always included on the final exam is it's um it's sort of geometric in nature, what you're trying to do is you have some subspace w and you, and maybe this lives.
Birne Binegar: In a higher dimensional vector space, but you have another vector V that you kind of want to figure out what its shadow is.
Birne Binegar: As you shine a flashlight down what it shadow is on this on the subspace that would be the component of the that lives inside w and then you also have a perpendicular component, the perp and in these compositions you're trying to take a vector V and decompose it.
Birne Binegar: into two pieces one part that lives inside your given subspace and another part that lives out decided.
Birne Binegar: What crucial to doing this is i'm.
Birne Binegar: need to find.
Birne Binegar: A basis for.
Birne Binegar: Your vector space that enables.
Birne Binegar: Such a decomposition.
Birne Binegar: Where the is equal to be w.
Birne Binegar: Plus.
Birne Binegar: The perp with the w living inside your subspace and that perpendicular part definitely not living inside that stuff space.
Birne Binegar: And fact these guys are determined uniquely.
Birne Binegar: dope.
Birne Binegar: Basically, what you do is you, you want to find a new basis for our three that's something different than the standard basis and the way you create it and you start.
Birne Binegar: With.
Birne Binegar: A basis let's call it the one and i'll just do it.
Birne Binegar: With an eye towards this problem where i'm working in a three dimensional space and i'm giving given a subspace that generated by two vectors.
Birne Binegar: And what I need is a basis for that subspace and, if you look at this these two factors, not only do they generate the entire subspace.
Birne Binegar: Definition but they're also linearly independent because the first component can never be used.
Birne Binegar: If I take a linear combination that these two vectors I can never get rid of the first component completely unless I set the coefficient of.
Birne Binegar: The first factor equal to zero, otherwise i'll have a non zero number added to zero number it's going to produce a non dual number.
Birne Binegar: And simulate in this last component, the only way I can get a linear combination of these two vectors to have zero in the last component is by taking the multiplier of the second factor equal to zero.
Birne Binegar: So these two vectors will never produce the zero vector unless we take a linear combination of both coefficients or equal to zero, so in the case at hand.
Birne Binegar: The vector one, what is it 111 no not offended 110110.
Birne Binegar: and
Birne Binegar: 011.
Birne Binegar: Are linearly independent.
Birne Binegar: And they generate.
Birne Binegar: w, so there are basis.
Birne Binegar: And so.
Birne Binegar: We need to we needed to find a basis for w and, in fact, from the problem statement as soon as you observed that these two guys are linearly independent these guys are going to be a basis for w.
Birne Binegar: But what we really need is a basis for the entire vector space, not the subspace because this particular vector here 111 does not live inside this subspace.
Birne Binegar: So.
Birne Binegar: We need.
Birne Binegar: An additional direction.
Birne Binegar: To get a basis.
Birne Binegar: For our three well.
Birne Binegar: Think of think of a cell if we had a we had two vectors in the blackboard so maybe this has been one.
Birne Binegar: And this is a direction be to using just these two directions, I can get any place I want on the blackboard by taking linear combination.
Birne Binegar: But if I want to fill out a three dimensional vector space, I need a third direction and the easiest way to find that additional direction is to make it a thoughtful to the two given directions so soon as I.
Birne Binegar: find a vector that's perpendicular to both of these guys, then I have an additional direction that I can now use to find a basis for the entire three dimensional space in which these vectors live.
Birne Binegar: So that's what i'm going to do i'm going to find a vector that's perpendicular to both these guys so that then, I have three linearly independent vectors that can span the entire three dimensional space.
Birne Binegar: So that amounts to looking at.
Birne Binegar: A set w perp equal a set of vectors X inside a three dimensional space, such that.
Birne Binegar: X is perpendicular.
Birne Binegar: To every vector.
Birne Binegar: Little w that lives inside that subspace capital w.
Birne Binegar: Well, turns out that this condition.
Birne Binegar: it's the same thing as the set of vectors inside our three such that X dotted.
Birne Binegar: With that first basis factors equal to zero and X dotted with a second basis factor is also equal to zero.
Birne Binegar: Because if you perpendicular to the basis vectors for subspace you're automatically perpendicular to every other vector.
Birne Binegar: So.
Birne Binegar: If he is perpendicular to B one and B is perpendicular to the second basis vector, then the is perpendicular to anything you create by taking linear combinations of these guys.
Birne Binegar: But the set of all possible linear combinations of you to basis vectors is what you're getting is every vector and site that subspace so the way to guarantee that your perpendicular to every vector inside w is just to make sure that you perpendicular to the two basis factors for w.
Birne Binegar: Well, that condition.
Birne Binegar: can be expressed by writing your basis vectors as the rows of the matrix.
Birne Binegar: And then.
Birne Binegar: Doing this matrix product, so the rule for.
Birne Binegar: multiplying matrices together issue take the rows of the first factor and you take its dot product with the columns at the second factor.
Birne Binegar: that's going to evaluate to first base it factor dotted with that first column in the second row it's going to be basic vector.
Birne Binegar: Also double X that's computing this matrix product, but what I want to happen in the end is that i'm looking for solutions where vectors X that are perpendicular to each of the basic factors, so I have to solve a linear system with this coefficient matrix.
Birne Binegar: Alright, so let's do that.
Birne Binegar: So we need to solve.
Birne Binegar: b one.
Birne Binegar: b two.
Birne Binegar: X one or simply X X one X two X three is equal to 000 if I.
Birne Binegar: yeah so I write down that matrix.
Birne Binegar: be one written horizontally, be to written horizontally that matrix is exactly 110011.
Birne Binegar: And since i'm trying to solve a linear system what i'm one is the north node space of the coefficient matrix.
Birne Binegar: So in fact space I want it's going to be the no space.
Birne Binegar: Of this matrix 110011 which is going to be the no space well let's do the reduction it's going to be one zero minus 1011 so what i'm this matrix is not quite in reduce your actual on form, but I can get it in reduce your actual informed by subtracting the second row from the first.
Birne Binegar: Alright, so what's this no space it's the solution set of this matrix times.
Birne Binegar: X one X two X three is equal to the zero vector, and so what we need is X one minus X three is equal to zero.
Birne Binegar: X two plus x is equal to zero that reading off the equations coming from this coefficient matrix.
Birne Binegar: And now I also look at this matrix to decide which variables i'm going to regardless fixed in which variables i'm going to regard is free.
Birne Binegar: it's the variable to correspond the columns with pivot that are fixed and if the variable corresponding to that third call.
Birne Binegar: at three that's going to be regarded as the free variable well now i'll have X one is equal to X three x is equal to minus X three, and so my solution vector looks like.
Birne Binegar: X three minus X three x three or.
Birne Binegar: X three times one minus one one.
Birne Binegar: Right yeah what was the purpose of finding this guy.
Birne Binegar: it's the additional direction I need that when I supplement these two directions with this guy, then I get a basis for the entire vector space are three.
Birne Binegar: And so.
Birne Binegar: Having.
Birne Binegar: Having found a basis for w and we've now supplemented with a basis for each orthogonal compliment, we have.
Birne Binegar: Now three doctors.
Birne Binegar: Now.
Birne Binegar: 110011.
Birne Binegar: To the original two vectors that were given and we've all we've already observed that there are basis for for w to get a basis for the entire vector space, we need a third direction and what we'll do is we'll use that direction.
Birne Binegar: So that one minus one, one that is going to be.
Birne Binegar: A basis.
Birne Binegar: For the entire face of three dimensional vectors.
Birne Binegar: So what we'll do finally do is we'll.
Birne Binegar: Try to expand.
Birne Binegar: Our given vector.
Birne Binegar: express the given vector which was 111.
Birne Binegar: i'm trying to write this guy as two pieces apart that lives inside w in a part that doesn't live inside w what i've got.
Birne Binegar: Since, since I have a basis for the space of all three dimensional vectors I can write this vector as a linear combination of my three basis factors.
Birne Binegar: And then, what will happen is because this piece because B one and B to our basis for w that's going to be your component of the that lives inside the subspace.
Birne Binegar: And then, this is going to be your perpendicular component and you will have achieved the orthogonal decomposition of your given vector with respect to the subspace w so long as you can figure out exactly what the number C one C two and see three are.
Birne Binegar: Well, how do we do that well.
Birne Binegar: This condition.
Birne Binegar: Let me write it in opposite order see one.
Birne Binegar: be one.
Birne Binegar: yeah what what I have is a linear system to solve and it's basically says that see one times.
Birne Binegar: 110 plus C two times 011 plus C three times one minus one, one has to equal the given vector 111.
Birne Binegar: Now this is just.
Birne Binegar: A matrix 1100111 minus one one times the column vector C one C to C three.
Birne Binegar: So, in fact, to get the number of C one C 23 what I have to do is solve a linear system this coefficient matrix times column vector of unknowns is equal to this deck this factor.
Birne Binegar: So we default that linear system we form the augmented matrix.
Birne Binegar: 0111 minus one one we put.
Birne Binegar: The vector we want to express in terms of the basis on the last call and we produce.
Birne Binegar: Let me write down the answer.
Birne Binegar: haven't.
Birne Binegar: Quite figured that out yet, but let's see the reduction is pretty simple here.
Birne Binegar: Have a wiki alright so.
Birne Binegar: Plenty time.
Birne Binegar: To reduce and matrix or subtract the second row from the first 1011 and then i'll get 01 minus zero which one negative one minus one is negative two and one minus one is zero and then here.
Birne Binegar: I have 0111.
Birne Binegar: Do more and more row reduction step.
Birne Binegar: And that will be I need to clear up this entry below the pivot there until will have.
Birne Binegar: 101101.
Birne Binegar: minus two zero and then, if I subtract the third row from the second row i'll get 00.
Birne Binegar: minus three.
Birne Binegar: minus one I guess.
Birne Binegar: start now just one.
Birne Binegar: It looks like.
Birne Binegar: i'm just gonna.
Birne Binegar: give you the answer to this one.
Birne Binegar: continue to reduce what you'll end up with is the matrix 100.
Birne Binegar: To 30 010 230 0011 third which is reduced row Echelon form.
Birne Binegar: Of the linear system, the augmented matrix correspond to the linear system, we were trying to solve and so we're going to have a unique solution X one is equal to two thirds X equal to two thirds and X three is equal to one third, and so.
Birne Binegar: X one or like call it see one equal to two thirds.
Birne Binegar: See to equal to two thirds see three equals one third is unique solution.
Birne Binegar: What do I needed those numbers for because I had a vector 111 that I was trying to express in terms of a particular basis, and what I wanted was these coefficients well i've just determined this coefficients.
Birne Binegar: i've interpreted this equation as a linear system i've written down the corresponding augmented matrix and i've reduced to reduce your form, and now I discovered the solution, and so what I know now is that the.
Birne Binegar: is equal to see one which which to third time's the first basis factor, which was 110 plus C two which worked out to be to third time's the second basis factor.
Birne Binegar: And then we have C three, which was one third, and the corresponding basis specter there was one minus one one.
Birne Binegar: All right, last step here is to.
Birne Binegar: take these first two terms they're a linear combination of the two basis vectors for w So when I add these two pieces together what I get is.
Birne Binegar: A vector and w.
Birne Binegar: i'll write it down explicitly it's going to be.
Birne Binegar: Two thirds.
Birne Binegar: For third.
Birne Binegar: Two third right two thirds plus zero which two thirds.
Birne Binegar: Two thirds plus two thirds is four thirds and zero plus two thirds two thirds.
Birne Binegar: And then we have this piece, which is going to be one third minus one third one third.
Birne Binegar: Then this is your VW.
Birne Binegar: And this factor here should be purpose and that's that's the decomposition.
Birne Binegar: Okay, so.
Birne Binegar: These problems are the composition what they, in the end, involve you is calculating some no spaces and but it.
Birne Binegar: yeah not study.
Birne Binegar: Not not only the example I gave in on Mondays lecture but also the this practical exam as well there'll be a problem, similar to this.
Birne Binegar: On the final exam just remember that the first thing you've got to do is find a basis for the complimentary subspace w, then you put.
Birne Binegar: Then you combine your to your basis for your subspace with your basis for the orthogonal subspace you combine those into a single basis for our three and then you can decompose any vector inside or three with respect to the a joint basis and I simply.
Birne Binegar: Expanding the vector in terms of the basis and then splitting that vector into two pieces apart that lives it's constructed just using the basis vectors for w and another piece that constructed using only the basis vectors.
Birne Binegar: Are the part of our three that lives outside of w that's how you achieve the decomposition.
Birne Binegar: Okay, so.
Birne Binegar: I have managed to cover the material.
Birne Binegar: From the part of the course that occurred after the second exam.
Birne Binegar: Let me do one last problem that we didn't cover on Wednesday, and that was his.
Birne Binegar: problem involving and linear transformation and i'm a hero i'll just.
Birne Binegar: i'll show you my solutions and how simple it is solve.
Birne Binegar: If you have.
Birne Binegar: A linear transformation.
Birne Binegar: Then, the first thing you'll need to do to find.
Birne Binegar: The matrix that represents, it is you'll have to look at how he acts on each of the standard basis vectors that just going to be a computation where you plug in.
Birne Binegar: For the component X one is equal to one X two is equal to zero X equals three you just plug into the formula for the result and you get.
Birne Binegar: One vector for each standard basis factor of your domain, the next step is to take these guys and don't write them, as I have here.
Birne Binegar: Here i've written them as horizontal vectors you need to write them as columns to get the right matrix to take your answers here right them is called, and that will give you the matrix that corresponds to your linear transformation.
Birne Binegar: That matrix.
Birne Binegar: My construction has the property that he, acting on a vector X it's the same thing.
Birne Binegar: As doing matrix multiplication by that matrix.
Birne Binegar: and
Birne Binegar: What that allows us to do is study linear transformations functions between two factor faces I studying instead they're associated matrices.
Birne Binegar: And back you need to do that, to answer Part B you're asked for a basis for the Colonel T.
Birne Binegar: The Colonel key is a salute as a set of vectors.
Birne Binegar: X inside that domain, so that starts off with three dimensional vectors such a ti of X.
Birne Binegar: is equal to the zero vector, but this is going to be the same thing.
Birne Binegar: As the north face.
Birne Binegar: Of the associate matrix because when you take a matrix and you multiply and unknown vector and you get the zero vector well that solution set it's exactly the north face of the matrix.
Birne Binegar: So you'll have to.
Birne Binegar: In this example we're gonna have to find the no space of this matrix.
Birne Binegar: So that would be the north base of the matrix.
Birne Binegar: What is it 101 minus one and one zero minus one.
Birne Binegar: To find that no space we reduce.
Birne Binegar: And let's see we'll get, I guess, all I have to do is is flip the two rows one zero minus 101 minus one, and from that, I can see that my solutions are gonna have to satisfy X one equals X three from the second row X two is going to have to equal X three, and that means my solution vector.
Birne Binegar: X has to be at the form.
Birne Binegar: X three times 111.
Birne Binegar: And so that's calculated here and so.
Birne Binegar: The kernel of tea is the north face of this matrix we've just seen it's generated by a single vector and so, in fact, the kernel of tea is going to be.
Birne Binegar: subspace of our three generated by that single vector and we're done.
Birne Binegar: Okay, so i've.
Birne Binegar: gone through all the problems on the practically examiner any questions.
Birne Binegar: If not well i'll let you guys go.
Birne Binegar: So again, the final exam is.
Birne Binegar: Monday at 10 o'clock in this room so i'll see you then good luck studying for my exam and also your other exams over the weekend, and I will see you on Monday.