Birne Binegar: Okay, I guess we'll start so.
Birne Binegar: plan for days to work away partially through a sample exam there's 10 problems on this practice exam the probability be.
Birne Binegar: 10 problems also on the final exam but i'm going to work through the slowly i'll probably get through the material that was covered on the first two exams today and then i'll save for Friday.
Birne Binegar: Questions about material we've covered since elastic damn which means determinants I conductors and eigenvalues diarization and orthogonal D compositions those topics i'll cover on Friday, as they appear on the final exam.
Birne Binegar: Okay So the first thing I want to stress is just like on the second exam I guarantee that they're going to be.
Birne Binegar: For problem on the essential definitions for this course, the notion of what a sub spaces, what a basis set of linear independent factors is and what a linear transformation is and what i'm expecting on the final exam is to you to give.
Birne Binegar: Correct definitions verbatim as best you can so.
Birne Binegar: Let me work out the.
Birne Binegar: Each of these problems, so a subspace as the name indicate this should be something like a small vector space and, indeed, what it is yes subspace of a vector space D is.
Birne Binegar: What it's a subset.
Birne Binegar: of your vector space so let's call that subspace w.
Birne Binegar: The such that when you have two conditions.
Birne Binegar: related to the fundamental operations of linear algebra first of all respect the scale of multiplication So if you have a number.
Birne Binegar: And you have a vector that starts off living in that subspace then necessarily what's going to happen is that scale or multiple of your vector by Lambda has.
Birne Binegar: continued to reside within that vector space within that subset and, second, of all you need if you have two vectors that start off.
Birne Binegar: living inside.
Birne Binegar: Your subset what you require is that their vector some.
Birne Binegar: stays inside that subset.
Birne Binegar: Okay, so there's one definition.
Birne Binegar: Second definition, going to be here on the final jam, is what is the basis and.
Birne Binegar: The idea of the basis that this is going to be the thing that allows us to coordinate I sub spaces and, more generally, vector spaces, what the definition.
Birne Binegar: doesn't really include that intuitive or formal idea so much as the tests that you need to make sure that you have a set of vectors that will actually do what you want, provide you with good coordinates what we say, is is a basis for a subset is a set of vectors.
Birne Binegar: let's say he calls.
Birne Binegar: The one through bk.
Birne Binegar: Such that.
Birne Binegar: First of all, you need to be able to reach every vector in inside your subspace by taking linear combinations of this list of vectors and so what you're going to one.
Birne Binegar: Is that.
Birne Binegar: For all vectors that live inside your subspace.
Birne Binegar: w is going to be capable of being represented as a linear combination of these vectors so see one time be one plus see to do, etc, until we reach the last basis vector with these coefficients see one two ck.
Birne Binegar: unique.
Birne Binegar: Well that's actually two conditions, the first of all, you have to be able to reach every vector inside your subspace just by.
Birne Binegar: Using formal expressions of this form, but you also have to require that the coefficients are unique this the uniqueness of each coefficients that allow us to use the number C 130 K is coordinates for a vector w that lives inside the subspace capital w.
alright.
Birne Binegar: A third definition, what is the set of linearly independent vectors well actually that's one of the requirements we would impose on a.
Birne Binegar: basis that all the factors have to be literally independent, but these definitions are related to test we can perform to make sure the objects will do what we want and linear independence here it's a simple test to make sure.
Birne Binegar: That, in fact, if you have a set of linearly independent vectors that you'll be able to uniquely right well, you have a set of vectors that.
Birne Binegar: that are capable of reaching every director and D and they're linearly dependent, then this condition will be automatic this uniqueness of these expansions will be automatic, but the formal definition is a set of vectors is let's call that set.
Birne Binegar: The one through bk.
Birne Binegar: A set.
Birne Binegar: Of vectors do you want to decay, is going to be linearly independent if.
Birne Binegar: And only if.
Birne Binegar: The only solution.
Birne Binegar: here's the equation, we want say X one B one plus dot dot plus x K, the K equals zero yes, all these coefficients have to separately banish.
Birne Binegar: X K is equal to zero right so here you're to regard these coefficients is unknown, this is the quick into the equation for the.
Birne Binegar: Numbers X one X Okay, the vector some that used to generate this equation, there are linearly independent if you only have one solution of that and that's the solution, where all the coefficients equal to zero now.
Birne Binegar: You have a basis and it's on that and the basis fact you're satisfied condition like this, then this record this is going to be automatic but.
Birne Binegar: These are two separate definitions, in fact, the textbook.
Birne Binegar: uses a different definition of basis the textbook says that a basis is a set of vectors so should every vector can be written this way and.
Birne Binegar: The vectors are literally independent, then this follows as a as a theorem that if you have a set of linearly independent vectors that such at every director and your subspace can be reached via such a linear combination, then these guys are are unique but i'm sticking to.
Birne Binegar: This this definition and this consequence of or or this notion of linear independence is being a separate idea.
Birne Binegar: And finally, we have.
Birne Binegar: A notion of a linear transformation.
Birne Binegar: And this is about functions between two vector space, so a linear transformation between two vectors basis BMW is a function.
Birne Binegar: or rule for taking elements of the element of w right it T the.
Birne Binegar: going to w such that well again since we're in linear algebra what we worry about is how scale multiplication and vector addition behave and, in fact, what would require is that when we apply the function to a scale or multiple.
Birne Binegar: Of a vector inside the domain, let me call it the then what should happen is that I could effectively delay that scale or multiplication until after I apply the function, so this should be true for.
Birne Binegar: All numbers Lambda and all vectors be that live inside that domain capital D and similarly for vector edition you want T D one.
Birne Binegar: Sorry arrow there, plus the second vector V2 should be equal to T that's the one plus T at the two, and that would be for all vectors V1 V2 in the domain of that function.
Birne Binegar: Okay, so.
Birne Binegar: memorize these definitions and be prepared to write them down verbatim and then you'll get what 20 points automatic.
Lisa Fisher: final exam.
Birne Binegar: Okay.
Birne Binegar: Next problem problem to on this practice exam is you probably see something like this again on the final exam I like this sort of problem because it doesn't involve any calculations per se, but you have to sort of understand the calculation that underlying the problem.
Is.
Birne Binegar: State, the answer so here what you're supposed to do is interpret each of these major sees as the augmented matrix of them have a linear system, but not the original augmented matrix but one where the augmented matrix has been reduced to.
Birne Binegar: Look, like everything every each of these major seasons and row Echelon form and, in fact, once you.
Birne Binegar: start with a linear system.
Birne Binegar: linear system, a X.
Birne Binegar: is equal to be then of course the way we solve it is we construct the augmented matrix and rewrote reduce.
Birne Binegar: And typically this procedure first you go to.
Birne Binegar: A row Echelon form.
Birne Binegar: And then.
Birne Binegar: This gets the pivots and.
Birne Binegar: This orders, the pivots of this augmented matrix correctly, but there's some more, we have to do in order to actually find a solution of this after you once you get it in relational informed and you go to reduce relational on board.
Birne Binegar: And then from this you get the equations.
Birne Binegar: and eventually you get the hyper planes solution.
Birne Binegar: Okay now this problem to.
Birne Binegar: Only addresses like the first half of the typical calculation here i'm only on this problem to.
Birne Binegar: What i'm imagining is I started off with a linear system, I broke down the augmented matrix and I reduced to halfway i've reduced to us a row Echelon born.
Birne Binegar: And I what i'm asking you in this problem is to look at this thing and tell me what the answer should look like, as well as where what kind of linear system this matrix and reduce all epsilon form came from.
Birne Binegar: Alright, so let's look at these guys, so this guy.
Birne Binegar: The coefficient part the augmented matrix as four rows and four columns so the original system had to come from the system with for equations.
Birne Binegar: and
For unknowns.
Birne Binegar: simulate this one it's also you look at the coefficient matrix there's four columns so there's for unknown and there's four rows so that also came from, or equations.
Birne Binegar: And for unknowns.
Birne Binegar: And the last one look like 12345 columns but for rose to the number of rows it's the number of equations.
Birne Binegar: And 500.
Birne Binegar: Okay, so you should be able to look at an augmented matrix and figure out what kind of linear system, it came from.
Birne Binegar: that's the first half of this and then the second half, will determine if they're if they're solution, and if there are solutions, how many free parameters, we occur.
Birne Binegar: Okay, so in preparation for that, let me underlying the pivots that occur in these.
Birne Binegar: Real Echelon forms of augmented matrix these.
Birne Binegar: here's the here's the guy that.
Birne Binegar: it's going to cause us problems if we actually proceed further and try to actually solve the linear system, because this third row.
Birne Binegar: It corresponds to a particular equation, it says that zero times X one plus zero times X two plus zero times X three plus zero times X is equal to one but that that third row, but then say that zero is equal to one which is a contradiction.
Birne Binegar: And so, in that second problem there's no solution because.
Birne Binegar: Amongst these four equations there's one that's definitely contradictory.
Birne Binegar: Alright, so that actually answers the question for Part B, but for Part A and C we don't have that issue, where we have a pivot in the last call, and so we off, we will have solution.
Birne Binegar: This one is also going to have some shins.
Birne Binegar: But.
Birne Binegar: to figure out the number of free variables, we have in our solution, what we do is we look at.
Birne Binegar: The matrix augmented matrix and row excellent form and count the number of columns without pivots on this first one you'll see that X three x four because they call correspond the columns out pivots these are going to be free and in this one.
Birne Binegar: Even though this all zeros in this column don't forget about the associated variable X one is going to be a free variable X two X three.
Birne Binegar: X two X three x four correspond the columns with pivot so they're not going to be three, but this last call this call doesn't have the pivot, so it is going to be free and similarly here so you're going to have this going to have a solution with two free parameters.
Birne Binegar: which are going to be X three x four and this one is going to be for equation to buy the note, and it has a solution with two free parameters.
Birne Binegar: X one X five.
Birne Binegar: And so that's the solution to the second problem.
Birne Binegar: Alright, so any questions so far.
Birne Binegar: All right now.
Birne Binegar: Next problem.
Birne Binegar: it's a problem when we actually solve a linear system.
Birne Binegar: So here, I have a system with look like three equations and three unknowns and what we're to do is we're solvable in your system expressing a solution said it's a hyper plane so i'm going to follow this procedure here first thing we do is we write down the augmented matrix.
Birne Binegar: So that amounts to reading off the coefficients and the numbers on the right hand side we're going to have coefficient of X one in the equation is zero then minus one to line zero second equation is to.
Birne Binegar: minus three zero there's no extreme second equation to and then one minus one minus one one all right that's our augmented matrix.
Birne Binegar: That we start off with you do directly from the original equations now we'll do the row reduction first row Echelon form.
Birne Binegar: So maybe the first thing to do is get a pivot up in this upper corner, this is the most convenient one putting a using one for a pivot it's always nice well let's interchange.
Birne Binegar: rose one in three and we'll get one minus one minus one and one.
Birne Binegar: second row is not changing to minus 302 and then zero minus 120 okay well.
Birne Binegar: I didn't have a pivot up in this corner, and this is what we're going to start So the first thing I did is make sure I got a pivot up there.
Birne Binegar: Now i'll use this pivot to clear out entries below it so i'll send row to row two minus two times the first row.
Birne Binegar: And that's going to produce one minus one minus one, one that broke not changing last rose not changing either to zero and let's see so.
Birne Binegar: Two minus two times, one is 03 minus two times negative one is going to be minus one zero minus two times negative one is going to two and then to minus two is going to equal zero all right, we got that and now, when we see.
Birne Binegar: We do one more row operation send row 303 minus two i'm going to get an augmented matrix now in a row Echelon four.
Birne Binegar: zero.
Birne Binegar: minus 1200000 so that's.
Birne Binegar: The role Echelon form of the original augmented matrix but we've only reached this stage of this procedure for completely solving the system.
Birne Binegar: what's wrong and what keeps us from being you know, reduce rush one one is first of all.
Birne Binegar: The pivots have to be equal to one so i'm going to have to reach scale this first row, the second row and multiply it by minus one that get that pivot equal to one.
Birne Binegar: And then I also have an entry sitting above pivot in the second row so i'll do two things first of all i'll send them row to the minus row two and i'll also send row one two row one.
Birne Binegar: minus.
Birne Binegar: The original row to there so that's correct.
Birne Binegar: This one if I subtract the second row from the first row i'm going to get one zero.
Birne Binegar: Negative negative three and then, once and then the second row, the only thing i'm doing here is multiplying by minus one.
Birne Binegar: Negative two zero in third row is not changing.
Birne Binegar: All right now i've reached reduce relational and.
Birne Binegar: I reached this stage of the process, and so the next stage is going to be to convert back to equations the first row is telling us that X one.
Birne Binegar: minus X three is equal to one second row is telling us that X two minus two X three sorry minus three x three minus two X three is equal to zero, and the last row doesn't tell us anything, it just says zero is equal to zero, but what I do next is I.
Birne Binegar: realized that this third column, because it doesn't contain a pivot it's going to be free it's that's going to be a free variable.
Birne Binegar: So I take these equations and I moved the three variables, to the right hand side, then I get X one is one plus three x three x two has to equal to X Ray.
Birne Binegar: Right now i've got the equations that simple, as I can write them now i'm going to write a corresponding hyperplasia solution.
Birne Binegar: So I try to write a typical vector X one X two X three but i'll use these two equations The sub to where I can for the variables that correspond the columns with it and that's going to lead to X one is equal to one plus three x three x two is two X three and last.
Birne Binegar: component is just X Ray and now the last step is to take this vector and expand it in terms of the three variables, so we have constant terms.
Birne Binegar: The only constant vector term here is this one there and then I have terms that are proportional X Ray and that's going to be three in the first component to in the second one last component and then this is the answer.
Birne Binegar: What i've done is i've displayed it a solution vector as living on a hyper plane.
Birne Binegar: file in fact it's a line it's a one dimensional hyperplasia your starting point and then you're allowed to depart from that starting point and arbitrary distance control by that factor in this particular direction so that's a one dimensional.
Birne Binegar: hyper planes.
Birne Binegar: were in general a hyper plane is a set of the form some starting point and then you're allowed to depart from the starting point and some fixed set of directions.
Birne Binegar: s one s K range in our and this this expression does assets for me just that we only have.
Birne Binegar: We have a particular constant vector here, and then we have some arbitrary function, this is a pre some arbitrary parameter, this is a pre parameter times a particular vector and that that's your corresponds to the one.
Birne Binegar: Okay, so.
Birne Binegar: that's really a fundamental calculation and the first exam was really building up so that this particular calculation that we effectively do over and over again in this course.
Birne Binegar: You can be able to do a simple problem like this to completion.
Birne Binegar: What i'll try to do as I did it in this Sam practice exam is once you get to the augmented matrix or try to make the robot actually proceed pretty quickly, but be prepared to perhaps interchange rows or deal with variables, for which.
Birne Binegar: The corresponding column contains all zeros etc.
Birne Binegar: You know the algorithm, then this is pretty straightforward and so just just follow the algorithm.
Birne Binegar: Alright, so.
Birne Binegar: The next problem.
Birne Binegar: Another application of reduction it's where we use row reduction to compute the inverse of the matrix.
Birne Binegar: Alright, so.
Birne Binegar: Here I give you a three by three matrix the way you find an inverse.
Birne Binegar: Do ambitious, you can try you doing your co factor expansion competing to co factor matrix and use the code factor, but really for three by three matrix the fastest way to compute it inverse is to just reduce.
Birne Binegar: The matrix I perform get by a joining to the given matrix the identity matrix and the idea is to take this to reuse row Echelon form and what will happen is that either.
Birne Binegar: The left hand side row reduces to the idea, actually matrix and when that happens, whatever happens to the identity matrix side that becomes the inverse or.
Birne Binegar: No inverse exists.
Birne Binegar: And more precisely, if the left hand side that's not real reduced to the identity matrix then no inverse is going to exist alright so let's do this row reduction.
Birne Binegar: I think that nicest thing to do is immediately.
Birne Binegar: Put a one up in this corner so i'll just interchange.
Birne Binegar: let's see my flip i'll have.
Birne Binegar: 010100001 and then.
Birne Binegar: 210 and minus 201 so all I did here is I interchange row one and two, because life is much simpler when you're pivots or equal to one makes it easier.
Birne Binegar: For example, then we knew these pivots and the lower rose because I just have to multiply the first row by negative to to get this.
Birne Binegar: In I combine row one in ritual to get that term drop out and if I multiply this robot to I can cancel out that guy and so that's what we'll do next will send row to to row two minus two row one and we'll say in a row three two row three plus two row one.
Birne Binegar: Then row one doesn't change.
Birne Binegar: But.
Birne Binegar: The second row is we're going to have to minus 201 minus two times zero which one zero minus 00 then i'll have one minus two times zero is one zero minus two times, one is negative two and zero minus 00 similarly.
Birne Binegar: Two plus negative two is 002 times zero plus one is one and then.
Birne Binegar: zero Plus 002 plus zero is to and zero plus one two times one zero it's going to be one all right good.
Birne Binegar: we've reached this form notice that the i'm.
Birne Binegar: On the right hand side.
Birne Binegar: We have the identity matrix that means sorry left hand side, you had the intention, he That means the right hand side is the inverse of a and so your answer is a the minus first 0101 minus 20021.
Birne Binegar: Okay.
Birne Binegar: Okay, so.
Birne Binegar: Here I want you to test to see whether or not a given subset of our three is actually a sub spit so what am I looking for well let's go back to.
Birne Binegar: The first problem or and remember the other sub spaces of set of vectors that closed under both scale or multiplication and vector addition, so what i'm going to have to do in this, this problem is check to see if these conditions are satisfied.
Birne Binegar: So.
Birne Binegar: here's the set.
Birne Binegar: w if the site of all vectors X, Y Z in a three dimensional vector space, such that there are some other components.
Birne Binegar: is equal to one.
Birne Binegar: So let's let.
Birne Binegar: let's like w equal the vector 100 because one plus zero Plus zero sum of these three components does equal one, that means that vector I wrote down here does live inside that subspace let's choose.
Birne Binegar: To check scale or multiplication let's just choose Lambda equals zero then Lambda times w is going to equal well it's going to equal 000 because each component of w is being multiplied by zero, but now, this sector.
Birne Binegar: We add up its component we get zero and not not the same as the condition, we need to live inside the set capital w.
Birne Binegar: And so.
Birne Binegar: This means that Lambda times w.
Birne Binegar: does not satisfy that condition required on the elements of the set capital w and so it's not closed.
Birne Binegar: Under scale and multiplication.
Birne Binegar: And for that reason it's not a subspace.
Birne Binegar: The show this much you're done with a problem but alternatively what you could have done is instead of looking at scale or multiplication you could have looked at vector addition.
And so.
Birne Binegar: Another way of solving the problem is this take two vectors say 100 and another vector.
Birne Binegar: 010 both of these guys live inside that set capital w because the sum of their components it's going to equal one, but when I add them together.
Birne Binegar: i'm going to get a vector with two non zero component.
Birne Binegar: And four which one plus one plus zero equals two, which is not equal to one, and so the some of the components of the summer w one w two is to not one, so this vector this combination that not.
Birne Binegar: does not lie in our subset.
Birne Binegar: Somewhat is not an element.
Birne Binegar: Of the subspace until the subset is not closed under vector edition, and so it's not a subspace.
Birne Binegar: Where it is well that's actually harder to prove because then, what you have to do is take somehow.
Birne Binegar: Take an arbitrary vector and make sure that no matter what number you multiply that vector as long as it starts off in that set set, then you have to.
Birne Binegar: write a sort of generic element of of the set and then make sure that that do scale or multiplication at it, what happened to that generic elements stays inside the set.
Birne Binegar: Or you have to generic elements of your set and you add them together they not say that's usually harder to prove because, instead of having to.
Birne Binegar: display a single counterexample you have to show abstractly that no matter how you choose Lambda, no matter how you choose w that Lambda times w is in the set and I typically don't do that on the final exam because it's closer to up.
Birne Binegar: Closer to approve when all I really are expecting you to know is that the set has to be closed, understand and multiplication and vector addition to look for contradictions that that that's easy that's easier to spot and.
Birne Binegar: What vector to choose, so that you get a contradiction that you can usually figure that out and that so that's typically the way I I asked the questions about whether it given subset is the subspace.
Birne Binegar: Okay.
Birne Binegar: So.
Birne Binegar: So so i'm sorry for.
Birne Binegar: For those who you're watching online i'm working on the on the whiteboard up quite a bit, if you want to see the solutions they're already posted on on canvas and so.
Birne Binegar: You can follow my discussion what's going on in a lecture here by looking at the solutions that are already posted on canvas the right at the top of our canvas page.
Birne Binegar: Okay, so.
Birne Binegar: let's move on.
Birne Binegar: Maybe what I can do that will make life easier for people who are watching online work directly on the paper here.
Birne Binegar: So, taking care of this one, this one.
Birne Binegar: In problem six what you're given are a set of four vectors inside a four dimensional vector space and you're asked determine if they're literally independent now.
Birne Binegar: The definition here.
Birne Binegar: i'm not expecting you to fly the definition directly, that is to make sure that a linear combination of these four vectors.
Birne Binegar: can only equals zero when all the coefficients equal to zero, rather, what I want you to do in this case is just think of the algorithm by which we determine if vectors are linearly dependent what we do is, we take the vectors and we write them as the rows of a matrix.
Birne Binegar: So if we set.
Birne Binegar: A equal to the matrix.
Birne Binegar: Sorry.
Birne Binegar: i'll set a equal to the matrix.
Birne Binegar: or use these four factors is rose one minus two to one.
Birne Binegar: The second row will have one minus 131 third row will have 0110 in the fourth row, we have two minus three, five and two.
Birne Binegar: Alright, so the point of this is that the the row space of this matrix.
Birne Binegar: A is exactly the span of the original vectors.
Birne Binegar: And so.
Birne Binegar: The question can be answered by looking at this matrix and figuring out what the dimension of that role space, it is.
Birne Binegar: If the dimension is for that will mean that each of the original rows with linearly and but if.
Birne Binegar: The we produce this matrix and we get a zero row with the bottom, that means we're not gonna have enough linearly independent rose to have a four dimensional, in fact, the number of non zero rows.
Birne Binegar: Of the row Echelon form of this matrix gives us precisely dimension of this row space, which is the same as the span of these original vectors.
Birne Binegar: So what i'll do is we'll simply reduce that matrix let me not do the calculation i've already done one row reduction so here i'll just give you the answer.
Birne Binegar: This matrix one to one etc, it reduces to.
Birne Binegar: A matrix of the form.
Birne Binegar: 1041011000000000.
Birne Binegar: All right now yeah.
Birne Binegar: So it's back these two vectors generate the entire row space, and so we generate the entire.
Birne Binegar: subset of our for that generated by these or vectors using linear combination.
Birne Binegar: Now, what we see is the entire face is generated by these two vectors and they're obviously linear independence, because I, if I add.
Birne Binegar: A multiple of this row to a multiple this row i'll never get the zero vector unless the first component is zero or first.
Birne Binegar: scale or multiple zero, otherwise we won't be able to get rid of that contribution nor raw unless the second scale or factors equal to zero, I won't be able to get rid of the contribution from.
Birne Binegar: The second slot So these are two independent vectors generate the same to linearly independent directors to generate the same subspace these are going to be basis vectors and they're just two of them until.
Birne Binegar: Amongst these four back doors there's only two linearly independent vectors and so.
Birne Binegar: they're not the original set.
Birne Binegar: Not linearly independent.
Birne Binegar: Let me state that here.
Birne Binegar: original set of vectors.
Birne Binegar: Is not.
Birne Binegar: linear independence and the way you spot, that is, you see these zeros It means that.
Birne Binegar: There were some way of adding these vectors together to produce zero row, so there were some linear combination of the original four set of vectors that could be used to create.
Birne Binegar: A linear combination.
Birne Binegar: That sums up to the 00 row So the answer is not linearly independent that answers Part A, and then the dimension, the subspace well here are your basis factors and so just going to.
Birne Binegar: You have to non zero vectors that's going to be a basis for the subspace it's generated by the original set of vectors.
Birne Binegar: Okay yeah.
Birne Binegar: yeah well what you.
Birne Binegar: Not necessarily the identity matrix but what you want to have is.
Birne Binegar: Better to say No zero row with the bottom.
Birne Binegar: Because I mean.
Birne Binegar: This but if I only had two a two rows, and this is the matrix it's not the identity matrix, but these two back these two rows are linearly dependent.
Birne Binegar: So, the main thing is to avoid these zeros if the zeros occur, that means you're not linearly independent.
Birne Binegar: And then, when you want to know the bait the dimension of the space that you generate like taking all possible linear combinations of the given list of vectors well that's going to be just to because.
Birne Binegar: These two vectors will be a basis for the subspace generated by those four factors.
Birne Binegar: Okay.
Birne Binegar: And so.
Birne Binegar: let's see what read okay i'm going to do one last problem today.
Birne Binegar: Because this and then i'll i'm more or less that covered the material that was already.
Birne Binegar: appeared on the first two exams well and i'll say for Friday discussion of determinants and calculating eigenvectors and eigenvalues.
Birne Binegar: let's just work out this last problem or you're given a function between two vector spaces and you were asked.
Birne Binegar: In fact I tell you that it's the linear transformation, so you don't have to check that all you have to do is construct the matrix that corresponds to this linear transformation and then find a basis for the Colonel the linear transformation.
Birne Binegar: So you start.
Birne Binegar: By forming the matrix that you can attach to such a function and.
Birne Binegar: What you do is you apply your transformation to each of the standard basis vectors of the domain, so the domain is our three so we have.
Birne Binegar: One standard basis vector i'm going to write it vertically.
Birne Binegar: Upon computation will have a second standard basis vector 010 and i'll compute what T does this vector and write the answer vertically.
Birne Binegar: And we have P applied to the last standard basis vector.
Birne Binegar: Right, the answer vertically okay upon computation what's going to happen, so the first component is going to be the difference between the last two components of the input vector and so i'll have.
Birne Binegar: zero there in the first slot in the second slot I have X one minus X three or one minute zero is one.
Birne Binegar: Alright, so i'm computing these things be careful because to get the right matrix.
Birne Binegar: Here it looks like the output it's going to be a horizontal vector remember the right the the output as a vertical vector.
Birne Binegar: And as long as you do, that this is pretty simple if we evaluate.
Birne Binegar: T on the second.
Birne Binegar: database of sector what we get is.
Birne Binegar: One one minus zero and zero minus zero, and then the third slot that's where the only non zero component effects three will get minus one from there and minus one from there.
Birne Binegar: Alright, so there's the matrix it's not quite in.
Birne Binegar: Were usual Echelon form and we're going to need that reduce rational on form the answer Part B so let's.
Birne Binegar: Note that the reduce row Echelon form of the matrix that we can attach to this linear transformation is one zero minus 101 minus one right all I did all I had to do to get that and reduce your actual on porn is interchange the first two rows.
Birne Binegar: Alright, so now we can answer Part B we've got.
Birne Binegar: A question coefficient matrix matrix and we've got it reduce your actual on form if we write down equations we're going to have X one minus X is equal to zero second row says that X two minus X three is equal to zero, or this simply says that X one is equal to X three.
Birne Binegar: X is equal to X three x three here reasonable i'm putting X Ray on the right hand side is because that's the variable that corresponds to a column about a pivot it's the free variable and the solution.
Birne Binegar: source solution factors are going to be able to form X three x one X three for X two X three for itself, or simply X three times 111 all right and there's your.
Birne Binegar: Every solution of t of X is equal to zero it's going to be a vector of the form X three times the specter 111.
Birne Binegar: And so that's your basis director.
Birne Binegar: For the Colonel.
Birne Binegar: Okay.
Birne Binegar: All right, so I think what i'll do now is just stop here, I only have two minutes left anyway yeah.
Birne Binegar: Ah that's right yeah yeah.
Birne Binegar: yeah if you have.
Birne Binegar: A column that's all zeros that doesn't mean we can ignore that column, it means that that the variable that corresponds to that column the zeros always going to be a free parameter, and you always have a term X three times.
Birne Binegar: The third only third component yeah there's no basis vector.
Birne Binegar: Okay okay so we'll continue the review on Friday what i'll do on Friday is i'll go over the last three problems on the exam which are not based on material we've covered since the second exam.