Birne Binegar: Okay, so.
Birne Binegar: today's review session for the final exam.
Birne Binegar: Real quickly here.
Birne Binegar: Let me, let me check something real quick.
Birne Binegar: Was that the chakras I wanted to give you the precise date of the exam so the the final.
Birne Binegar: it's going to be Monday.
Birne Binegar: At 10am.
Birne Binegar: In this room so Ms.
Birne Binegar: 445.
Birne Binegar: And what i'll do today is.
Birne Binegar: go over practices damage very really close to the actual exam what probably managed to do today is just get through the problem.
Birne Binegar: They cover that portion of the course it's already been covered, on the first two exams, but the calculation done there, of course, important, in that last end of the course where we looked at determining factors.
Birne Binegar: And orthogonal decomposition.
Birne Binegar: So it's for free, do we carefully material already covered again anyway.
Birne Binegar: One thing I want to stress so that's.
Birne Binegar: Just like on the second exam I told you that we're gonna definitions.
Birne Binegar: damn and you should say that, precisely so to on the final exam it's going to be the same for definitions, because they're effectively, the most important ideas of the course the concept of the subspace of basis linear independence and what a linear transformation is so let me step through.
Birne Binegar: These definitions.
Birne Binegar: To prepare for the final exam memorize the definition such I am about to write them now and just be able to restate these on the final exam.
Birne Binegar: Okay, so the first definition is what is the subspace.
Birne Binegar: we're looking ever since the first exam we've constantly been counting.
Birne Binegar: Talking about subspace is either at the solution set of a homogeneous system, the span of a set of vectors maybe the kernel of a linear transformation, there are many ways at some spaces arrived, the basic definition is that a subspace of extra space D is first of all it's a subset.
Birne Binegar: let's call it w up the space where all your vectors live such that.
Birne Binegar: Well, the idea.
Birne Binegar: Throughout linear algebra is to make sure everything behaves wealth respect the scale or multiplication and vector addition.
Birne Binegar: The context of a subspace what we want is anytime you have a number, so a real number Lambda and a vector w arrow vert indicating it vector let's put it inside that subset.
Birne Binegar: What we want is.
Birne Binegar: For any number any vector that lives in the scientists upset we want the scale and multiple that vector by Lambda to stay inside that subset.
Birne Binegar: So that's closure on their scale or multiplication and, similarly, if we start off with two vectors.
Birne Binegar: That beginning w.
Birne Binegar: Once we add them together.
Birne Binegar: as vectors will get a new vector and what we want require of it is it still inside that subs that subset so.
Birne Binegar: subspace of the vector space be is a subset w such that whenever Lambda is in our and w's in capital w and the scale of multiple little w.
Birne Binegar: land is stays inside the set and if I add two vectors inside this It started off inside the set their vector some light inside the set so that's the definition of a subspace.
Birne Binegar: Next definition is a basis, so a basis for stuff base is a set of vectors.
Birne Binegar: All right now here it's just it's going to be a small set of backers Let me give you the intuitive idea.
Birne Binegar: Flying why we're interested in bases, but the definition is slightly different than.
Birne Binegar: What we're looking for the definition that basically designed to be a test to make sure that a given list of vectors will actually do what we want a basis to do.
Birne Binegar: What we want a basis to do is to provide us coordinates for all the vectors inside the subspace if we think of these elements of w.
Birne Binegar: As a lower dimensional substation V, then we don't need and say be as an n dimensional subspace we don't need all and coordinates of every doctrine D to fill out a smaller dimensional subspace.
Birne Binegar: The idea of a basis is it gives us a minimal set of directions that's capable of reaching every point in the subspace and also capable of giving us good coordinates for those points.
Birne Binegar: So here's the definition that really is a test to make sure that a given list of vectors is actually going to fulfill I need so a basis for subspace w is going to be a set of vectors let's say the w.
Birne Binegar: wk i'm going to use a different letter the end up to the K.
Birne Binegar: Such that.
Birne Binegar: Well, first of all, we want to be able to reach every vector and and subspace w by taking linear combination these guys, so you want every.
Birne Binegar: w.
Birne Binegar: And w.
Birne Binegar: can be express.
Birne Binegar: s.
Birne Binegar: w equals and here's the utility of the basis, some constant time set first basis vector similar terms, until we reach the last basic vector we want to be able to write every vector inside our subspace in exactly this fashion, and moreover.
Birne Binegar: The coefficients.
Birne Binegar: C one C keys K must be unique.
Birne Binegar: If we what we actually one of these these vectors do you want to bk if they're being a smaller set of vectors.
Birne Binegar: Such that we can reach every point in our subspace by taking linear combinations of the given list, but we have too many of these guys they're not going to be.
Birne Binegar: Independent of one another there's nobody multiple paths to a single point so when you think of basis.
Birne Binegar: A good good example to keep in the back of your mind is the standard basis of standard basis safe for our three the vector.
Birne Binegar: going in the direction the vector going the y direction that vector going into Z direction those three directions allow you to reach any point in a three dimensional space.
Birne Binegar: If you had more than three vectors, then there will be multiple ways of writing down a vector and such an expansion and if there are multiple ways, then the number C one.
Birne Binegar: c K, are not going to determine a given w uniquely these numbers these coefficients end up being the coordinates by which we specify individual vectors inside of subspace.
Birne Binegar: Okay, so what i'm what i've written down.
Birne Binegar: A set of factors, such that every vector can be presses a linear combination of vectors of those lists with unique coefficients that's the definition of what a basis is and effectively gives you a test to make sure you have a basis, you have to make sure that when you.
Birne Binegar: choose a vector inside w that it can only be written one way.
Birne Binegar: As a linear combinations to get them list now.
Birne Binegar: separate from that is the notion of linear independence, it gives me, you know Christ an alternative way.
Birne Binegar: Well, a simpler check for seeing that, in fact, I do have a basis if I try to apply the definition directly, I have to check every vector inside w to make sure I can.
Birne Binegar: write it in only one way, but it turns out that it's sufficient to just check the zero vector inside w and make sure there's only one way of writing the zero vector so we phrase that a little bit differently, we call that that notion of linear independence, we say that if.
Birne Binegar: If the only solution.
Birne Binegar: Of let's set up an equation X one times the vector V one plus x two times the vector V two plus x K times the vector DK suppose we have a linear combination of vectors that sums the zero if the only solution of this equation is.
Birne Binegar: That first coefficient has to equal zero the second coefficient has to equal zero and all the way up to last component has to equal zero so if we only have one solution of this then.
Birne Binegar: Then.
Birne Binegar: The given list of vectors.
Birne Binegar: Are linearly independent.
Birne Binegar: Okay, and then the way that condition is used is a bit if you have a linearly independent set of vectors then.
Birne Binegar: anytime you can express a vector in terms of a linear combination of linearly independent factors there's only one choice for such a an expansion and so rather than if you want to test to see if a given this is the base is a basis.
Birne Binegar: Rather than checking every vector inside w it's the Feist and still make sure that only the zero vector just look at the zero vector and make sure there's only one solution.
Birne Binegar: Of this equation that a linear combination of you've given list of vectors produces zero if there's only one solution of this then then your vectors are linearly independent and they will then provide a basis for any subspace that they generate.
Birne Binegar: All right, then there's one last definition.
Birne Binegar: This one concerns functions between do vector spaces, we say a linear transformation between two vectors stages dfw is a function.
Birne Binegar: that's going to take me from one vector space to the other such that.
Birne Binegar: Well, here it's kind of analogous to that first definition what we demand for our special class of functions, is it behave well respect this are two fundamental vector operations, what we want is with respect to scale or multiplication if we start off with a number.
Birne Binegar: And we choose a vector inside our domain.
Birne Binegar: Then, what we want.
Birne Binegar: Is that.
Birne Binegar: He applied to the scale of multiple Levi Lambda has to be have the same value as the scale of multiple by Lambda of the value of T at the point the vector V that's.
Birne Binegar: compatibility with scale or multiplication and we have a similar rule analogous to to upon problem one if we start off with two vectors.
Birne Binegar: living in the domain of this function well, we want it said.
Birne Binegar: When I add the two vectors together and then apply the function, I get the same result is if I apply.
Birne Binegar: My function to the first factor and then delayed adding it to the second factor until after I applied the the function to the second vector.
Birne Binegar: Alright, so those are the four definitions that are guaranteed to appear on the final exam.
Birne Binegar: memorize these definitions and just be prepared to write them down and you'll have 20 points on on the final exam.
Birne Binegar: That should not be too hard, but.
Birne Binegar: shouldn't be too hard to accomplish memorizing and being able to regurgitate the definition, but my hope is is that having these definitions precisely on the exam issue.
Birne Binegar: goes through the rest of the exam and I asked you for basis, you can always turn back yeah.
Birne Binegar: Your exam to the first page and look at what your answers going to look like and maybe that will give you hints as to what you need to do.
Birne Binegar: In particular, for bases, I often when I asked questions find a basis oftentimes i'm presented with answers like.
Birne Binegar: Oh i'm given a subspace rather than giving a given a basis has an answer what we're looking for if i'm asking you for basis, probably just a small list of vectors that are capable of generating the entire subspace.
Birne Binegar: Alright, so.
Birne Binegar: The second question on the exam is another question it's not going to involve any calculation at all, but.
Birne Binegar: it's included on the final exam because it sort of test you on your understanding of things, so in this problem to you're not actually required to calculate anything.
Birne Binegar: Rather, I just want you to sort of recognize what happens in the middle of the calculation and how you can interpret that so.
Birne Binegar: This problem to if.
Birne Binegar: it's concerned with augmented matrix ease that you get when you use when you try to solve a linear system, so let me remind you how we solve a linear system.
Birne Binegar: So here we have a bunch of equations maybe i'll step back from this, we might even have a bunch of equations.
Birne Binegar: A one m X m equals one a to one X one similar terms a to an xm equal to.
Birne Binegar: A say we have any equations in MN owns a and m m equals the end so here's a system of linear equations so any equations and unknowns.
Birne Binegar: And the way we handle, that is, first of all, we think of it as a matrix equation, and we construct.
Birne Binegar: From the coefficient matrix in the in homogeneous term on the right hand side we construct an augmented matrix the way we've solved the system if we then reduce this matrix first to.
Birne Binegar: Its row Echelon one.
Birne Binegar: But that's not such that tells us proceeding this far, will tell us some qualitative properties of the solution of the original system but it's not quite the solution, yet once you get to row Echelon warm you continue reducing until you reach reduce rotational on one.
Birne Binegar: And then, once you reach reduce watch long form this is supposed to be the augmented matrix of the solution until the idea will then be the convert back to equations.
Birne Binegar: And then finally express.
Birne Binegar: A solution vector.
Birne Binegar: As a point on the hyper plane.
Birne Binegar: Now i'm going to do this go through the steps of solving a linear system in problem three what i'm going to pointing out is that problem two deals with this process.
Birne Binegar: But what is concentrating on.
Birne Binegar: Is the form of this matrix and reduce all Echelon form and this sort of information that you can tell about the original in your system just from looking at this matrix midway through the process of actually writing down the solution.
Birne Binegar: What I want, for you to know for each of these is what kind of linear system did it come from and are there solutions and finally what's the dimension of the solution, set how many free parameters, will there be in the solution set all that information.
Birne Binegar: We can extract from the row Echelon form of the augmented matrix that's what you're both to interpret each of these matrix shares so, for example.
Birne Binegar: That first.
Birne Binegar: My for right directly on the slide here.
Birne Binegar: This first augmented matrix it's got a total of four rows so the original and in your system had to involve four equations.
Birne Binegar: On the left hand side is the coefficient matrix a and what you each column at the coefficient matrix corresponds to a particular variable since I have four columns I must come from a system of equations in for unknowns.
Birne Binegar: So the number of equations is coming from the number of rows and the number of unknowns is coming from the number of columns on the left hand side.
Birne Binegar: Here, we also have four rows so for equations.
Birne Binegar: And we have four unknowns.
Birne Binegar: And in see we have.
Birne Binegar: Still for equations because we have four rows but now we have five columns on the right hand side so there'll be five nodes.
Birne Binegar: All right, so from the augmented matrix in relational and form, I can.
Birne Binegar: ascertain properties of the original in your system that that came from.
Birne Binegar: Then, in the second part, which is to describe the solution space, and so the first thing I look for is are there solutions.
Birne Binegar: and
Birne Binegar: Let me look at be first that augmented matrix corresponding the part B if you're looking at that third row it corresponds to a particular equation what it says is that.
Birne Binegar: you write it out, it says zero times X one plus zero times X two plus zero times X three has to equal the number one well that just means that zero is equal to one which is a contradiction.
Birne Binegar: So amongst.
Birne Binegar: The four equations that are implied by this augmented matrix one of them is contradictory it implies zero is equal to one if one of the equations that's contradictory, the whole set is contradictory and so in Part B there's no solutions.
Birne Binegar: But, and that happens, whenever you end up with a pivot in this last call so all zeros on the left in a non zero number on the right that occurs in a matrix in row Echelon form well if that occurs anywhere, and you know, like minute matrix it means.
Birne Binegar: You have a contradictory equation amongst your equation set alright, so no solution, there is nothing left to be done in Part B, but for problems, a and see that phenomenon doesn't occur.
Birne Binegar: i'm never have a.
Birne Binegar: bunch of zeros and a nonzero number in the last call so both A and C are going to have solutions and then back so this has solution.
Birne Binegar: And this one also has a solution.
Birne Binegar: But now I can count the number of free parameters.
Birne Binegar: The free parameters of an.
Birne Binegar: augmented matrix that has a solution it's going to correspond to the variables corresponding the columns without pivots so here, I have a pivot no pivot in the last two columns so.
Birne Binegar: What we have is a solution with two three parameters.
Birne Binegar: And those three parameters are going to be corresponding to the last two columns so X three x four will be a free parameters and i'm.
Birne Binegar: In the third problem there's a column without a pivot.
Birne Binegar: And there's a call that a pivot but columns two, three and four contain pivots So those are going to mean that X one X five X of the variables, corresponding to those columns they're going to be free.
Birne Binegar: So in problem see you have for equation five unknown, it has a solution with two free parameters.
Birne Binegar: Three parameters or in this case, going to be X one X five.
Birne Binegar: So.
Birne Binegar: important thing to note in this example, just because I have zeros in this first call doesn't mean I nor.
Birne Binegar: I nor X one if you write down equations for this matrix you'll see that you have no equation involving X one.
Birne Binegar: That doesn't mean you forget about it it's still going to exist as a free parameter and your solution is just that you can adjust X one arbitrarily because apparently no conditions at all are placed on that variable so it will remain free.
Birne Binegar: OK, so the first page is it purely formal.
Birne Binegar: or interpretive.
Birne Binegar: In the second and the third problem on the.
Birne Binegar: Practice exam what I have to do is i'm going to ask you to look at a linear system and work its solution, all the way to the point where you reach the hyper plane.
Birne Binegar: So let's carry that out on for this given system of three equations and three are known.
Birne Binegar: For what we do.
Birne Binegar: Is we form the augmented matrix by reading off the coefficients and the number on the right hand side.
Birne Binegar: So from the first equation, we have zero times X one minus one times X two plus two times X three it's going to equal zero second equation, we have two times X one minus three times X two.
Birne Binegar: X Ray does not appear, and that has to equal to, and then we have one minus one minus minus one coming from the third equation.
Birne Binegar: All right now what we'll do next, is all real reduced to relational on form well mobile one of the pivots to occur in the staircase but right now, what we have our pivots below pivots and in fact pivots.
Birne Binegar: Proceeding off to the left of pivots and the rose above it.
Birne Binegar: But maybe the first thing straight now it's all prior to put a pivot up in this upper left hand corner let's interchange.
Birne Binegar: row one and row three.
Birne Binegar: That will produce one minus one minus one one and then second row it's not changing.
Birne Binegar: Third row is becoming the old real one.
Birne Binegar: All right now I got a pivot in this upper left hand corner and I can use it to clear out entries below it and then i'll get the matrix relational and form, at least up to the first row.
Birne Binegar: So.
Birne Binegar: I guess thing to do is defend row to.
Birne Binegar: To row to minus two times row one i'm trying to get rid of this guy.
Birne Binegar: And that's going to produce well one minus one minus one one first row is not changing under this operation, nor is the last row.
Birne Binegar: But what happened to the second rose, I have two minus two times, one is 03 minus two times negative one it's going to end up being minus one and zero minus two times negative one is going to be plus two and.
Birne Binegar: Two minus two times zero is going to be zero.
Birne Binegar: Okay, good now we've got it in rotational on form up to the second row.
Birne Binegar: We need.
Birne Binegar: To do.
Birne Binegar: Send row three two row three minus.
Birne Binegar: two.
Birne Binegar: will get one minus one minus 110 minus 120 and 0000 because in fact play the second row is exactly equal to the third row and we'll get a zero at the bottom of that matrix.
Birne Binegar: And this stage we've reached.
Birne Binegar: relational on form we still have still have to go to reduce rational on form, before we start to convert back to equations.
Birne Binegar: So what's wrong with this matrix why isn't it in row excellent.
Birne Binegar: Well, to be in reduce your epsilon form, you need the pivots to equal one so i'm gonna have to multiply the second row by negative one, so that its pivot is one instead of minus one.
Birne Binegar: And then I also need to clear out the entries to the blood pivots and the second this pivot here there's an entry above it all need to clear that out.
Birne Binegar: So what we do next it we send row to to it's negative that will fix that or we normalize that constant and i'll send row three two.
Birne Binegar: Sorry i'll send her a one to one.
Birne Binegar: minus row to so that I can clear out this entry in the.
Birne Binegar: One to slot and that then going to produce a matrix one Okay, so when I subtract the second row from the first i'll get zero.
Birne Binegar: Negative one minus two it's going to be negative three and.
Birne Binegar: One minus zero is going to equal one.
Birne Binegar: second row we have zero now i'm also gonna.
Birne Binegar: Take the second roll you know multiply by minus one minus 200000 all right now i've reached.
Birne Binegar: The matrix in reducing Echelon form it's in Russia line form there the pivot all pivots or equal to one and above and below pivot I have nothing but zeros good, so the matrix in reduce show Echelon form is the augmented matrix of the solution.
Birne Binegar: So let's write down the solution equations the first row says that X one minus three x three.
Birne Binegar: is equal to one second row says that X two minus two X three is equal to zero, and the third row doesn't tell us much it just says zero is equal to zero, when I convert the augmented matrix back into equations.
Birne Binegar: there's one more thing we want to do with this matrix and reduce all Echelon.
Birne Binegar: Reduce rational and form is decide which variables are independent, which variables are fixed by the equation, in which we are going to be leftover and regarded as three variables.
Birne Binegar: The rule we use this, we take the variables corresponding two columns without pivot we interpret those as three variables, so in this set of equations X three is going to be a free variable i'm going to put it on the opposite side of the equation.
Birne Binegar: So what will get is X one is equal to one minus one plus three x three l flip sides as I move it to the opposite side of the equation X two is going to equal to X three and now i'm finally in a position to write down a solution vector.
Birne Binegar: It would have.
Birne Binegar: without any restrictions on the variables, it would just have three components, but I demanding that this vector actually satisfy the original linear system.
Birne Binegar: And that means that X, one can be must be expressed in terms of X Ray is one plus three x three secondary bill.
Birne Binegar: can be expressed in terms of the free variable X three as can't X three itself, then the last step is to expand this vector solution in terms of the free variable so we have a constant vector coming from the constants in each component.
Birne Binegar: And then we're going to have.
Birne Binegar: terms that are proportional the X three.
Birne Binegar: In the first component it's that three the second component it's two and last component one times X Ray OK, and now i've got the hyper plane expression for the solution of the original in your system.
Birne Binegar: Any solution vector.
Birne Binegar: can be achieved by starting at the point 100 and then straying from that point along the direction of the vector 321 by an arbitrary now, this is a free parameter so it's allowed to be any real number.
Birne Binegar: So that's actually a hyper plane more generally.
Birne Binegar: A hyper plane.
Birne Binegar: it's a set of vectors that you construct by taking a given starting point and then stray from that starting point.
Birne Binegar: And as many directions, as you want, but miss example K directions and then you allow these coefficients.
Birne Binegar: To range over the entire real line that's what a hyper plane is it's basically a flat surface embedded in a higher dimensional space with.
Birne Binegar: In this instance K K directions, from which we can stray from an initial starting point.
Birne Binegar: This expression has exactly that form in fact it's a line line is the special case of a hyper plane it's where you started a point and you head off in only one direction, from that point.
Birne Binegar: Okay, so our solution is.
Birne Binegar: The solution is align it a one dimensional hyper plane.
Birne Binegar: Okay, so that's some.
Birne Binegar: Really kind of crucial calculation that we've done over over in this course.
Birne Binegar: oftentimes I did this in the case of a hand I did this for non homogeneous system of equations where somewhere on the right hand side I had some non dual numbers but you'll follow the same procedure, even when you have a homogeneous system.
Birne Binegar: It just simplifies a little bit so similar.
Birne Binegar: kalki calculation.
Birne Binegar: For what we call homogeneous in your system.
Birne Binegar: That we have eight X is equal to the zero vector.
Birne Binegar: In this case, you basically follow the same procedure, but nothing ever happens in this last call, so we tend to just work with the.
Birne Binegar: Real reduce the matrix a rather than the augmented matrix a with the zero vector are joined to it so yeah and in fact we do this calculation a lot, because the solution set of this.
Birne Binegar: is completely characterized by the matrix and we call it the no space of the matrix.
Birne Binegar: Okay, oh that's an aside, Madame.
Birne Binegar: Okay, so let's move on to the next problem.
Birne Binegar: In this fourth problem you're asked to compute the inverse of the matrix um.
Birne Binegar: there's two ways of doing this one is to compute the co-factor matrix of a ticket transpose and then divide that by the determinant but.
Birne Binegar: I don't think i'm going to give you an example where that method needs to be applied, we can also find in verses of matrices simply by doing row reduction, so let me show you how that works in this case.
Birne Binegar: What you do is you form.
Birne Binegar: You take the given matrix and you would join to it, the identity matrix at the same size and then you reduce.
Birne Binegar: All the way to reduce row Echelon form and one of two things is going to happen.
Birne Binegar: Either you get a matrix we're on the left hand side you get the identity matrix and if that happens.
Birne Binegar: On the right hand side you'll get the desired inverse or you're going to get something else.
Birne Binegar: In that left hand side and, in which case it doesn't make any difference if there's no inverse.
Birne Binegar: In other words, when you read this this matrix and you get something other than the identity matrix on the left, that means that the inverse of the original matrix does not exist.
Birne Binegar: Usually the way this will happen if you look at the bottom of that matrix and you'll see a zero and that identity matrix does not contain the zero row.
Birne Binegar: Okay, so let's let's work it out for this particular problem we'll start with 210100 minus 201 and.
Birne Binegar: All join to that.
Birne Binegar: The three by three identity matrix.
Birne Binegar: Alright, so.
Birne Binegar: Maybe the first thing we'll do to get the rub reduction going is to.
Birne Binegar: I like pivots when they're already equal to one so first thing i'm going to do is i'm just going to interchange row 120 and then i'll have 0101202120.
Birne Binegar: minus 201001 all right, so all I did this first step of row reduction is, I wanted to get a convenient pivot up in this corner, so I interchange two roads.
Birne Binegar: All right.
Birne Binegar: Now.
Birne Binegar: Now, what I can do is I can use this pivot in the upper corner here to clear out entries low, so I can send row to row two minus.
Birne Binegar: Two times one and i'll send wrote three two row three plus two times one.
Birne Binegar: And what will come out of that will be 1200102 minus two is 01 minus two times zero is one zero minus two times zero is 01 minus two times zero is one.
Birne Binegar: Zero minus two times, one is minus two and zero minus two that means zero and what happened to the third row is i'll get 00 minus 001 minus two times zero is one.
Birne Binegar: And then, what happens here is zero.
Birne Binegar: And negative.
Birne Binegar: yeah sorry positive to.
Birne Binegar: i'm taking pause two times positive two times the first row and i'm adding it to the third row and then i'll get one alright, so what happened here.
Birne Binegar: During this role reaction i've converted this matrix into a matrix of this form here's the three by three identity matrix that means I can identify the right hand side is a the mind first.
Birne Binegar: So all you have to do the finish the problem is simply write a the minus first the 0101 minus 200 to one and that your answer.
Birne Binegar: Okay, so it's.
Birne Binegar: You know the algorithm and it's just a real reduction calculation and like this example i'll try to design the problems on the exam so the real reduction is fairly straightforward and fast.
Birne Binegar: Alright, so.
Birne Binegar: problem five is.
Birne Binegar: that's a problem where you given a set of.
Birne Binegar: Sorry you're.
Birne Binegar: Given a set of vectors set of three dimensional vectors and you're asked to determine if it's a substance or not.
Birne Binegar: What does that mean.
Birne Binegar: well.
Birne Binegar: What we're looking for is a set of vectors such that it's closed under scale and multiplication and closed under vector edition.
Birne Binegar: So let's look at that set and try to decide if it's closed under, for example.
Birne Binegar: Is w.
Birne Binegar: closed.
Birne Binegar: Under scale or multiplication.
Birne Binegar: If it's not then it cannot be a subspace so let's choose a number zero.
Birne Binegar: Lets you choose a vector w equal I wanted to sit inside this set but it's pretty easy to see that this vector 100 when I added components.
Birne Binegar: I get the number one, so the three components of this vector they add up the one that means.
Birne Binegar: they'll be you is inside this set.
Birne Binegar: What I want to check is to make sure that the scale of multiplication of that element of the set by the number zero still lives inside that set.
Birne Binegar: So when we do Lambda times w well, of course, anytime you scale and multiply by zero.
Birne Binegar: you're just going to get the zero vectors result, but now zero Plus zero Plus zero when I add up the components of this vector I get zero, and that does not equal one, so this vector Lambda times w doesn't satisfy that condition to be a member of capital w and so it's not closed.
Birne Binegar: Under scale or multiplication and so it's not a subspace.
Birne Binegar: All right, and that's the answer but there's another way, you could have shown.
Birne Binegar: shown minutes, not a subspace.
Birne Binegar: The definition of Sub base requires it to be closed under both scale or multiplication in Dec tradition so alternatively what you could have done.
Birne Binegar: Is you said i'm choosing a number in the vector and making sure scale and multiplication is not his clothes choose two vectors.
Birne Binegar: w one let's set it equal to 100 and w to set it equal to 010 so In both cases, one plus zero Plus zero equals one, so it lives inside w and here we have zero plus one plus zero equals one, so it also lies.
Birne Binegar: With the factors live inside.
Birne Binegar: The given set question is.
Birne Binegar: Does.
Birne Binegar: Their vector some.
Birne Binegar: live inside that set well, that would be 100 plus 010 equals 110 but now.
Birne Binegar: When I add the components of this factor one plus one plus zero I get to.
Birne Binegar: Not.
Birne Binegar: The value, one that I need in order to live inside this set and so it's.
Birne Binegar: Not in w and so it's w.
Birne Binegar: is not closed.
Birne Binegar: Under vector.
Birne Binegar: based.
Birne Binegar: On the final exam I mean, I had a question like first lecture oh.
Birne Binegar: What if what if it was a sub base would you do the same thing and so i'm sorry to say that it if it is a subspace it's harder directly.
Birne Binegar: justify that because you have to show that both that is closed under skill and multiplication, no matter how you choose a vector w and he left show that it's closed under back tradition, no matter how you choose vectors D one.
Birne Binegar: w one w two it's easier to disprove than it is to prove and so probably what i'll do on the final exam is give you a subspace and ask you to.
Birne Binegar: Effectively, the show that it's not a subspace by showing that it's not closed under scale the multiplication or that it's not closed under vector edition.
Birne Binegar: Okay.
Birne Binegar: Oh.
Birne Binegar: And in my solutions here, I gave you two answers you can either show that it's not closed under back tradition and deduce that it's not a safe space or you can show that it's not closed under steel and multiplication and therefore not subspace you don't have to prove both.
Birne Binegar: Alright let's see problem six problem six.
Birne Binegar: deals with linear independence.
Birne Binegar: But.
Birne Binegar: be aware, for these kind of problems you're not actually going to use the definition of linear independence rather you're going to use this test.
Birne Binegar: For linear independence that we've been employing.
Birne Binegar: Basically, we we can we answer this question by doing a quick role reaction.
Birne Binegar: So what we'll do in this for this example to determine if these factors are linearly dependent well.
Birne Binegar: Basically, what you want to do is make sure there's no way of adding these factors together and producing the zero vector without taking all the coefficient separately equal to zero, but I won't do that directly what i'm going to do instead is i'm going to form a matrix.
Birne Binegar: Using the given list of vectors as rose so one minus 2211 minus 1310110 and two minus 352 and i'm going to reduce.
Birne Binegar: And that's all ticket see my solutions, I think I go all the way to reduce your actual form.
Birne Binegar: Probably only have to go to row so one form.
Birne Binegar: That matrix row reduces 210410110 and then, I have two zeros at the bottom, but that matrix.
Birne Binegar: Now this actually this real reduction actually shows that the original set of vectors we're not linearly independent.
Birne Binegar: This row reduction process, what do you do well, for the most part, you don't do much either just change the order of the Rose or you multiply a row by a scale or factor, the operation use most often is you take a row, and you replace it by it, some with a multiple of another row.
Birne Binegar: and
Birne Binegar: When you do this operation over and over again, what you do is you replace rose by linear combinations of you given list of rows.
Birne Binegar: So if you start off with a given list of actors and you're reducing each of these roads it's basically a linear combination of the four factors that you started off with it's just that you manipulated them over and over again, but you're still just taking linear combinations.
Birne Binegar: Like so of the original rose and what's what's happened here is that you found some linear combination of the original rows that will produce the zero and you've also found that you've found two different linear combination probably that produced the zeros and so.
Birne Binegar: there's gotta be a way of taking these four vectors something them up in a linear combination and producing a zero row well, that means that the original rose.
Birne Binegar: If you see these arrows appearing in the row reduction, that means that the original vectors we're not linearly independent because there were some linear combination of these guys that produce a zero row.
Birne Binegar: So.
Birne Binegar: So in this case the non zero rose.
Birne Binegar: or sorry the zero rose.
Birne Binegar: indicate.
Birne Binegar: The original vectors.
Birne Binegar: are not.
Birne Binegar: linearly independent.
Birne Binegar: Right and that answers.
Birne Binegar: Part A, Part B.
Birne Binegar: The span of these vectors.
Birne Binegar: it's just the the row space of this matrix and what i've calculated here, if you remember how we find a basis for the road spaces rewrote reduce the matrix and we grab the non zero rows.
Birne Binegar: So, in fact, the set of all possible linear combinations of these four factors well, they can be that set can be generated by just these two vectors and these factors are linearly independent because I won't be able to cancel the first component.
Birne Binegar: Unless I set up by try to take a linear combination, the first row with the second row I cannot prove zero without taking the first component.
Birne Binegar: coefficient equal to zero because I can't cancel it was something near and Nora can I cancel the second component about taking the multiple at the second row equal to zero, so these two are.
Birne Binegar: Two vectors that are literally independent and they generate the entire space that's generated by the original form factors that means this subspace is two dimensional because it has to basis factors, and so the answer there.
Birne Binegar: Is because you have to non zero rose.
Birne Binegar: means that the row space.
Birne Binegar: Of the matrix a is two dimensional.
Birne Binegar: and that in turn.
Birne Binegar: Sorry, this it really 1230 already on i'm sorry if I run a boy anyway, this is probably.
Birne Binegar: More than halfway through the exam.
Birne Binegar: i'm sorry man four minutes over.
Birne Binegar: At any rate, up because you have to non zero the bros the row space a two dimensional now to equal to the span of the original vectors.
Birne Binegar: Okay, so on Friday what i'll do is i'll finish up the rest of this practice exam.