Birne Binegar: Okay, so you notice i'm wearing a mask again, not because i'm sick nor because there was a big surgeons in stillwater rather's because I have friends that are coming from France, next week and we're both trying to keep things safe for their visit so hope you don't mind this.
Birne Binegar: What we'll do today is review for the second exam and i'll start by.
Birne Binegar: Reviewing the two topics that are basically.
Birne Binegar: The substance of the exam we linear sub spaces, which are special sets that are relevant to linear algebra and linear transformations which are the functions that are relevant to linear algebra but then, of course.
Birne Binegar: Those are the two main topics, but attached to these or are.
Birne Binegar: a bunch of Sub spaces and how.
Birne Binegar: What we need to understand, in order to use up spaces, how to coordinate is them and that's where the other topics, like the notion of bases and linear independent comes in, but in the end, when you actually take the exam.
Birne Binegar: What you'll find if you look at the practice exam you're basically most that, aside from knowing the definitions, most of your problems are just going to be.
Birne Binegar: Recognizing what subspace we're talking about, and then the procedure by which we find a basis for that space.
Birne Binegar: Okay, so.
Birne Binegar: Let me go over the topics first.
Birne Binegar: So i'm.
Birne Binegar: Working with vectors all the time in this course, but what we specialize to now or special subsets of vectors and vector space, say, I have a bunch of vectors that put them in a set called w.
Birne Binegar: requirement we want for interesting set is that with respect to the two fundamental vector space operations be set to be well behaved.
Birne Binegar: Meaning that if I add two vectors inside this set I can stay inside the set, just like to buy add to three dimensional vectors I stay inside of three dimensional space and similarly will want it to behave well for respect the scale of multiplication.
Birne Binegar: To more precisely what we need is if we start off with number any number any vector inside the set once we do the scale or multiplication we still inside that set or if we take.
Birne Binegar: A summit two years they start off living in that set we combine them to get to an to produce a new vector that that new vector stays inside that set, though, though.
Birne Binegar: that's a vector that have tho those two properties are called sub spaces.
Birne Binegar: And there's a fundamental objects in linear algebra and so once I introduced subspace is the next thing I did is try to describe how we talk about.
Birne Binegar: Individual vectors that live inside of subspace typically these are going to be lower dimensional surfaces embedded inside a larger dimensional space.
Birne Binegar: So if i'm subspace which is typically a hyper plane that passes through the origin it's if it's a lower dimension than the space in which it lives.
Birne Binegar: than the original coordinates are not really apt for describing points in it, if I had a two dimensional surface in bed in a 10 dimensional space.
Birne Binegar: I don't want to have to use 10 parameters to find all the vectors on that two dimensional surface, so what the basis idea of a basis is and it's.
Birne Binegar: You can think about a number of different ways, the main purpose of it is to find good coordinates for the barriers that live inside of subspace but what they represent are.
Birne Binegar: Independent directions that live inside that space, you need enough independent direction, so you can reach every point on this in the subspace but.
Birne Binegar: You want to use these guys to get coordinates so not only need to reach every point in your subspace you want to make sure there's unique way of getting it because that's how we're going to get coordinates so more precisely.
Birne Binegar: Maybe not.
Birne Binegar: More precisely what we need instead effectively two conditions, first of all, you have to be able to reach every vector.
Birne Binegar: In your subspace by taking linear combination and, moreover, you want these numbers see one two ck the coefficient of such an expansion, you want those to be unique.
Birne Binegar: Get the idea, then it's going to use these K numbers, rather than the dimension of the original vectors use these K numbers as coordinate for the elements that live inside of subspace.
Birne Binegar: All right now the.
Birne Binegar: This definition reflects what we're going to use a basis for we use bases to coordinate coordinate ties sub spaces, but.
Birne Binegar: There are really two separate conditions, one is to make sure you can write every back through this way and separately that they're unique, this is actually this uniqueness property that make them useful for finding coordinates.
Birne Binegar: And then distinct from this.
Birne Binegar: Or at least with with things so initially this this notion of linear independence all right, forget about bases.
Birne Binegar: If you simply have a set of vectors and the only way, you can get them to add to zero is by taking all the coefficients or zero what we say that the vectors are linearly independent.
Birne Binegar: Now, this is a nice mathematic condition to check it's a single equation and but which we should have a solution well this this can be checked mathematically.
Birne Binegar: And in fact what is telling you is that the vectors you start off with are really point independent directions.
Birne Binegar: Just think about two vectors if I have two vectors and they're parallel well, then I can get them to cancel one another, I just point the vectors in the opposite direction and they'll end up canceling but if they actually point in different directions.
Birne Binegar: multiplying one vector by in a number just makes it longer shorter maybe it reverses its direction that it doesn't change its direction.
Birne Binegar: and multiplying another vector by number makes it longer or shorter potentially reverses that direction, but it doesn't stop it from being independent of the other direction.
Birne Binegar: So what this condition is really saying is that these guys are really independent directions.
Birne Binegar: And so what you can do is effectively, you can replace this second condition, by demanding that these.
Birne Binegar: That these coefficients are unique if these guys are independent directions, then you you're already guaranteed that these coefficients you're going to be unique you won't be any way to get them to cancel yeah if they're actually if this condition is true.
Birne Binegar: So, so the way to understand bases is they provide coordinates and the way you understand linear independence, this is the way you check to make sure that you really have independent directions to which to build your space.
Birne Binegar: All right now.
Birne Binegar: Once we have a set of vectors that sufficient to generate space, and I also know they represent independent directions, then in fact I know those vectors are also a basis so.
Birne Binegar: I think the way the textbook defines a basis that they say it's a set of vectors that generate a space and are linearly independent.
Birne Binegar: And then, what they do is they prove that the coefficient in the expansion of any vector in the set in terms of these factors is always you need so they kind of reverse.
Birne Binegar: My logic my logic is to get the coordinates first and then.
Birne Binegar: Introduce linear independence is a simple way to check to make sure you have independent directions, but other textbooks and other lecturers will say well we'll start off with these two conditions that's the definition of what a basis is and then will prove that the expansion is unique.
Birne Binegar: For the exam a stick to my the way i'm doing it.
Birne Binegar: The definition of a basis is every vector can be in the subspace can be represented as a linear combination, and second, of all that linear combination is unique.
Birne Binegar: That will that will give us a coordinate and then just understand this notion of linear independence this a way to make to check to make sure that you really have independent directions.
Birne Binegar: Okay, and then.
Birne Binegar: Some other things to know about bases is first of all, every time you have anytime you have a subspace it's guaranteed to have some basis.
Birne Binegar: And in fact every basis for it it's going to have the same number of vectors think of three dimensional space there's three independent directions, you need to.
Birne Binegar: You get a basis for three dimensional space, but you have many different choices for how you orient your coordinate axes, those were all correspond to choices of bases, but the number of factors.
Birne Binegar: in any way it's going to stay the same, and the number of factors that you require for a basis it's actually what we've referred to as a dimension of the space so three dimensional space there's three three distinct directions up maybe maybe sideways and in and out.
Birne Binegar: Those three independent directions will generate the space and they give us a dimension of our face.
Birne Binegar: All right now.
Birne Binegar: Thus far, i've given you well one theorem that relates some definitions, but everything else has been definition on the exam.
Birne Binegar: You will be required to state definitions so there's a practice exam posted you'll see the definitions presented there in the solution i'm also presenting the definitions here on these like these lecture flight what I.
Birne Binegar: want to stress to you is that you're going to need to know these definitions verbatim what I suggest you do is.
Birne Binegar: Take 15 minutes sometime over the weekend and make sure you can write we reproduce these definitions without looking at any notes, and you can copy the verbiage, exactly as I write it.
Birne Binegar: down that way, not that 15 minutes of effort will guarantees you maybe 25 points on the exam yeah it's very important for me that you understand the definition and and and not sort of.
Birne Binegar: Using the many different ways, many different applications of the definitions you that you can actually distinguish an application of a definition from the definition itself so, for example, the serum.
Birne Binegar: In this course, is not a definition once we know where the basis is as defined on the preceding slide This turned out to be a consequence of.
Birne Binegar: Of of the definitions so stepping back a little bit or i'll actually i'll show you in a minute when we actually look at the practice exam i'll show you what you would need to do to answer questions on definitions, but we're almost through.
Birne Binegar: The material to be covered.
Birne Binegar: Once we had the notion of Sub spaces, then I tried to give you examples of Sub basis and, particularly, ones that we can calculate so.
Birne Binegar: I have given you three substations that one can attach to any matrix we start off with an n by n matrix and then we can get sub spaces from this matrix.
Birne Binegar: three different ways, depending on how we think about this matrix we can think of it as a bunch of coefficients putting slot we can think of it as a collection of robe actors or as a collection of calm directors and depending on how we thinking about that matrix we get different subspecies.
Birne Binegar: For example, the row space, because the matrix is being a bunch of robe actors and then just take all possible linear combinations of those robe actors.
Birne Binegar: that's going to create a subspace typical way you create a sub space is just by taking the span of a given list of vectors.
Birne Binegar: But notice that now the subspace because every row is going to have exactly elements in it don't want em you're going to get end up with a bunch of factors that are all n dimensional vectors so we get a subspace of our am.
Birne Binegar: Thinking about the matrix is a collection of columns.
Birne Binegar: We get a different sub space and it lives in a different sector space it's still constructed and essentially the same way by taking all possible linear combinations i'm calling vectors.
Birne Binegar: But now my columns each column has exactly in entries So these are all in dimensional vectors and so my subspace will live in, or in.
Birne Binegar: Now, what you have to understand about these guys is not only what they are, but how do we find a basis for these two substances.
Birne Binegar: Well, what we've seen is a single row reduction calculation will yield the answer to both.
Birne Binegar: The way you get a basis for this row space, if you take the matrix and you reduce it to a row Echelon form.
Birne Binegar: Now what might happen during that row reduction process and some zeros get created the because rose end up canceling one another one you try to remove pit well.
Birne Binegar: You ignore the zeros and the relational and you grab only the non zero rose and those vectors will give you a basis for this row space.
Birne Binegar: For the column space, you can use the same row reduction process, but the way you get the basis factors is difference.
Birne Binegar: In for the column space you reduce the matrix to a rational on form you look at where the pivots live.
Birne Binegar: That and then the pivots what you do is you transport them back to the original matrix and you grab the columns.
Birne Binegar: So, if my real actual on form, for example, had a pivot in the third column that would need the third column of the original matrix was actually a basis vector for the columns face and you get all the pieces factors that way, all the basis vectors for the column space.
Birne Binegar: Then.
Birne Binegar: there's a third way of thinking about this matrix, and that is as the coefficient matrix for linear system.
Birne Binegar: And when you think about the matrix that way.
Birne Binegar: This matrix exactly being all the coefficients have some system of linear equations in a variable X well, the solution set up this homogeneous linear system it's going to be a subspace and it's called a null space of the corresponding matrix.
Birne Binegar: The way we get a basis for that subspace is, we can actually use the same row reduction calculation.
Birne Binegar: described on the preceding fly you take the matrix you reduced to rational and form, but in this case, because you actually want the solution, you have to row reduce all the way to reduce real Echelon.
Birne Binegar: And so it back you actually have to solve this linear system and a refined The basis for this set what it is, is once you've written down the solution of this equation it's a hyper plane so you've got the solution express this up.
Birne Binegar: a bunch of free parameters times constant vectors you just grab those constant vectors that occur in that expansion and those will be the basis factors for your your no space.
Birne Binegar: Alright, so um let me see if I can do a.
Birne Binegar: Quick example.
Birne Binegar: Suppose.
Birne Binegar: A equals 111210011 and hit me put another.
Birne Binegar: Relative yeah your question okay so suppose this matrix in here's the big supposition because i'm not going to do the computation to pose this robe reduced to.
Birne Binegar: Suppose ever used to that and I probably not, but I just want to start with a matrix, and this is suppose that the row Echelon form of the matrix I started off with.
Birne Binegar: The way I would get bases for the row space for COM space and the no space given this calculation would be as follows, first of all, the row space.
Birne Binegar: Of a it's a non dual roles in the row Echelon form so it's just going to consist of 1011.
Birne Binegar: comma and then 0100 right, then those that will be a basis.
Birne Binegar: For the road space, these two vectors are going to generate the entire row space and they're linearly independent alright so that's.
Birne Binegar: basis for one subspace the basis for the common space.
Birne Binegar: Well, the same calculation is done, except what we do, once we release it is different for the column space you look at where the pivots live and then you go back to the original matrix and grab the corresponding columns and so.
Birne Binegar: This is this given matrix it's just going to be well here's a pivot I go back to the original matrix and I grabbed that call 120.
Birne Binegar: here's a pivot living in a row Echelon form, I go back to the original matrix and I grabbed the corresponding column.
Birne Binegar: Right and there's.
Micah M: there's a basis for the column space.
Birne Binegar: One more subspace.
Birne Binegar: The no space.
Birne Binegar: Of the matrix or what I have to do is I actually have to solve.
Birne Binegar: X is equal to zero well we're more than halfway to the solution once we've done this row reduction, so all we have to do is complete, the solution of this.
Birne Binegar: Writing down the solution of this equation using this matrix and reduce show Echelon form, so this matrix tells me that X one.
Birne Binegar: plus x three plus x force going to have to equal zero short reading off the equation that would come from the first row second row is going to tell me that X two.
Birne Binegar: has to equal zero.
Birne Binegar: And then.
Birne Binegar: We still haven't written down the solution, yet we want to write it as a hyper plane, so I use the form of the augmented matrix and reduce your national on wi fi the three variables it's going to be in this example X three x four because those are columns that don't contain pivot.
Birne Binegar: that's going to suggest that I expressed variables that correspond to columns width it fixed variables, there are going to be expressed as minus X three minus X for just moving the free variables, to the right hand side and here we'll still have X is equal to zero.
Birne Binegar: All right now, now that I utilize the equations coming from the argument matrix and reduce your actual on form, I write down a solution director now I guess i'll do it up here X is going to equal.
Birne Binegar: it's going to equal let's see its first component has to be equal minus X three minus X for its second component has to equal zero and the third and fourth component there free parameters so i'll leave them with that.
Birne Binegar: And then, when the last thing I do if I expand in terms of the free parameters minus 1010.
Birne Binegar: And plus four.
Birne Binegar: minus 1001 all right, then these two vectors these content factors that are being multiplied by the free parameter they're the basis vectors for the null space, so that no space is actually equal to the span.
Birne Binegar: Of.
Birne Binegar: looks like minus 1010.
Birne Binegar: and
Birne Binegar: minus 1001 and so these are your basis factors.
Birne Binegar: Alright, so any questions on that.
Birne Binegar: Being being able to take a matrix and finding basis for both it's ro space colon space and no space that's really the principal calculation, you have to do on the exam.
Birne Binegar: And then.
Birne Binegar: there's one more topic, one more definition to know, and that is well some spaces are very special function or sub spaces are set special sets.
Birne Binegar: that are relevant to linear algebra linear transformations are the functions that are relevant linear algebra their functions between two vector space.
Birne Binegar: That behave, naturally with respect to scale and multiplication in depth tradition, you are the rules.
Birne Binegar: Whether I scale or multiply vector before I apply the function or after I apply the function I should get the same result and or whether I add vectors before.
Birne Binegar: I map over into the new vector space or after I move over to the new vector space I should get the same result now most functions are not going to have that property, but if they do we call it a linear transformation.
Birne Binegar: And it turned out that these are the notion of a function that's most relevant to linear algebra.
Birne Binegar: Oh, in fact, now.
Birne Binegar: One way to create a linear transformation.
Birne Binegar: One way to create a function that has those properties is just use matrix multiplication.
Birne Binegar: And so, in fact, if you have a matrix.
Birne Binegar: A citizen n by n matrix.
Birne Binegar: Well, an n by n matrix can multiply a column vector with em components, because you if you have this match between inner dimensions.
Birne Binegar: And n by n matrix times, an invite one column vector that multiplication is defined and it's going to produce a matrix that has enrolls in one call, as a result, well what you can do is take a take an n dimensional vector and the result is going to be an n dimensional vector.
Birne Binegar: And so, given a matrix you have a way of going from n dimensional vectors to n dimensional vectors and in fact it's a linear transformation of the sort, and what did what you would find it, but you move move the key here, you would define the linear transformation, the matrix multiplication.
Birne Binegar: Now this theorem says the opposite is also true if you have you can not only use a matrix to define a linear transformation, but a linear transformation, you can use.
Birne Binegar: You can define a matrix that basically reproduces the action of that that function, but it does so via matrix multiplication rather than by evaluation at a particular point.
Birne Binegar: And so.
Birne Binegar: What it also does is allows us to transport questions about linear transformations between vector spaces, we can now deal with them through the.
Birne Binegar: Use of this numerical matrix it's just going to be a bunch of numbers in closed in flight square braces so.
Birne Binegar: it's going to be naturally thing we can apply addition and subtraction and do calculations, while this to understand these function between two vector spaces book more on the variables, rather than just a matrix of numbers.
Birne Binegar: So.
Birne Binegar: The idea of this matrix a subscript T the matrix that corresponds to a linear transformation is so that we have a calculable model which to understand functions between.
Birne Binegar: Between Dr spaces and we use that matrix let me write it out it's functional form.
Birne Binegar: Oh, if you have.
Birne Binegar: is going from an n dimensional space.
Birne Binegar: To n dimensional state and it's.
Birne Binegar: A linear transformation.
Birne Binegar: here's how you construct the associated matrix.
Birne Binegar: You like the T is going to act on em dimensional vector so let's apply it to the first standard basis factor.
Birne Binegar: Of this n dimensional space oh you'll get an answer it's going to be an n dimensional vector but right the answer out vertically.
Birne Binegar: That will get to dimension to this matrix to be appropriate, it will end up being a matrix with as many columns, as we have dimensions here and as many roles, as we have dimensions here.
Birne Binegar: The last one is going to be key flying to the last standard basis factor.
Birne Binegar: Of this domain.
Birne Binegar: And we take the answer we write it vertically that gives us a concrete numerical matrix that basically contains all the information, and in fact analysis to reproduce the action of this of this linear transformation using this matrix this matrix is going to have the property.
Birne Binegar: Whatever T does to an n dimensional vector can be replicated by regarding this m dimensional vector as a column vector and then multiplying it by this matrix.
Birne Binegar: So you have that relationship.
Birne Binegar: Okay.
Birne Binegar: So.
Birne Binegar: The last thing we did.
Birne Binegar: Yes, we defined some sub spaces.
Birne Binegar: attached to a linear transformation.
Birne Binegar: spaces were not.
Birne Binegar: defined.
Birne Binegar: Using algebraic ideas linear algebraic ideas, rather we using function theory ideas, so the first one.
Birne Binegar: Was the range of a linear transformation.
Birne Binegar: And now that was defined.
Birne Binegar: You saw a function theoretic way what we do, if you have a linear transformation.
Birne Binegar: From an n dimensional space.
Birne Binegar: To an n dimensional space, what we can do is collect together all the points in the co domain.
Birne Binegar: That can be reached by applying this function, so you want why to equal.
Birne Binegar: He applied to act.
Birne Binegar: For some.
Birne Binegar: vector that lives inside the domain rm.
Birne Binegar: and
Birne Binegar: Once this was defined it was immediately verified that it's also a subspace it's closed under scalable station and vector addition.
Birne Binegar: A second subspace.
Birne Binegar: Was.
Birne Binegar: A kernel of a linear transformation, it was defined using function theory as a set of point, this time in the domain.
Birne Binegar: lecture we start off and the condition we want is that.
Birne Binegar: You want the vectors in the domain, they get sent to the zero vector in the co domain or.
Birne Binegar: Do we want all the point, whose value when we apply this function to it, end up being the zero vector.
Birne Binegar: Alright, so these.
Birne Binegar: Two sets are defined using function theory see how where the function takes point but they're calculable because each of the sub spaces.
Birne Binegar: can be calculated using the auxiliary matrix this range is exactly the same thing.
Birne Binegar: As the column space at the associated matrix.
Birne Binegar: And we know how to we know by this formula, how to compute that matrix and then from the preceding discussion about sub spaces associated with major cities we know how to find a basis for the COM space.
Birne Binegar: and similarly on this panel.
Birne Binegar: it's a subspace defined by.
Birne Binegar: The way he acts as a function but it corresponds to the same set of factors, so no space a fee associated matrix.
Birne Binegar: So the calculation that we did define bases for column space and no spaces or matrices is now can also be used to find bases for.
Birne Binegar: substations they are now attached to a linear transformation, rather than the matrix, but what we do.
Birne Binegar: We wanted to know what the range of a linear transformation is first thing I would do is I would calculate this matrix.
Birne Binegar: Then I would take that matrix and I would reduce it to reduce your last one form figure out where the pivots live go back to the original matrix and grab those columns and now I have a basis, not only for the column space, but a basis for this.
Birne Binegar: subspace associated with the linear transformation thought of purely on on functional terms.
Birne Binegar: and similarly Colonel.
Birne Binegar: Once we we want to find a basis for this kernel calculate the associated matrix and then use the matrix calculation to figure out a basis for this no space.
Birne Binegar: Alright, so that's basically what you're going to be doing on this exam.
Birne Binegar: Identifying substations often i'll usually state what they are, but you're gonna have to figure out how to find bases and typically what that's going to involve and she find a matrix you reduce and then.
Birne Binegar: Depending on which subspace attached to that matrix you're interested in you'll either grab the non to roast roast days grab the.
Birne Binegar: columns of the original matrix corresponding the pivots of a ritual informed to get a basis for the columns based and for the Colonel you actually have to solve a linear system, a subscription fee of X is equal to zero.
Birne Binegar: All right, so that's the material that we've covered.
Birne Binegar: Let me.
Birne Binegar: restart my share screen.
Birne Binegar: hold on a second here.
Birne Binegar: Okay well i'm gonna put up this practice exam.
Birne Binegar: i'll try to do as much as this of this as I can, let me just quickly go over the first page so.
Birne Binegar: What you see on the practice exam and I promise you, you what you will see on the exam on Monday, probably identical to that on the first part, I want you to memorize those definition and just be able to write them down verbatim and, in fact, what version do I want you to.
Birne Binegar: write, let me show you.
Birne Binegar: All right, so here's the Posted solutions to the practice exam so for a subspace remember that you need be able to.
Birne Binegar: You simply have to state that it's closed under stealer multiplication and vector edition, and here are the two conditions that you're going to have to write down to make sure.
Birne Binegar: That I understand exactly what you mean by closure understand the multiplication and vector edition.
Birne Binegar: start with a number in the vector you do scale multiplication you have to stay inside the set start with two vectors begin living in that you add it together, you have to stay inside that set.
Birne Binegar: here's the definition of a basis here's the definition linear independence so for this.
Birne Binegar: This first problem on the final exam simply practice that you can actually replicate these definitions and that will then you'll get full credit on those.
Birne Binegar: The second problem.
Birne Binegar: let's look at it, this might be the first time you see a question like this on the exam you might be shocked, but it's really not that hard.
Birne Binegar: All i'm asking you in that second problem.
is to.
Birne Binegar: figure out if the unit circle.
Birne Binegar: satisfies these conditions is it a subspace So if I take a number and an element on the circle, and I do the scale multiplication do I stay on the circle.
Birne Binegar: If that's always true and, moreover, whenever I add vectors on the circle I stay on the circle, if that were always true, then I would.
Birne Binegar: It would be a subspace, but in fact this example the counter example to have subspace it fails to be closed under scale of multiplication and it fails to be closed under back tradition and showing either one of those conditions, showing that.
Birne Binegar: If you have a vector in a number and you do the scale of multiplication you end up outside the set well if that happens, and then it can't be a subset.
Birne Binegar: Or if you add two vectors that live in that live inside this subset and you end up with a vector that lives outside that said it's it's not going to be a subspace so what you could do in this present example yes.
Birne Binegar: show that you take a number like zero and you take a vector be like one zero well that lives inside the set right.
Birne Binegar: Because one squared plus zero squared is going to equal one and that's the condition the some of the squares of the components has to be equal to one to live inside the set.
Birne Binegar: One squared plus zero squared does equal one.
Birne Binegar: And that's why I know that this vector lives inside the set Okay, so I get it, a number of vector let's do the scale of multiplication well it's going to equal to zero vector but now.
Birne Binegar: When I some the squares of the components here.
Birne Binegar: I get zero.
Birne Binegar: Not one, so this new vector doesn't satisfy the condition to live inside this subset so still a multiplication has taken me outside of S so s not closed.
Birne Binegar: Under scale or multiplication.
Birne Binegar: And that means is not a subspace.
Birne Binegar: You could argue differently, you could also started with say be one equals one zero.
Birne Binegar: Be two is equal to 01 notice it well one squared minus zero squared equals one, so it inside ass in this case zero squared plus one squared equals one so.
Birne Binegar: The vector lives in five this substate when I add them together.
Birne Binegar: I get back to you with to non zero component in one squared plus one squared that's not.
Birne Binegar: Equal wanted to rather than one and so.
Birne Binegar: One plus B to it's not an F, and so I can either show that s is not closed under scale and multiplication that would be sufficient.
Birne Binegar: To do show that it's not a suspect or I can start off with two vectors that live inside, yes, show that when I add them I get it back to that lives outside that and then I also proved that it's not a subspace.
Birne Binegar: So you don't necessarily have to prove both if soon as you find that this property is not be either property is not satisfied, then you know it's not a subspace.
Birne Binegar: Alright, so.
Birne Binegar: A problem like to on the exam.
Birne Binegar: may look unfamiliar with it, I don't know if there were any web assign problems that have you do this explicitly, but it should be simple enough to do.
Birne Binegar: And then let's see.
Birne Binegar: Let me go to.
Birne Binegar: The practice exam.
Birne Binegar: Another problem might look unfamiliar at first.
Birne Binegar: it's this third problem where.
Birne Binegar: I asked you to find a basis for subspace but that that subspace it's not stated as a row space a column space or no space of a matrix, nor is it related to any particularly in your transformation.
Birne Binegar: How do I find a basis for a subspace I create by simply taking a set of vectors and taking all possible linear combinations of them well.
Birne Binegar: The idea is that.
Birne Binegar: Even though.
Birne Binegar: The subspace is not the row space of a matrix we can interpret it that, as long as we return.
Birne Binegar: The three doctors is row vectors so what you doing a problem was that problem three.
Birne Binegar: What you do is you form.
Birne Binegar: The matrix.
Birne Binegar: hey what i'll use for its rose is the given list of vectors.
Birne Binegar: 11111 minus two one.
Birne Binegar: and
Birne Binegar: 303 I guess.
Birne Binegar: Because.
Birne Binegar: The row space.
Birne Binegar: is exactly the same thing as the span.
Birne Binegar: Of these three vectors oh i'm in your shoes use the location of the problem.
Birne Binegar: it's going to be exactly equal to w just the span of these three row vectors so what I do to answer problem three is I find a basis for the Rose face of this matrix meaning, I wrote it gives it to a rational on form and I grabbed the non zero row So the answer.
Birne Binegar: A question.
Birne Binegar: By reducing the matrix.
Birne Binegar: And grabbing the non two rows.
Birne Binegar: So in the solutions like done that, like if I.
Birne Binegar: got my solution here.
Birne Binegar: That matrix.
Birne Binegar: Really rapidly.
Birne Binegar: reduces to this.
Birne Binegar: 1110 minus 30000 I think I think just to row operations to clear out this pivot you're clear out this pivot and I think you're already at this matrix this is enroll Echelon form.
Birne Binegar: And that's efficient you really only need to reach real I saw on form to get a basis for the aerospace and so here you're not into a rose.
Birne Binegar: 1110 minus three zero, and so a basis for this space it's going to be the same thing as the base basis for the space of this matrix which is just going to be.
Birne Binegar: This way the span of 111 and.
Birne Binegar: Zero minus three zero tho those are your basis factors.
Birne Binegar: All right.
Birne Binegar: let's see I guess the.
Birne Binegar: Next problem on exam let me see, I have eight minutes left.
Birne Binegar: Let me skip this problem, what we may come back to this one, and yet I want to do.
Birne Binegar: yeah I guess problem problems, five and six.
Birne Binegar: Show.
Birne Binegar: going from an m dimensional space.
Birne Binegar: To an n dimensional space.
Birne Binegar: X equals.
Birne Binegar: X.
Birne Binegar: Is a linear transformation.
Birne Binegar: All you have to do is make sure that the properties of a linear transformation hold when a function is defined by matrix multiplication so, for example to show that he Lambda X is equal to Lambda tx.
Birne Binegar: that's what we'd want to check.
Birne Binegar: That is compatible with scale or multiplication so what you would do is you calculate both side, this would be key times Lambda or athlete what the function does is it does matrix multiplication so it's gonna be a time clamp X right hand side, we have a Lambda times X.
Birne Binegar: Right so did the matrix multiplication before I apply the skill of multiplication or do the skill of multiplication before I multiply by the matrix but either way this is true, because matrix multiplication always can use to scale a multiplication.
Birne Binegar: And second of all you want to show that t of X Plus, why does that equal, the same thing as to X plus t of why.
Birne Binegar: Well, just write down both sides, this is going to be a matrix a apply to the summer vectors Expo that's why, and here we have a applied to X.
Birne Binegar: And we're adding to it a times, why Well, this is also true because matrix multiplication always distributes over sums of vectors that's true and now.
Birne Binegar: I following the chain of equality, she got it if you prove the identity and you won't have to do so, this problem is just it just writing down.
Birne Binegar: Using this formula writing down both sides of the equation, and then observing and it's true.
Birne Binegar: Okay.
Birne Binegar: All right, and then.
Birne Binegar: You do explicitly the last problem.
Birne Binegar: Here the last problem on the exam were given a linear transformation that.
Birne Binegar: going from a four dimensional space.
Birne Binegar: To a two dimensional space is taking a vector with component X one X two X three x four and it sending it to.
Birne Binegar: A new vector with two components, the.
Birne Binegar: X one.
Birne Binegar: minus X three.
Birne Binegar: X four, I think.
Birne Binegar: And then.
Birne Binegar: Two X three.
Birne Binegar: minus two X.
Birne Binegar: Two X two.
Birne Binegar: minus two X three.
Birne Binegar: X four.
Birne Binegar: In this problem, which I guess is number six.
Birne Binegar: or three sub problems monitors we're gonna find the matrix associated with this linear transformation and then we're going to find bases for drange and its nose and it's Colonel so here's the matrix.
Birne Binegar: What i'm gonna do is i'm going to apply this function that each of the standard basis vectors for our for.
Birne Binegar: Okay, so what happens when I choose.
Birne Binegar: The first standard basis factor, he one where if components zero only the first component is non zero when I apply it to this vector okay.
Birne Binegar: One in the first component plus zero Plus zero and the second component.
Birne Binegar: i'm going to get zero minus two times zero Plus zero i'm gonna get back alright so very important here, even though it looks like I should be.
Birne Binegar: Seeing where this vector goes after substituting in numerical values and computing the resulting factor, the thing that you have to remember to do on the exam it's the right result as a vertical vector not as a horizontal director.
Birne Binegar: Now we can do the same for each of the other standard basis vectors the second one will be what X two is one and the other components are equal to zero, so what will get zero minus 02 and in the second component right below it's going to be two minus zero Plus zero.
Birne Binegar: third component it's going to be where X three is the only non zero entry, what will get is let's say minus one, and the first component minus two in the second component.
Birne Binegar: And then we have the last standard basis factor we're only export is non zero and we'll get to because X whores one in this case and one.
Birne Binegar: Alright, so.
Birne Binegar: that's how rapidly, you can calculate the associated matrix.
Birne Binegar: To complete this problem you're asked to find basis.
Birne Binegar: For.
Birne Binegar: The range of that linear transformation and you're also asked for a basis.
Birne Binegar: For the call up for the.
Birne Binegar: Colonel that linear transformation.
Birne Binegar: All right now, what happens is the range is the same thing if the column space of the associated matrix and the Colonel of the linear transformation is the same thing as a column space.
Birne Binegar: Of the associated matrix so the finish the problem we're gonna have to reduce this matrix to reduce show Echelon form and then from that calculation grab.
Birne Binegar: The vectors corresponding to a basis for the columns base and the vectors corresponding to a basis for the sorry this should be the north face matrix.
Birne Binegar: let's do that real quickly, I think this matrix.
Birne Binegar: quickly going to reduce 212 minus one to simply divide this.
Birne Binegar: I do, that correctly.
Birne Binegar: I shut shut zero here right yeah uh yeah cuz X two doesn't occur.
Birne Binegar: In the first component here so that should have been zero there alright so that's really nice, because now, as soon as like divide the second row by 201.
Birne Binegar: minus one one half i've got it sorry this is zero here i've got a matrix and we do show Echelon for.
Birne Binegar: That equal to reduce ro s one form okay so as I promise on the exam the robe reduction, you should be really simple, but now.
Birne Binegar: define a basis for this column space.
Birne Binegar: We look at where the columns live in the reduced relational on form.
Birne Binegar: Or look where the pivots live I guess it's these two guys.
Birne Binegar: let's call them in this column they both contain on leading on zero entries in a row So these are the pivot columns I go back to the original matrix and I grabbed the corresponding columns so what i'm going to get here is.
Birne Binegar: The basis for.
Birne Binegar: Calm face and that's going to be the same thing as.
Birne Binegar: Here the pivots go back to the original call my grab that guy.
Birne Binegar: One zero and here's a pivot go back to the original matrix and grab the corresponding column 02.
Birne Binegar: All right, and there's your basis.
Birne Binegar: For the north face what we have to do is once we've reached this matrix and reduce cholesterol on form we still have to solve the linear system, a subscript key X is equal to zero.
Birne Binegar: But we've done most of the calculation we've done all the rowing reduction, we need to do this first row tells me that X one minus X three plus two X four has to equal zero and the second row tells me that X two minus X three.
Birne Binegar: plus one half X four equals zero.
Birne Binegar: that's going from.
Birne Binegar: The augmented effectively the augmented matrix of the solution to the the actual equations of the solution, then we also use this matrix to interpret.
Birne Binegar: The variables that correspond the columns with pivot as fixed variables that can be expressed in terms of the remaining variables X three x four, so what you do as we did on the first exam as you use the equations to express fix variables in terms of three variables.
Birne Binegar: To it's going to equal X three minus one half X four and then, once you have these equations you write down a solution, better.
Birne Binegar: And that solution vector is going to be X equals.
Birne Binegar: Well, for X one we can use.
Birne Binegar: X three minus two X or for the second component, we can use X three minus one half X or and then the third and fourth component to our free variables.
Birne Binegar: last step.
Birne Binegar: is then to expand this solution vector in terms of the free parameters.
Birne Binegar: 1110.
Birne Binegar: plus x four.
Birne Binegar: Times minus two minus one zero and one then here are your basis factors.
Birne Binegar: There basis vectors because first of all, every solution vector can be written as a linear combination of those factors, and second, of all these factors are linearly independent because.
Birne Binegar: A corresponding to this variable X three is this component here, and the first factor that can't be canceled with corresponding entry in the in the second vector so the only way I can get the third component.
Birne Binegar: Of a vector some like this the equals zero is by taking X three x equals zero.
Birne Binegar: and, similarly, in order to get the fourth component to banish when I do this, some I have to take X is equal to zero.
Birne Binegar: Well, this linear combination only produces the zero vector when the two coefficient or equal to zero so they're linearly independent, so I got two vectors that generate the entire space and they're linearly independent to their basis.
Birne Binegar: OK, so the main thing you're going to need to know how to do, besides the definitions, to be able to reduce a matrix and those will be simple calculation, but then knowing what to do with that matrix in the end you do different things for different sub spaces.
Birne Binegar: Okay, well, good luck studying this weekend, and I will see you on Monday.