Birne Binegar: Okay sorry sorry about that delay let's see here.
Birne Binegar: There we go okay sorry for the technical issues here we're back up and running okay so.
Birne Binegar: subspace is a.
Birne Binegar: Once you're living in a vector space and subspace it's going to be a special set of subsets that closed with respect to the fundamental vector space operations scale multiplication and vector addition.
Birne Binegar: And then.
Birne Binegar: we've uncovered that there's two basic prototypes for subspace one way, you can make us up spaces like taking it in a bunch of factors and then considering.
Birne Binegar: All possible linear combinations of those factors now fights very construction it's going to be closed under these two operations, because these two operations are used to make every single vector in that set.
Birne Binegar: Until and back this is the first type of subspace where we actually have.
Birne Binegar: achieved finding a method for finding a basis for it, but the basis in general is not going to be the original set of vectors.
Birne Binegar: Usually you're going to have too many vectors in that set, and so, if you have a sore too many vectors describing your subspace your goal will be to eliminate some of the vectors so you get a good set that's capable of giving you good coordinates.
Birne Binegar: A second way of producing a subspace is like solving linear equations homogeneous linear equations we're on the right hand side, you have the zero vector.
Birne Binegar: Alright, so what we discussed on Mondays lecture was how to find a basis for subspace of this first type, so let me remind you quickly what a basis is it's a special set of vectors.
Birne Binegar: All living inside the subspace but it's it's relatively small set you only want to have as many vectors that you need to reach every point of this.
Birne Binegar: Of this subspace So if you have a set a list of vectors.
Birne Binegar: Maybe four or five depends on the actual size of the subspace then we'll say you have a basis if.
Birne Binegar: whenever you have a vector that lives inside that set that you can express it as a linear combination if you're given list with unique coefficients see one two ck so what's.
Birne Binegar: What utility of the basis is because it gives us coordinates what are our coordinates or vectors that live inside of subspace what we do is we use these unique numbers.
Birne Binegar: To specify a given vector that lives inside of subspace and those will be good coordinates two vectors will be different if they have different coordinates.
Birne Binegar: But the coordinates are somehow a representation of the vector living and living in a different subspace the factors if essence the subspace of, say, are 10.
Birne Binegar: Each of these doctors is going to be a 10 dimensional vector but I may only need K parameters, maybe K, is for maybe only need four parameters to reach every coin in that subspace well that's the utility of the basis that gives you a.
Birne Binegar: Smaller set of vectors that still live in the subspace are capable of reaching every point in the subspace.
Birne Binegar: and capable of providing you with good coordinates for every point in the subspace.
Birne Binegar: Alright, so here's the picture I want you to keep in mind the three dimensional boxes you're supposed to imagine issue general vector space space for all the vectors live.
Birne Binegar: These blue lines represent an arbitrary arbitrary set of vectors living in this space that's confined to this flat plane, which is have to represent a subspace sitting inside our end.
Birne Binegar: On this flat surface, we got a whole bunch of actors, but the basis sectors are going to be even more special they're going to be.
Birne Binegar: In this example there's only two of them, because there's only two independent directions inside this.
Birne Binegar: The subspace now you don't necessarily have to take these red vector as your basis vectors actually almost any to have the original set effect you're.
Birne Binegar: still provide you with a basis we're trying to do, though, is getting rid of all the superfluous actors and so that we end up with good core.
Birne Binegar: Okay So how do we find a basis for the case where we have just given a list of vectors and we want to find a minimal set of vectors that are capable of reaching every point in this space and such that the.
Birne Binegar: Every point in the space will be uniquely represented by a linear combination of them well what you do, as we saw on Monday is we take these vectors and we write them as the rows of the matrix.
Birne Binegar: Then we reduce that matrix to a row Echelon form, so we get the pivots occurring in a downward staircase on the upper left to lower right.
Birne Binegar: Then what happens then is those row vectors even in the row Echelon form continue to generate the entire.
Birne Binegar: Space it's generated by the original vectors but if you're in the course of role reaction you produce some zeros there's gonna be some.
Birne Binegar: vectors and aren't really in the relational and phone that aren't going to contribute the zero vector aren't going to add anything to a linear combination.
Birne Binegar: So this process of rubber ducky and jelly what what it's doing it's it's removing the redundant directions in the original set and reducing things down to a minimal set of still capable of generating the entire subspace.
Birne Binegar: And so, when we find a basis for a list of vectors if we write the factors as a Rosa the matrix reduced to if relational on form.
Birne Binegar: Then the non zero rose will be a basis, not only for the road space of the row excellent form you'll be a basis for the road space of a, which is the same as the original space we started off in.
Birne Binegar: Okay, so all doing examples of this.
Birne Binegar: Towards the end of the lecture one.
Birne Binegar: I want to talk about the other case of a basis and how we find.
Birne Binegar: Other case of a substation how we find a basis for that and then i'll do not a lot examples, at the end of today's lecture.
Birne Binegar: Okay, so that's when your subspace is presented to you as a span of a list of vectors What if you're given a subspace I specifying it as a solution set of a homogeneous linear system.
Birne Binegar: Here we're going to get a set of vectors living inside our and it's going to be a subspace, how do we get a basis for that.
Birne Binegar: Well, it turns out that if you know how to solve this equation, you know how to find the basis as well.
Birne Binegar: And what i'll do is rather than trying to explain this formula, let me give just give you an example minute you how you solve set systems and then i'll show you where the basis POPs out.
Birne Binegar: Okay, so let's look at this system of to equate comes in for.
Birne Binegar: The homogeneous system, because on the right hand side, we have only zeros.
Birne Binegar: here's the coefficient matrix when I pen to it a column vector form from the numbers on the right hand side, I get the augmented matrix at the system we saw the system by reducing this augmented matrix to.
Birne Binegar: Well it's reduced so Echelon form so i'm not going to do the explicit row reduction first thing mom would have done, maybe subtract the first row from the second row that clear out that pivot but they're made it may have been other operations as well.
Birne Binegar: Reducing the original augmented matrix to reduce relational on form it's the first step in finding the solution set to the system.
Birne Binegar: Once you reach reduce relational inform you convert back to equations get this guy is effectively the augmented matrix of the solution.
Birne Binegar: So from this first row we see X one plus x for half the equals zero or simply X, one has to equal minus X, for the second row is telling us that X one minus X three is equal to zero, or the X is equal to three what I did here is.
Birne Binegar: I thought that the equation correspond to each line and this augmented matrix but i've also made a distinction between three variables.
Birne Binegar: and fixed variables So here we have a pivot here we have a pivot so that means X one X two the two variables that correspond to those columns these are going to be fixed.
Birne Binegar: And meanwhile the third column, and the fourth column, corresponding to variables X three x four, these are going to be the free variables in the solution.
Birne Binegar: don't in fact that's why I use the equations coming from the augmented matrix and reduce all Echelon form well I use those equations to express the fixed variables in terms of the free variables.
Birne Binegar: Alright, so next step is then to write down a typical solution vector what i'll do is I substitute where I can where the fixed variables so in the first slot I right minus X four instead of X one second slot I use X three instead of X two.
Birne Binegar: And then the last step here is to take this.
Birne Binegar: This vector and see how it could sit on a hyper plane, well, here it is you just expand in terms of the three parameters.
Birne Binegar: So you don't have X three has no component and the first slot one coefficient one in the second slot coefficient one, and the third slot know coefficient last slide similarly for X for wiki pull that out, and now, what we have is every vector.
Birne Binegar: Every vector that lives in the solution that's going to lie in the span of these two vectors right, these are X three x four arbitrary numbers so by very nice vector.
Birne Binegar: These numbers like it every vector that would occur in the set of all possible linear combinations of these two constant vectors.
Birne Binegar: Moreover, not only is every solution vector linear combination of these two vectors notice, also, that if you set the only way, you can have the solution vector equal to zero vector is by having.
Birne Binegar: X three equal to zero and X four X equals zero see unless the X three is equal to zero.
Birne Binegar: This some of these to this linear combination of two vectors it's gonna be non zero because there's nothing in the third slot that can be used to cancel the X three times one.
Birne Binegar: And so, if the sun is going to equal zero then we're going to need X equals zero, and similarly looking at this last component.
Birne Binegar: A multiple this vector plus a multiple this vector it's not going to equal zero unless X force equals zero because there's nothing here that can cancel that one.
Birne Binegar: So what what shifted observed from this is not only as soon as we represent the solution of this linear system as a hyper plane.
Birne Binegar: We get it presented as living in the span of a particular set of vectors and, moreover, those vectors are literally independent.
Birne Binegar: Because we can't add them and produce the zero vector without taking both coefficients equal to zero, and they generate the entire.
Birne Binegar: set of solutions, so we have a pair of factors that are literally independent and generate the entire subspace they were basis, so, in fact, these guys these.
Birne Binegar: Constant vectors that we've been finding all along, since ever since we started solving homogeneous linear systems turns out that those constant vectors are exactly the basis vectors the basis vectors for the solution set.
Birne Binegar: So.
Birne Binegar: So the answer to the problem post.
Birne Binegar: find the basis for the solution side of this equation, well, we did have to solve the equations to do it, but as soon as we reach the solution set expresses a hyper plane, then we ought we simultaneously found a basis for the entire solution set.
Birne Binegar: Okay, and it's just those these two factors here I stay our basis.
Birne Binegar: There exactly the vectors that occur when we expand a solution vector in terms of the free parameters.
Birne Binegar: Okay, so.
Birne Binegar: I had to basic prototypes house sub spaces occur either fan of a set of vectors or or solution set of a homogeneous linear system.
Birne Binegar: I also on Monday talked about substations that we're taught attached to a an arbitrary n by n matrix and.
Birne Binegar: The base does still have the same basic prototype we have the row space of the matrix that's defined as well the vectors subspace you get by taking all the road vectors of the matrix.
Birne Binegar: and taking all possible linear combinations of them, though it's a subspace up the first time, and so is the column space.
Birne Binegar: The common space.
Birne Binegar: is also formed by instead of working with the road vectors of the matrix you work with the column vectors and you.
Birne Binegar: find all possible linear combinations of these guys that's going to create a sub space.
Birne Binegar: And then the third subspace that we can attach to a given matrix is going to be well you use the matrix as a coefficient matrix of a homogeneous linear system and the subspace then attached to a is the solution set of the corresponding set of equations.
Birne Binegar: Okay, so what i'm going to do next is tell you how you find bases for these three sub spaces well.
Birne Binegar: First, one is just a span of a set of vector So what do we do well.
Birne Binegar: Normally we do is we write the vectors as a rose of the matrix reduce and then grab the non zero rose at the rational and born, but in this case.
Birne Binegar: The road vectors of a are already the road vectors of a so I just wrote reduce a to run roof it's real actual form or any form and I grabbed the non zero rose that will provide being a basis for the row space.
Birne Binegar: To find a basis for the last days well what we do have to do is solve the linear system X is equal to zero expand our solution set in terms of the free parameters and then grab the.
Birne Binegar: The concept vectors that are being multiplied by the three parameters of the solution, as an on the last slide.
Birne Binegar: Then we have the last one, which is a basis we if you want to find a basis for the common space of a matrix well.
Birne Binegar: Unfortunately.
Birne Binegar: About a theorem them about the state you'll have to do with a completely separate reduction in fact what I hope you notice is that if you know you need to find bases for.
Birne Binegar: These two substations you only need to do one reduction right just row reduce a all the way to reduce your national on form, then you will be able to solve.
Birne Binegar: This and get the basis vectors for the null space, but you'll also having produced them.
Birne Binegar: The real reduce your rational on form is still going to be in row Echelon form so like he also grabbed the non zero rose up the redo matrix in reduce your isilon porn and use that as a basis for the row space to pay, but the column space, we have to treat a little bit differently, because.
Birne Binegar: If a is equal to.
Birne Binegar: A one one a one m a n one and and.
Birne Binegar: And just putting in the corner entries of this matrix.
Birne Binegar: Where we found a basis for the row space, the span of all the throwback jerseys was by reducing this to a row Echelon form but that.
Birne Binegar: Well, we can off it doesn't tell us anything about the column the span of these COM vectors and but what you could do.
Birne Binegar: You find a basis for the columns spaces, you could take the columns and convert them into rose by doing the transpose operation, so this first column a one one through a and one.
Birne Binegar: I taking this column i'm now writing it as a row i'm taking the last column in writing it as the last row of new matrix so this one and a and m.
Birne Binegar: yeah.
Birne Binegar: Alright, so that's I take a matrix I convert it to a transpose but now the column space.
Birne Binegar: Of the original matrix it's just the same thing it's the row space.
Birne Binegar: Of a transport transpose up the matrix so what I would have to do, I can still it's still a span it's still subspace created by taking all possible linear combinations of a given list of factors.
Birne Binegar: But our procedure for finding the basis for Spanish, the right to guys's arose in the matrix that's exactly what i'm doing here, and so what you would do, then, is you would take this transpose matrix.
Birne Binegar: And you put it in its row Echelon form.
Birne Binegar: And then you grabbed onto your own rose.
Birne Binegar: Once you get this matrix in row Echelon form a non zero rows are going to be a basis for the span of the Rose, which is the span of the original columns of the original matrix.
Birne Binegar: All right now.
Birne Binegar: That means that.
Birne Binegar: I now have the means to find bases for each of these three sub spaces, but it involves two separate reduction right.
Birne Binegar: Reducing a to its reduce relational on form it's going to give me.
Birne Binegar: This basis for the road space it's also going to give me a basis for the no space of a matrix but the column space.
Birne Binegar: To get that at this point of the game, we have to reduce the different matrix, we have to take the transpose of the matrix before we do the role reduction in order to get the basis vectors.
Birne Binegar: Alright, so i'm going to improve upon that next slide.
Birne Binegar: here's the theorem.
Birne Binegar: If you have any n by n matrix than a basis for the column space can be a pain Now this is going to be different we're gonna not reduce the transpose of original matrix i'm going to reduce the original matrix itself to intro Echelon form.
Birne Binegar: But what i'm gonna do is once I reach this rule Echelon form it's now different i'm going to figure out which columns of the real Echelon forum have pivots.
Birne Binegar: And then, what I do is I go back to the original matrix and I grabbed the corresponding calls.
Birne Binegar: And then, what will happen is that the.
Birne Binegar: yeah.
Birne Binegar: Let me give you an explicit.
Birne Binegar: Suggesting here is that if you had a matrix.
Birne Binegar: A.
Birne Binegar: let's say the row Echelon form.
Birne Binegar: Of a once you've got it through Echelon form say suppose you end up a matrix look like this.
Birne Binegar: To one maybe something like that I don't know i'm suppose at your role Echelon form well the statement and let's suppose the original matrix was 3111210001 okay so.
Birne Binegar: Suppose.
Birne Binegar: Suppose when you reduce this matrix you get this i'm not going to work with this matrix directly until the very end.
Birne Binegar: What you do to find a basis for the columns space.
Birne Binegar: You take the original matrix you reduces to each row I will inform and you figure out where the pivots live.
Birne Binegar: Alright, so I got a pivot first call the third column, and the fourth call, what I do you get a basis for the column space is I go back to the original matrix and I grabbed that first column I grabbed the third column, and I grabbed the fourth column, and then.
Birne Binegar: The statement would then be having done this role reduction, I can now conclude that the column space of my matrix it's going to have.
Birne Binegar: A basis for that it's going to equal.
Birne Binegar: Here I have a pivot I go back to the original matrix and grab the corresponding call 310.
Birne Binegar: Here you have the pivot in the row excellent form, I go back to the original matrix and grab the corresponding call so that would be 211 and then here's a pivot in the role Echelon form, I go back to the original matrix and I have 102.
Birne Binegar: And then that would be how you get a basis.
Birne Binegar: For the column speak to this matrix.
Birne Binegar: Just using the row reduction of a not the reduction of its transpose but the reduction of a itself.
Birne Binegar: Reduce throw Echelon form figure out where the pivots live but remember to go back to the original matrix at those are the call and you're going to grab.
Birne Binegar: Okay.
Birne Binegar: Alright, so what i've been described in is how to find bases for all three sub spaces associated with a matrix like doing a single row reaction by reducing your given matrix all the way to reduce show.
Birne Binegar: us the non to Rosa that reduce your national on form as a basis for the row space use the columns.
Birne Binegar: The columns of the original matrix that correspond to the columns of The reduced Echelon born to have pivots.
Birne Binegar: Use columns of The those columns but the original matrix to get a basis for that conversation, the original matrix.
Birne Binegar: And then, finally, you also use to reduce relational on form up your matrix a to write down the solutions of this and then grab the constant vectors that appear as you present the solution as being points on a hyperlink.
Birne Binegar: The constant vectors be multiplied by the three variables in your expression solution, those are going to be your basis sectors.
Birne Binegar: OK now let's do some examples.
Birne Binegar: here's a matrix to three rows and columns.
Birne Binegar: And I won't do the calculation, but my claim is is that if you apply elementary row operations this matrix it's reduced real Echelon form it's going to be exactly this this matrix notice, we have a zero row with the bottom here.
Birne Binegar: That that essentially means there were some way of picking linear combinations of the rows of a and producing a zero vector.
Birne Binegar: That tells you that the original rows are not linearly independent rather they were linearly dependent and we've removed dependence relation effectively month.
Birne Binegar: or down to a pair of factors which are still going to generate the row space of a but now we have a smaller list of generators, in fact, there are a basis for the growth space and.
Birne Binegar: So, all we have to do to get a basis for the roast base of a Israel reduce it to a row Echelon form or even a very neutral Echelon form grab the two non zero rose, and that will be a basis for the row space.
Birne Binegar: What about the column space, the column space where you find that is you look at the matrix enroll action long form and you figure out where the pivots live alright, so they live in the first column in the second column, so I go back to the original matrix.
Birne Binegar: And I grab I hope it's.
Birne Binegar: Showing on screen here I grabbed that first column, because there was a pivot in the relational important and I grabbed the second column it's a second basis factor, and now I have a basis for the column space.
Birne Binegar: The last subspace was the solution set.
Birne Binegar: Of the corresponding linear system.
Birne Binegar: So.
Birne Binegar: From this matrix in reducing national on form, you can see that.
Birne Binegar: X one is going to have to equal minus X three minus two X for that coming from the first row and X to it's going to have to equal.
Birne Binegar: To X three plus two X four i'm writing down the equation, then moving taking the terms that correspond to the free variables i'm moving them to the right hand side.
Birne Binegar: And then.
Birne Binegar: Then I write down a solution vector so I guess the next step that I skipped here in the slide would be the right solution vector well for the first component is going to be minus X three minus two X for.
Birne Binegar: The second component is going to be two X three plus two X, for the third component and the fourth component there free variables, so I don't change them and then I expand in terms of the free variables so i'll have.
Birne Binegar: X three times minus 1210.
Birne Binegar: plus x four times minus two in the first slot to in the second slot nothing in the third slot one in the fourth slot.
Birne Binegar: And you see now we've got every point in the solution set expressed as a linear combination of these two vectors so the solution set is the span of these two vectors and, moreover, these two vectors are linearly independent why.
Birne Binegar: Well, corresponding to this X three.
Birne Binegar: grandmother, here we have this non zero entry here and there's nothing over here that can cancel it and simulate corresponding to X four.
Birne Binegar: We have an entry in that vector that cannot be counseled by anything from the first factor until there's no way I can get these two vectors to.
Birne Binegar: linear combinations of these two vectors that equal to zero vector without taking the both X three and the parameter X force separately equal to zero, which means these vectors are linearly independent.
Birne Binegar: Alright, so the upshot of this is that once you've written down the hyper planes solution for the linear system you've also identified basis sectors for that solution set it's the constant vectors that occur when you expand the solution in terms of the free parameters.
Birne Binegar: Okay, so.
Birne Binegar: neck example I gave you a matrix and I found bases for all three of the attached sub spaces.
Birne Binegar: Alright, so the next thing I want to talk about.
Birne Binegar: we'll do we'll do some more examples later in the lecture I think is.
Birne Binegar: Some important facts about they see.
Birne Binegar: The first one is is that.
Birne Binegar: Every subspace except.
Birne Binegar: The zero subspace has a basis, I mean it's true you can maybe think of the zero vector is the basis element, but actually.
Birne Binegar: The zero vector is not even linearly independent of itself right by only had a single vector and I wanted to know if it was linearly independent the question would be, can I solve X one V one like well there's no other vectors to add to it, can I solve this equation.
Birne Binegar: Without taking X one equal to zero, and of course you can, if.
Birne Binegar: If D is just zero vector then, no matter what I multiply zero by i'm still going to produce a zero vector So if you have a subspace that consist of only the zero vector that's the one case where you don't have a basis, but in any other case where you have.
Birne Binegar: Multiple non zero vectors as a basis, then every subspace is going to have a basis.
Birne Binegar: So long as it's not.
Birne Binegar: The single point at the origin.
Birne Binegar: And then.
Birne Binegar: Second theorem about the bases, is that if you have a basis for subspace you could have different bases, all you really need.
Birne Binegar: For implicate a basis for the vectors that live on this blackboard or two vectors that hit head off an independent directions.
Birne Binegar: And then we only have two vectors i'll be able to reach every point on the blackboard by taking linear combination these guys geometrical you just.
Birne Binegar: corresponds to realizing this point as a corner of a parallelogram that I formed from us from these original vectors I could have started off with.
Birne Binegar: Two different directors as well, and these two vectors so long as they're pointing in independent directions on the blackboard are still going to provide me move the basis.
Birne Binegar: So you have many choices of bases many choice of directions that are sufficiently independent, you can reach every point in your subspace sheet by a unique linear combination, but what even though you can have multiple bases, they all have to have the same number of vectors.
Birne Binegar: And i'm.
Birne Binegar: combining these two facts that every vector that.
Birne Binegar: Point at the origin has every subspace except.
Birne Binegar: The origin alone as a basis, and the fact that every basis has the same number of vectors that gives us the definition of dimension.
Birne Binegar: The dimension of the sub base is the common number guaranteed by the second theorem a common number of basis vectors in any basis for your subspace.
Birne Binegar: Then, in fact, this even works when our sub spaces, the zero vector.
Birne Binegar: So.
Birne Binegar: The you have a sub space that's consistent only the zero vector it has no basis vectors so by this definition, its dimension is zero.
Birne Binegar: Okay, so that that sort of justifies what i've been saying all along.
Birne Binegar: That.
Birne Binegar: Individual vectors have dimensions zero, but when you create such a vectors by taking linear combination that's where you get infinitely many vectors and that's where you're dealing with them non trivial subspace.
Birne Binegar: Okay, so.
Birne Binegar: let's see if we can count the dimensions of Sub spaces, that we can attach to a matrix so let's look at the roof space first way we get.
Birne Binegar: The basis factor for the row space is we reduce the matrix to a row Echelon form and we grabbed the non to rows so the dimension is going to equal, the number of basis vectors you need to create the subspace.
Birne Binegar: Well, in this for the Rose face we just basis vectors correspond to the non zero rows of a row Echelon form but put another way, if you have a non zero row in a row excellent form.
Birne Binegar: Arbitrary matrix maybe here it's enroll Echelon form, so we might have a non serial number here, but as soon as you have a non zero row, you have a non zero first term until it's going to have a pivot and so, basically, you can count.
Birne Binegar: The number of faces factors for the roast faced by counting simply counting the number of pitch you see in the role I salon form they're just gonna I mean you're effectively just counting number of non zero rows but you're also simultaneously counting the number of pivots.
Birne Binegar: On the other hand, the columns face well it's dimension is going to be the number of faces factor if you need to produce that some space, but the way we do, that is.
Birne Binegar: For the column space we wrote do some matrix to it through Echelon form, and then we grab the columns of the original matrix that correspond to columns of The roadshow on form that have pits.
Birne Binegar: So, although I don't use the row Echelon form as basis vectors as far as counting the number of basis vectors they're still good in the row Echelon form.
Birne Binegar: Say that's true excellent form of a let's say a some other matrix 132211000 same size as the original matrix but what this would tell us that the first basis factor for the column space of a it's going to live in the first column so so this would be the first faces factor.
Birne Binegar: For the column space of a and then whatever I guess the next 1234 so I have to go over four and then I better put another column there, and this would be the second.
Birne Binegar: basis factor, because if you go to the roadshow on form you figure out where the pivots live and you go back to the original matrix to get the basis factor.
Birne Binegar: But what you're doing, though, you have one basis one vector over here for each column in the row exelon form that has a pivot and so, for each pivot you have a basis factor.
Birne Binegar: So what that tells us you look at these last the qualities, the dimension of the roast database equal, the number of pivots in a row Echelon born the dimension but calm space of the matrix it's the same it's the same as the number of pivots in a row Echelon on.
Birne Binegar: The Left, let me show you why that's a little bit surprising.
Birne Binegar: If you have a matrix a with.
Birne Binegar: him columns and rows.
Birne Binegar: Then the row space.
Birne Binegar: up a what where you sit live.
Birne Binegar: Each row has em components so it's going to be a subspace.
Birne Binegar: of an n dimensional vector space meanwhile.
Birne Binegar: The call and space of a.
Birne Binegar: Well, if you look at the column each column is going to have.
Birne Binegar: End components so we'll take the span of all the columns we're going to get a subspace of our end so the row space and column space or sub spaces, but they live in different vector spaces.
Birne Binegar: So that's why this has a little bit of a miracle this vector space row face lives inside our am the column space live inside another vector stakes aren't and yet the dimensions of those two spaces this to substations are the same.
Birne Binegar: So the columns yeah yeah um so yeah.
Birne Binegar: These two vector space is the vector space it's in which the sub spaces live are different, they have different dimension but.
Birne Binegar: Because they came from the same matrix they actually have corresponding substations also have the same dimension.
Birne Binegar: Alright, so that motivates a definition, I think I give on the next slide which is.
Birne Binegar: The rank of a matrix you have an n by n matrix.
Birne Binegar: Then the rank that matrix is going to be the common dimension, the row space and the columns base of the matrix.
Birne Binegar: What it is it's the number of basis factors, you have for the row space or equal to the number of basis factors, you have for the column space, but both of these two numbers basically a reducible to the number of pivots you have an any row Echelon for so whether you.
Birne Binegar: Using the pivots to and figure out which columns of the original matrix to use its basic factors or you're using the matrix general Echelon form is non to a rose to get basis vectors for the Rose space.
Birne Binegar: In both cases you're going to get.
Birne Binegar: A basis vector one for each pivot that occurs in the row Echelon porn.
Birne Binegar: Alright, so.
Birne Binegar: You now should have the wherewithal to actually begin to do the homework problems that are in.
Birne Binegar: Problems said, what are we up to, I think, six, I guess, so a new website problems that it's been posted.
Birne Binegar: What I want to point out here is that both the textbook and web assign refer to the dimension of the no space of a matrix as senility in a.
Birne Binegar: But what this is course is number of free parameters in the solution set a X is equal to zero or it's equal to the number of columns and the.
Birne Binegar: usual Echelon form that you not have pivots that's equal to number of free parameters, so the novelty of a basically the dimension of the solution, set of a times X is equal to zero.
Birne Binegar: and
Birne Binegar: Let me do this.
Birne Binegar: Again suppose a row uses to.
Birne Binegar: Make 10000111000000 reduces to that, then.
Birne Binegar: uh well, actually, let me.
Birne Binegar: Let me describe the system, the if we had a homogeneous linear system X is equal to zero, and then this would be the reduce row Echelon form.
Birne Binegar: Of that augmented matrix so it'd be the augmented matrix at the solution to this original in your system.
Birne Binegar: we're used to counting.
Birne Binegar: columns without pivots its corresponding to three parameters in the solution.
Birne Binegar: And columns with pivots are fixed variables.
Birne Binegar: So what I wanted to point out on this slide is that not only.
Birne Binegar: The pivot here account to pivots I know that the dimension of the road space of the original matrix is going to be too, and I also know the dimension of the column space of the original matrix is equal to two.
Birne Binegar: Because I have to, I have to pivots and this reduce relational and form, but I also know.
Birne Binegar: That, I have to free parameters in the solution, so my solution set it's going to be two dimensional and, in fact, if you look at how the numbers add up, we have total number of columns as for.
Birne Binegar: total number of columns.
Birne Binegar: And this example is.
Birne Binegar: For so i'm talking about the a pod not the zero column of the augmented matrix so the total number of columns is for these you can divide up into with pivots.
Birne Binegar: And without pivots.
Birne Binegar: And then.
Birne Binegar: If they have pivot then.
Birne Binegar: Depending on what you want to do, you can take any any road, it has a pivot and you can produce a basis factor.
Birne Binegar: For the road space of the matrix.
Birne Binegar: Or you can take a column that contains pivots goes back to the original matrix and grab the corresponding column so pivot in the real actual form is also going to.
Birne Binegar: Lead you to a basis factor.
Birne Binegar: For the column space.
Birne Binegar: Okay, and then the columns about pivots they correspond to three parameters in the solution.
Birne Binegar: And the solution set and.
Birne Binegar: What that tells us then.
Birne Binegar: Number of columns that.
Birne Binegar: Well, the number of pivots you have is actually going to equal the rank so number of columns.
Birne Binegar: With pivots.
Birne Binegar: The rank of a.
Birne Binegar: Number of pivots.
Birne Binegar: that's going to equal.
Birne Binegar: What the book calls Nokia a.
Birne Binegar: it's the dimension of the solution, set.
Birne Binegar: up.
Birne Binegar: A X is equal to zero, and what this.
Birne Binegar: These numbers tell you, then, is that the novelty of the a the rank of a always equals the number of columns a.
Birne Binegar: And so that's what i've stated here.
Birne Binegar: The novelty being the dimension of the subspace we get by solving the homogeneous linear system it's dimension is going to be equal to the number of free variables it just equal to the number of columns about pivot in any real Echelon form.
Birne Binegar: And so.
Birne Binegar: Number of pivots number of columns about pivots equals the number of columns a you always have.
Birne Binegar: This correspondence between the dimensions of the subspecies of a matrix.
Birne Binegar: Whether i'm not talking about the road space of a matrix or the columns face of a matrix it's dimension, plus the dimension of no space of the matrix is always going to be the same as the number of columns and the original matrix so yeah.
Birne Binegar: dimension of space Okay, so our rule for finding the basis well as that, let me give you an example 123011211 suppose that this thing.
Birne Binegar: i'm not going to do the road reduction suppose that thing Rover uses to 0101012 suppose it and maybe 000 down here suppose, this was the.
Birne Binegar: Reduce row excellent form of the matrix so this calculation, where I picked an arbitrary matrix and I calculate it reduce your electron form that's going to give me bases for all three sub spaces, the way you get a basis.
Birne Binegar: For the row space.
Birne Binegar: You look at the rifle on form and you grab the non two rows so you're just going to use 101 and 012.
Birne Binegar: To get a basis.
Birne Binegar: For the call mistakes and here's where your question comes in.
Birne Binegar: The procedure, I told you use to find this is use the same row production calculation, but figure out where the pivots live, but then don't use these columns.
Birne Binegar: This is what you look at first, but the idea is you go back to the original matrix.
Birne Binegar: And you grab those guys.
Birne Binegar: i'm not proving that in this course yeah so it's a trick.
Birne Binegar: You do you do have a way of doing it.
Birne Binegar: But by other things I proved you take them matrix calm and you write them it's rose, and you produce and then you could use that technique to get a basis but.
Birne Binegar: This this method I didn't prove into a class and i'm not going to try, because it's it it's it's hard but it works, it does work the way you get a basis for the column space is you figure out where the pivots live you go back to the original matrix and you grab those calls.
Birne Binegar: And, but because of the way I produce this list of basis vectors I get one for each pivot, and so the counting States the same.
Birne Binegar: think it may be kind of abstract.
Birne Binegar: i'm thinking is true if you have a set of linearly independent vectors and they generate the same subspace they're going to be a basis.
Birne Binegar: And so I think well, not the way the proof goes issue show that the vectors that you find this way are linearly independent and they generate the column space.
Birne Binegar: But, and so you have to show two things you show that one that you can't add these vectors and and.
Birne Binegar: come up with the zero vector without setting all the coefficients equal to zero and separately, that the show that the directions are enough to reach every point in the in the span of of the original set of factors.
Birne Binegar: But we're not going to do that, but yeah it's done more abstractly and in fact I think you do see the proof, if you take advanced linear algebra.
Birne Binegar: Okay i'm so i'm sorry i've run a couple minutes over.
Birne Binegar: point out that.
Birne Binegar: Okay.