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Current time:0:00Total duration:13:40

Showing that the candidate basis does span C(A)

Video transcript

two videos ago we asked ourselves if we could find the basis for the column space of a and I showed you a method of how to do it you literally put a in reduced row echelon form so this vector this matrix R is just the reduced row echelon form of a and you look at its pivot columns so this is a pivot column as a one and all zeros this is a pivot column one in all zeros and the one is the leading nonzero term in its row and this is a pivot column let me circle them these guys are pivot columns and this guy is a pivot column right there and you look at those in the reduced row echelon form of the matrix and the corresponding columns in the original matrix will be your basis so this guy this guy so the first second and fourth so the first second and fourth columns these so if we call this this is a1 this is a2 and let's call this a four this would be eight three and this would be a five so we could say that a 1 a 2 and a 4 are a basis a basis for the column span of a and I didn't show you why two videos ago I just said this is how you do it you have to take it as a bit of an article of faith now in order for these to be a basis two things have to be true they have to be linearly independent and I showed you in the very last video this though after the second in our kind of series dealing with this vector I showed you that by the fact that let's say that this guy is r1 this guy is r2 and this guy is r4 it's clear that these guys are linearly independent they each have a one in a unique entry and the rest of their entries are zero so because all of the other and this is true I'm looking at three pivot columns right now but it's true of the definite we had n pivot columns that each pivot column would have a 1 in a unique place and all of the other pivot columns would have 0 in that entry so there's no way that the other pivot columns with any linear combination of the other ones could never add up to each of them so this these are definitely linearly independent and I showed in the last video that if if we know that these are linearly independent we do that they are given that our has the same null space as a we know that these guys have to be linearly independent I did that in the very last video now the next requirement for basis we checked this one off is to show is to show that a 1 a 2 and a n that a 1 a 2 and a n that their span equals the column space of a now the column space of a is the span of all 5 of these vectors so I have to throw a 3 in there and a 5 so to show that just these three vectors by themselves span our column space we just have to show that look I can represent a 3 and a 5 as linear combinations of a 1 a 2 and a 4 if I can do that then I can then I can say well you know then these guys are redundant these are guys are redundant then the span the span of a 1 a 2 a 3 a 4 and a 5 doesn't need the a3 and the a5 terms that we can just reduce it to this because these guys these guys can be represented as linear combinations of the other three if that's these guys are redundant and if we can get rid of them then we'll show if we can if we can show that these guys can be represented linear combinations together then we can get rid of them and then the span of these three guys would be the same as the span of these five guys which is of course the definition of the column space of a so let's see if we can do that so let's go back to our original our original let me fill in so these each of these column vectors a 1 3 a 5 and then each of these column vectors let me label them r1 r2 r3 r4 and r5 now let's explore let's explore the null spaces again or not even the null spaces let's just explore the equations ax is equal to let me write it this way a instead of X let me write X 1 X 2 X 3 X 4 X 5 five is equal to zero right that's just that's well that this is how we define the solution set of this all the potential X ones through X 5 so all the potential vectors X right here that represents our null space and then also let's explore all of the R times X 1 X 2 X 3 X 4 X 5 s is equal to 0 this is the 0 vector in which case it would have to have 4 entries in this particular case would be a member of RM so these equations can be rewritten I can rewrite this as what were the column vectors of a there were a 1 a 2 3 a 5 so I can rewrite this as x1 times a1 plus x2 times a2 plus x3 times a 3 plus x4 times a 4 plus x5 times a 5 times a 5 is equal to 0 right that's we've learned that was from our definition of matrix vector multiplication and I just rewrote you know this is just a bunch of column vectors a 1 through a 5 I drew it up here a 1 3 5 and I can just rewrite this equation like this similarly I can rewrite this equation as the vector R 1 times X 1 or X 1 times R 1 plus X 2 times R 2 plus X 3 times R 3 plus X 4 times R 4 plus X 5 times R 5 is equal to 0 now we know that when we put this into reduced row echelon form the X the X variables that are associated with the pivot columns are so what are the X variables listed with the pivot columns well the pivot columns are r1 r2 and r4 so that's r1 r2 and r4 that the X variables associated with them they call that we could call them pivot variables and then the ones that are not associated with our pivot columns are free variables so the free variables in this case so x3 and x5 are free variables free free variables and that applies to a the so this solution set all of the sets all of the vectors X that satisfy this equation also satisfy this equation and vice versa they're the exact same null space the exact same solution set so we can also call this we can also call this X 3 and this X 5 is free variables as free variables now what does that mean we've done multiple examples of this the free variables you can set them to anything you want so X X 3 in this case X 3 and X 5 you can set it to any real number you can set to anything you want set to anything and then from this reduced row echelon form we express the other pivot variables as functions of these guys maybe X 1 X 1 is equal to you know ax 3 plus B X 5 maybe X 2 is equal to C X 3 plus D X 5 and maybe X 4 is equal to e X 3 plus F X 5 that's that's almost that comes directly out of the literally multiplying this guy times this equals 0 you would get a system of equations that you could solve for your pivot variables in terms of your free variables now given this I want to show you that you can always you can always construct one of your in your original vector in your vision in your original matrix so if we go to our original matrix you can always construct one of the vectors that are associated with we could call them with the free columns you could always construct one of the free vectors using the linear combination of the ones that were associated with the pivot columns before and how do I do that well let's say that I want to find some linear combination that gets me to this freak on gets me to a three so how could I do that let me rearrange this equation up here so what do I get if I get if I subtract X to a three from both sides of this top equation I get minus I'm sorry that's X 3 a 3 if I subtract X 3 a 3 from both sides of the equation I get minus X 3 a 3 is equal to X 1 a 1 plus X 2 a 2 plus I don't have the 3 there plus X 4 a 4 plus X 5 sorry X is in the vector X 5 a 5 all I did is I this this like a salmon-colored statement here is just another rewriting of this equation right here and all I did is I subtracted this term right here X 3 a 3 from both sides of the equation now X 3 is a free variable X we can set it to anything we want and so is X 5 so let's let's set X 3 is equal to -1 so we set X 3 is equal to minus 1 then this term right here becomes a 1 right because I was minus X 3 and let's set X 5 equal to 0 as X 5 is equal to 0 so if X 5 is equal to 0 this term disappears and I did that because X 5 was a free variable I can set them to anything I want now I've written a 3 as a linear combination of I guess you could call it my my potential basis vectors right now or the vectors that were so this is a 1 a 2 and a 4 a 1 a 2 and a 4 they're the vectors in the original matrix that were associated with the pivot columns now in order to show that I can always do this we have to show that there's for this combination there's always some there's always some X 1 X 2 and X 4 that satisfy this well of course there's always some X 1 X 2 s that satisfy this we just have to substitute our free variables X 3 is equal to minus 3 and X 5 is equal to 0 into this into these equations that we get from our system when we did it with the reduced row echelon form so you get you know I mean you don't I don't know if I even have to go through you in this case you get x1 is equal to minus a plus 0 X 2 is equal to minus C so on and so forth so you can always do that you can get your you can get you can always express the vectors you can always express the vectors that are associated with the non-pivot columns as linear combinations of the vectors that are associated with the pivot columns what I just did for a3 you could just as easily have done for a 5 by subtracting this term from both sides of the equation setting X 5 to negative 1 and setting X 3 to 0 so that the three term disappears and you could run the same exact argument so given that I've hopefully shown you or at least helped you see or or made you comfortable with the idea that the vectors let me do them in a nice vibrant color these magenta color vectors here that are associated with the three columns that are associated with the free columns or with the free variables right the free variables where X 3 and X 5 X 3 and X 5 and that those were these columns right here that they can always be expressed as linear combinations of the other columns because you just have to manipulate this equation set set the coefficient for whatever you're trying to find a linear combination for equal to minus 1 and set all the other free variables equal to 0 that you're not solving for and then you can get a linear combination of the of the of the vectors that are associated with the pivot columns so given that we've shown you that look these free vectors and I'm using my terminology very loosely but these ones that are associated with the non-pivot columns can be expressed as linear combinations of these guys so they're unnecessary the span of this is equivalent to the span of this the span of this is the column space space of a so the span of this is the column space of a so the last video I showed you that these guys are linearly independent and now I show you the span of these guys is the column space of a so now you should be satisfied that these vectors that are associated let me do it in a color that that vector that that column vector this column vector and this column vector that are associated with the pivot columns in the reduced row echelon form of the matrix in do indeed represent a basis for the column space of a anyway hopefully you didn't find that too convoluted