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Current time:0:00Total duration:8:33

Showing relation between basis cols and pivot cols

Video transcript

in the last video we saw a method of figuring out what the basis for column space is and we use these exact examples where if this was matrix a I just took it and I put it in reduced row echelon form and I figured out which of these columns in my reduced row echelon form of a our pivot columns and it turned out to be the first one the second one and the fourth one and then the method is you say look the corresponding columns in a so the first one the second one the fourth one formed my basis for my column space and since they form the basis and you want to know that if you want to know the dimension of your basis of your column space which is also called the rank you just say well there's 3 in there so it has a rank of 1 2 3 in this video I want to discuss a little bit about why this worked why were we able to just take the corresponding columns why did linear independence of these three guys imply linear independence of these three guys why was the fact that I can represent these guys this guy right here is a linear combination of these three or this guy is a linear combination of these three why does that imply that I can well that can construct this guy as a linear combination of my basis vectors so the first thing I think we that wasn't too much of a stretch of the imagination in the last video was the idea that these pivot vectors are linearly independent so r1 r2 and r4 and everything I'm doing I'm kind of applying to this special case just so that it's easier to understand but it should be generalizable in fact it definitely is generalizable that all of the pinned pivot columns in reduced row echelon form are linearly independent and that's because the very nature of reduced row echelon form is that you are the only pivot column that has a 1 in that respective row so the only way to construct it is with that vector you can't construct it with the other pivot columns because they're all going to have 0 in that row and when I say it's linearly independent I'm just saying the set of pivot columns so let me say this in general the set of pivot columns pivot columns for any for any reduced row echelon form made is linearly independent and it's just a very straightforward argument because they each column is going to have one in a very unique place all the other wrote all of the other pivot columns we're gonna have a zero in that same place and so you can't take any linear combination to get to that one because zero times anything minus or plus zero times anything can never be equal to one so I think you can accept that now that means that the solution to the solution to our one our C one times R one plus C 2 times R 2 plus let me say C 4 times R 4 the solution to this equation because these guys are linearly independent we know that this only has one solution and that's C 1 C 2 and C 4 is equal to zero that's the only solution to that so another way we could say it is if we write R times some vector X some vector well I'll just write it times this particular X where I write it as C 1 C 2 0 C 4 and 0 is equal to 0 so this will be some special member of your null space it's a particular solution to the equation this is equal to 1 2 3 4 zeros because we have four rows here now if this if we just expand this out if we just multiply 1 times C 1 plus 0 times C 2 minus 1 times C minus 1 times 0 plus 4 times 0 you'll get or actually let me do a better way to explain it this multiplication right here can be written as and we've seen this multiple times C 1 times R 1 plus C 2 times R 2 plus 0 times R 3 so we could just ignore that term plus C 4 C 4 times R 4 plus 0 times plus 0 times R 5 that's our 5 right there all of that equal to 0 so the only solution to this because we know that these three columns are linearly independent or the set of just those three columns those three pivot columns are linearly independent the only solution here is all of these equal to zero that's exactly what I said right up here so the only solution here where if these two are 0 is that these guys also all have to equal 0 if I already constrain these two now the one thing that we've done over and over again we know that the solution set of this equation the solution set of our X our X is equal to 0 is the same as a solution set of a X is equal to 0 now how do we know that or what do I mean well the solution set of this is just the null space the solution set is just the null space of our it's all of the X that satisfy this equation and we know that that is equal to the null space of a because R is just a in reduced row-echelon form so this is the null space of a is all of the X's that satisfy this equation now the only the only version of this that satisfied this equation was when c1 c2 and c4 are equal to 0 so that tells us that the only version of this the only version of this c1 c2 0 c40 that satisfies this equation or this equation is when C 1 C 2 and C 4 is equal to 0 or another way of saying that if this is this is vector a 1 a 2 a 4 right here if you multiply this out you get c1 let me do it over here you get I do it in blue you get C 1 times a 1 plus C 2 times a 2 and then 0 times a 3 plus C 4 times a 4 plus C 4 times a 4 is equal to 0 now these guys are going to be linearly independent if and only if the only solution to this equation is they all equal to 0 well we know that the only solution to is that they all equal zero because anything that's a solution to this is a solution to this and the only solution to this was if I if I go ahead and I constrain these two terms to being equal to zero the only solution to this is all of these C's have to be zero so likewise if I constrain these to be zero the only solution to this is that C 1 C 2 and C 4 have to be 0 so those guys have to be 0 which imply which imply that these three vectors a 1 a 2 and a 4 so that implies that the set a 1 a 2 and a 4 are linear in linearly independent so we're halfway there we've shown that look because the pivot columns here are linearly independent we can show and they have the same solution set the null space of the reduced row echelon form is the same as the null space of our original matrix we were able to show that the only solution to C 1 times this plus C 2 times this plus C 4 times this is when all the constants are 0 which shows that these three vectors or a set of those three vectors are definitely linearly independent now the next thing to prove that they are a bassist is to show that look all of the other vectors all of the other column vectors can be represented as multiples of these three guys and I realize just for the sake of it's just for the sake of clarity or maybe not boring you too much I'll do that in the next video so in this one we saw that look if these if the pivot columns are linearly independent they always are all pivot comes by definition are linearly independent or the set of pivot columns are always linearly independent when you take away the non-pivot columns then the corresponding columns in your original vector are also linearly dependent and the next one will show that these three guys also span your column space