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Current time:0:00Total duration:12:48

we've seen in several videos that the column space column space of a matrix is pretty straightforward to find in this situation the column space of a is just equal to all of the linear combinations of the column vectors of a so it's equal to oh another way of saying all of the linear combinations is just the span of each of these column vectors so if you know we call this one right here a 1 this is a 2 a 3 a 4 this is a 5 then the column space of a is just equal to the span of a 1 a 2 a 3 a 4 and a 5 fair enough but a more interesting question is whether these guys form a basis for column for the column space or even more interesting what is the basis for the column space of a and in this video I'm going to show you a method for determining the basis and along the way we'll get an intuition for maybe why it works and if I have time actually I probably won't have time in this video in the next video I'll prove to you why it works so what you do to figure out the the basis so we want to figure out the basis for the column space of a remember the basis just means that vectors span vectors span see a clearly these vectors span our column space I mean the span of these vectors is the column space but in order to be a basis the vectors also have to be linearly let me just write linearly independent and we don't know what whether these guys or what subset of these guys are linearly independent so what you do and I'm just really going to describe the process here as opposed to the proof is you put this guy in reduced row echelon form so let's do that so let me see if we can do that let's keep our first row the same 1 0 let me do it actually on the right side right here so let's keep the first row the same 1 0 minus 1 0 4 and then let's turn our let's replace our second row with the second row minus two times the first row so then our second row 2 minus 2 times 1 is 0 1 minus 2 times 0 is 1 0 minus 2 times negative 1 0 minus so that's 0 plus 2 0 minus 2 times 0 is just 0 and then 9 minus 2 times 4 is 1 enough now we want to zero out this guy well it seems like a pretty straightforward way just add replace this row with this row plus the first row so minus 1 plus 1 is 0 2 plus 0 is 2 5 minus 1 is 4 1 plus 0 is 1 minus 5 plus 4 is minus 1 and then finally we got this guy right here and then maybe we can just in order to 0 him out let's replace him with him minus the first row so 1 minus 1 is 0 minus 1 minus 0 is minus 1 minus 3 minus negative 1 that's minus 3 plus 1 so that's minus 2 minus 2 minus 0 is minus 2 and then 9 minus 4 is 5 so we did one round we got our first we got our first pivot column going now let's do another round of row operations well we want to zero all of these guys out luckily this is already 0 so we don't have to change our first row our second row so we get 1 0 minus 1 0 for our second row becomes 0 1 2 0 1 and now let us see if we can eliminate this guy right here and let's do it by replacing our blue row our third row with the third row minus 2 times the second row so 0 minus 2 times 0 is 0 2 minus 2 times 1 is 0 4 minus 2 times 2 is zero one minus two times zero is 1 minus 1 minus 2 times 1 is minus 3 all right now this last guy we want to eliminate him and then we want to turn this into a zero we just let's replace this fourth row with the fourth row plus the second row so zero plus zero is zero minus 1 plus minus 1 is zero minus 2 plus minus 2 is zero minus 2 plus zero is minus 2 and then 5 plus 1 is 6 5 plus 1 is 6 we're getting close so let's see we have this let's look at our pivot entries we have this as a pivot entry that's a pivot entry and this is not a pivot entry and let's see because it's following obviously another this guy is a pivot entry right here or will be we need a 0 this minus 2 out so let's I think we'll be done so let me write my first row just the way it is because everything above it is 0 so we have to worry about it so my first row I can just write as 1 0 minus 1 0 4 i can write my second row 0 1 2 0 1 I can write my third row as 0 0 0 1 minus 3 and now let's replace my fourth row let's replace it with it plus 2 times the second row so 0 plus 2 times 0 0 plus 2 times 0 0 plus 2 times 0 minus 2 plus 2 times 1 is just 0 6 plus 2 times minus 3 that's 6 minus 6 that's just 0 and there we have we've we've actually put our matrix in reduced row echelon form so let me put a little brackets around it it's not so bad if you just kind of go in and just do the manipulations and sometimes when you kind of get a headache thinking about doing something like this but this wasn't too bad so this is the reduced let me just say the reduced row echelon form of a let me just call that matrix R so this is matrix are right there now what do we see about matrix R well it has three pivot entries or three pivot columns let me square them outer circle amount column one is a pivot column column two is a pivot column and column three is a pivot column now and we've done this in previous videos you can you can well there's two things that you can see these three columns they're clearly linearly independent how do we know that well this guy's got a one where and that's just with respect to each other if we just took a set of let's call this r1 r2 and this would be our three this would be our four right here it's clear that the set r1 r2 and r4 it's clear that this is linearly independent and you say why is that well look r1 has got a 1 here while the other two have a 0 in that entry right and this is by definition of pivot entries pivot entries have zeros or pivot columns have zeros everywhere except for where they have a 1 so any if for any pivot column it will be the only pivot column that has zeros there or it will be the only pivot column that has a 1 there so there's no way that you can add up combinations of these guys to get a 1 you could say 100 times 0 minus 3 times 0 just going to get a bunch of zeros so no combination of these two guys is going to be equal to that guy by the same reasoning no combination of that and that is going to equal this this is by definition of a pivot entry by no when you put it in reduced row echelon form it's very clear that any pivot column will only will be the only one to have one in that place so we very clear that these guys are linearly independent now it turns out and I haven't proven it to you that the corresponding columns in a so the correspondence are one but at a before we put in reduced row echelon form that these guys right here so a 1 a 2 and a 4 are also linearly independent so a 1 let me recycle the a 1 a 2 and a four so if I write it like this a 1 a 2 and a 4 let me write it set notation these guys are also linearly independent which I haven't proven but I think you can kind of get a sense that these row operations really don't change the sense of the matrix and I'll do a better explanation of this but I really just wanted you to understand how to develop a basis for the column space so they're linear independent and so the next question is do they span do they span our column space and in order for them to spend obviously all of these five vectors if you have all of them that's going to span your column space by definition but if we can show and I'm not going to show it in this video but it turns out that you can always represent this the free the non-pivot columns so you can always represent the non-pivot columns as linear combinations of the pivot columns and we'll show and we've kind of touched on that in previous videos where we find the solution for the null space and all of that so these guys can definitely be represented as linear combinations of these guys I haven't shown you that but if you take that as kind of as on faith then you don't need that column that column to span if you if you did then or I guess a better way to think it you don't need them to span although they are part of the span because if you needed this guy you can just construct him with linear combinations of these guys so if you wanted to figure out a basis for the column space of a you literally just take a into reduced row echelon form you look at the pivot entries in the reduced row echelon form of a then that's those three and then you look at the corresponding columns to those pivot columns in your original a and those form the basis because any linear combination of them or linear combinations of them can be used to construct the non-pivot columns and they're linearly independent so I haven't shown you that but for this case if you want to know the basis it's just these three vector sorry it's just a 1 a 2 and a 4 and a 4 right there and now we can answer another question so let me say so the basis so a 1 a 2 and a 4 form a basis basis for the column space of a because you can construct the other two guys with linear combinations of our basis vectors and they're also linearly independent now and the next question is what is the dimension of the basis or what is the dimension not the dimension of the basis what is the dimension of the column space of a well the dimension is just the number of vectors in any basis for the column space and all bases have the same number of vectors for any given subspace so we have one two three vectors so the dimension of our column space is equal to three and the dimension of a column space actually has a specific term for it and that's called the rank so the rank of a which is the exact same thing as the dimension of the column space it is equal to three and another way to think about it it's the number the rank of a is the number of linearly independent column vectors that you have that can span your entire column space or the number of linearly independent column vectors that can be used to construct all of the other column vectors but hopefully this didn't confuse you too much because the idea is very simple take a put it into reduced row echelon form see what your pivot your see which which wrote columns our pivot columns the corresponding columns are going to be basis are going to be a basis for your column space if you want to know the the rank for your matrix you can just count them or even if you don't to count those you could literally just count the number of pivot columns you have in your reduced row echelon form so that's how you do it in the next video I'll explain why this worked