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Current time:0:00Total duration:25:13

what I want to do in this video and will probably occur over several videos is really integrate everything we know about matrices and null spaces and column spaces and linear independence so I have this matrix here this matrix a and I guess a good place to start is let's just figure out there its column space in its null space and the column space is actually super easy to figure out it's just the span of the column vectors of a so we can write from the get-go right that the column space the column space of our matrix a let me do it over here I can write the column space of my matrix a is equal to the span the span of the vectors 1 2 3 1 1 1 4 and 1 1 4 1 4 1 and 1 3 2 I'm done that was pretty straightforward a lot easier than finding null spaces now this may or may not be satisfying to you and there's a lot of open questions is this a basis for the space for example is this a linear independent set of vectors how can we visualize this space and I haven't answered any of those yet but if someone says hey what's the column space of a this is the column space of a and we can answer some of those other questions but was this linear in if this is linearly if these if this is a linearly independent set of vectors then this would be a basis these vectors would be a basis for the column space of a we don't know that yet we don't know whether these are linearly independent but we can figure out if they're linearly independent by looking at the null space of a remember these are linearly independent if the null space of a only contains the 0 vector so let's see we figure out what the null space of a is and remember we can do a little shortcut here the null space of a is equal to the null space of the row the reduced row echelon form of a and I showed you that when I first when we first calculated the null space of a vector because when you perform these you could this essentially if you want to solve for the null space of a you create an Augmented matrix and you put the Augmented matrix in reduced row echelon form but the zeros never change so essentially you're just taking a and putting in reduced row-echelon form so let's do that so let me I'll keep row 1 the same 1 1 1 1 and then let me replace Row 2 with Row 2 minus Row 1 Row 2 minus Row 1 so what do I get - no I actually I want to I want to zero this out here so Row 2 minus 2 times Row 1 actually even better because I eventually want to get a 1 here so let me do 2 times Row 1 minus Row 2 so let me say 2 times Row 1 and I'm going to minus Row 2 so 2 times 1 minus 2 is 0 which is exactly what I want there 2 times 1 minus 1 is 1 that's nice to have right there 2 times 1 minus 4 is minus 2 2 times 1 minus 3 is minus 1 all right now let me see if I can zero out this guy here so what can I do I see I could let me let me take because I'm just and I could do any combination anything that that essentially zeros this guy out but I want to minimize my number of negative numbers so let me take this this third row minus 3 times this first row so I'm going to take minus 3 times that first row and add it to this third row so 3 minus 3 times 1 is 0 4 minus 3 times well these are just going to be a bunch of threes 4 minus 3 times 1 is 1 1 minus 3 times 1 is minus 2 and 2 minus 3 times 1 is minus 1 now if we want to get this into reduced row echelon form we need to target that 1 there and that one there and what can we do so let's keep my middle row the same my middle row is not going to change 0 1 minus 2 minus 1 and to get rid of this one up here I can just replace my first wrote my first row minus my second row because there now this won't change I've 1 minus 0 is 1 1 minus 1 is 0 that's what we wanted 1 minus minus 2 is 3 right that's 1 plus 2 1 minus minus 1 that's 1 plus 1 that is 2 fair enough now let me do my third row the third row well let me just add or let me just subtract my let me replace my third row with my third row subtracted from my first row they're obviously the same thing so if I subtract this row from if I subtract the third row from the second row I'm just gonna get a bunch of zeros 0 minus 0 is 0 1 minus 1 is 0 minus 2 minus minus 2 is 0 and minus 1 minus minus 1 that's minus 1 plus 1 that's equal to 0 and just like that we have it now in reduced row echelon form so this right here is the reduced row echelon form of a that's straightforward now we the whole reason why we even went through this exercise is we wanted to figure out the null space of a and we already know that the null space of a is equal to the null space of the reduced row echelon form of a so if this is the reduced row echelon form of a let's figure out its null space so the null space is the set of all vectors all vectors in r4 because we have four columns here 1 2 3 4 the null space is the set of all vectors that satisfy this equation where there's going to have we're going to have 3 zeros right here that's the 0 vector in r3 because we have three rows right there and you can figure it out this times this has to equal that 0 that dotted with that essentially is going to equal that 0 that dotted with that is equal to that 0 I say essentially because I didn't define a row vector dot a column vector I've only defined column vectors dotted with other column vectors but we we've covered that in previous video where you could say this is the transpose of a column vector so let's just take this and write a system of equations with this so we get 1 times x1 so this times this is going to be equal to that 0 so 1 times x1 that is x1 plus 0 times x2 let me just write that out plus 0 times x2 plus 3 times x3 plus two times x4 is equal to that 0 is equal to 0 and then in let me I'll do it in yellow right here I have 0 times x1 0 times X 1 plus 1 times x2 plus x2 minus 2 times X 3 minus X 4 is equal to 0 and then this gives me no information zeros times all of this is equal to 0 so just turns it to 0 equals 0 so let's see if we can solve for our pivot entries or our pivot variables what our pivot entries this is a pivot entry that's a pivot entry that's what reduced row echelon form is all about getting these entries that are 1 and they're the only nonzero term in their respective columns and that every pivot entry is to the right of a pivot entry above it and then the columns that don't have to attend trees these are called these columns represent the free variables so this column has no pivot entry and so the when you when you take the dot product this column turned into this column in our system of equations so we know that x3 is a free variable x3 is free we can set it equal to anything likewise x4 x4 is free is a free variable x1 and x2 our pivot variables because they're corresponding columns in our reduced row echelon form have pivot entries in them fair enough so let's see if we can simplify this into a form we know and we've seen this before so if I solve for x1 this 0 I can ignore that 0 I can ignore I could say that x1 is equal to minus 3 x3 minus 2 X 4 right I just subtracted these 2 from both sides of equation and I can say that X 2 is equal to 2 X 3 2 X 3 plus X 4 and if we want to write our solution set now so if I wanted to in the null space of a so the null space of a the null space of a which is the same thing as the null space of the reduced row echelon form of a is equal to all of the vectors let me do a new color maybe we'll do blue it's equal to all of the vectors x1 x2 x3 x4 that are equal to so what are they going to be equal to x1 has to be equal to minus 3 x3 minus 2 X 4 so just to be clear these are free variables because I can set these to be anything but and these are pivot variables because I can't just set them to anything I can when I determine what my x3 is in my x4 is are they determine what my x ones and my x2 is have to be so these are pivoted variables these are free variables I can make this guy PI and I can make this guy minus 2 we can set them to anything so x1 is equal to let's see let me write it this way they're equal to x3 let me do it in a different color do x3 like this so it's equal to x3 times some vector plus x4 plus x4 times some other vector so any solution set in my null space is going to be a linear combination of these two vectors and we can figure out what these two vectors are just from these two these two constraints right here so it's we can let me do it in a neutral color x3 x1 x1 is equal to minus 3 times x3 so minus 3 times x3 minus 2 times x4 straightforward enough x2 is equal to 2 times X 3/2 times x3 plus x4 what's x3 equal to well x3 is equal to itself whatever we set x3 equal to that's going to be X 3 so X 3 is going to be 1 times x3 plus 0 times x4 does not going to have any X 4 and X 3 is going to be kind of an independent variable where it's going to be free we can set it to whatever it is so we set it and then that's going to be our x3 in our solution set X 4 it's not going to have any x3 in it it's just going to be 1 times x4 and so our null space is essentially all of the linear combinations of these two vectors right this can be this is just any real number and x4 is just any member of the real space so all of these the set of all of the valid solutions to ax is equal to 0 where did I write that did I even write that down now I haven't even written that anywhere the valid the set of all ax is equal to 0 where this is my X it equals to all of the linear combinations of this vector and that vector right there and we know what all of the linear combinations mean it means so my null space let me write it this way is equal to the span of these two guys the span of minus 3 2 1 0 and let me scroll over a little bit and minus 2 1 0 1 now let me ask you a question is is this is the is the is the are the columns in a are they a linearly independent set are they a linearly independent set let me write that down so if we write these vectors right there this is these are the vectors the column vectors of a alright so let me write that down so are the column vectors of a so what were they let's see 1 1 3 2 and remember them 1 I notice 1 2 3 1 2 3 1 1 4 1 4 1 and 1 3 2 so this is just the column vectors of a Q just write a is just this much of columns but my question is is this a linearly independent set linearly independent and here you might you you might immediately start thinking well when we said that something is linearly independent so linearly independence let me just write it like this linear independence implies that there's only one solution we this I think two videos ago that there's only one solution one solution to one solution to ax is equal to zero and that is the zero solution that X is equal to the zero vector or another way to say that is that the null space of my matrix a is equal to just the zero vector that's what linear independence implies and it goes both ways if my null space is just a zero vector then I know it's linearly independent if my null space includes other vectors then I am NOT linearly independent now my null space of a what does it include is it just the zero vector well no it includes every linear combination of these guys it includes actually an infinite number of vectors it's not just one solution obviously zero vector is contained here if you just multiply both of these if you pick zero for that and that it's contained but you can get a whole set of vectors so because this the span of this guy so you can kind of you know the null span of a the null space sorry the null space of a does not just contain the zero vector so it has you know more than just zero than just more than just zero so that what does that mean well that means that there's more than one solution of this and that means that this is a linearly dependent set linearly dependent dependent set and what does that mean well the very beginning of the video I said what's the column space of a and we said oh the column space of a is just the span of the column vectors right I just wrote it out like that and I said well it's not clear whether this is a valid basis for the column space of a and what's a basis a basis is a set of vectors that span a subspace that span a subspace and they are also linearly independent and we just showed we just show that these guys are not linearly independent not linearly independent so that means that they are not a basis for the column space of a they do span the column space of a by definition really but they're not a basis they need to be linearly independent for them to be a basis so let's see if we can figure out what a basis for this column space would be and to do that we just have to get rid of some redundant vectors if I can see if I can show you that this guy can be represented by some combination of these two guys and I can get rid of that guy he's not adding any new information same with that guy who knows so let's see if we can figure this this piece of the puzzle out so we know already we know already that X 1 let me write it this way that X 1 times maybe I'll just kind of leave you hanging and continue this in the next video but we know that X 1 x 1 2 3 plus X 2 times 1 1 4 plus X 3 times 1 4 1 plus X 4 times 1 3 2 we know that this is equal to 0 now if we're able to solve for x 4 in terms of let's say I can solve them in terms of let's let me just think that I can solve for my my the the vectors that are associated with my free variables using the other vectors so let me see if I can do that and then you'll see it's actually pretty straightforward so let's say I want to figure out I want to solve for X 4 so I want if I subtract this from both sides of this equation I get what I get - let me put it this way let me set X 3 equal to 0 it was a free variable I can do that so if I set X 3 is equal to 0 then what do I get here I get if I said X 3 equals 0 this guy disappears and if I if I subtract this from both sides of this equation I get X 1 X 1 times 1 2 3 plus X 2 times 1 1/4 is equal to I'm just setting X 3 equal to 0 that was a free variable so I'm setting X 3 equals 0 so this whole thing disappears so that is equal to minus X 4 over R times 1 so three two now I set x3 equal to zero let me set x1 and Safari let me set x4 to be equal to minus 1 x4 is equal to minus 1 if X 4 is equal to want minus 1 what is minus x4 well then this thing will just be equal to 1 and I'll have X 1 times 1 2 3 plus x2 times 1 1 4 will equal this fourth vector right here and can I always find things like this well sure I can actually find the particular ones if X 3 is equal to 0 and X 4 is minus 1 let me copy and paste this over that I have up here let me copy and paste this sometimes it doesn't edit copy edit paste there you go okay let me scroll down a little bit this is what we got when we figured out our null space right there so if I'm setting remember these are the free variables if I set X 3 equal to 0 and X 4 is equal to minus 1 what is X 1 then this will imply that X 1 well it's equal to 2 minus 3 times X 3 that's just 0 minus 2 times X 4 if X 4 is minus 1 minus 2 times minus 1 X 1 it will equal 2 and then what will X to be equal to X 2 is equal to 2 times X 3 which is 0 plus X 4 so it's equal to minus 1 so I just showed you that if I set this equal to 2 and this equal to minus 1 I have a linear combination of this vector and this vector that can add up to this fourth vector and you can even verify it 2 times 1 minus 1 is equal to 1 2 times 2 minus 1 is equal to 3 2 times 3 is 6 minus 4 is equal to 2 so it checks out so I just showed you using really our definitions looking really looking at what what were our free variables versus our pivot variables we were able to show you it kind of just very simply solve for this third this fourth vector in terms of these first two so we know if we go back to the set that this fourth vector is really unnecessary really not adding anything to I guess the span of the set of vectors because this guy can be written as a combination of this guy and this guy now let's see if this guy this third guy can we can do the same exercise this is also dictated by a free variable so let's see if I can write him as a combination of these first two well we'll do the exact same thing instead of setting X 3 equal to 0 and X 4 equal to 9 is 1 let's set let us set X 4 is equal to 0 X 4 is equal to 0 because I want to cross that out and let me set X 3 is equal to minus 1 if X 3 is equal to minus 1 so this equals minus 1 what is our what is this equation reduce to we get X 1 times 1 2 3 plus X 2 times 1 1 4 is equal to if this is minus 1 times 1 4 1 and then we add it to both sides of this equation we get plus 1 times 1 4 1 and once again we can just solve for our X 1 and X 2 if X 4 is 0 and X 3 is minus 1 then X 1 X 4 is 0 so X 3 is just minus 3 times X 3 so X 1 would be equal to 3 right minus 3 times minus 1 and what would X 2 be equal to X 4 is 0 we can ignore that X 2 would be equal to minus 2 so this would be 3 and then this would be minus 2 let's see if it works out 3 times 1 minus 2 is 1 3 times 2 minus 2 is 4 3 times 3 minus 8 is 1 it checks out so I'm able to write this vector as little that was associated with a free variable as a linear combination of these two so we can get rid of him from our set so now I've shown that this guy can be written it as a linear combination of these two this guy can be set written as a linear combination of these two so the span of all of those guys should be equal to the span so let me write it this way I cross that guy the column space of a I can now rewrite it was before it was the span of all of those vectors it was a span of all of the column vectors V 1 V 2 V 3 and v4 now I just showed you that v3 and v4 can be rewritten in terms of v1 and v2 so they're redundant so that is equal to the span of v1 and v2 which are just those two vectors 1 vector 1 2 3 and vector 1 1 4 now is this are any of these guys redundant can i express one of them as a linear combination of the other and essentially when I'm talking about the linear combination of only one other vector it's just multiplying it by a scalar well let's think about that and you can you can there's multiple ways you can show this but the easiest way is well look to go from this entry to that entry I'm just multiplying by 1 but if I multiply this whole vector times 1 then I'm going to get a 2 here and I'm gonna get a 3 here so it won't work any let me put it this way any if I'm if I want to represent this guy's a scalar multiple of that guy so any scalar multiple of 1 2 3 is going to be equal to is going to be equal to 1 C 2 C 3 C right and so we're saying this guy has to be represented somehow like that if we say that this guy is somehow a scalar is somehow it can be represented by that guy so that would have to be equal to 1 1 4 when you look at this top entry it implies that C would have to be equal to 1 when you just C is equal to 1 but when you look at this second entry you think that C would have to be equal to 1/2 so you get a contradiction over here C would have to be equal to 4/3 so there's no C where this will work there's no multiple of C and you can work that both ways so there's no way that you can represent one of these guys as a linear combination of the other and you can actually prove other ways more maybe more formally that this is linearly independent but given that this is linearly independent I think you're satisfied with that we can then say that the vectors the set of vectors 1 2 3 or let me put it this way oh yeah 1 2 3 and 1 1 4 this is a basis this is a basis for the column span and of a now I'm going to I'm going to let you go in this video because I think I've gone well over time but what I'm going to do in the next few videos is now that I've established that this is a basis for the column span of a week an attempt to visualize it because we can say that the column span of a the column span of a is equal to the span of these two vectors and we can think about what the span of those two vectors are we're going to see that it's a plane in r3 span of 1 1 4 and this is a quick reminder I said a couple times when I say it's a basis all I'm saying is that these guys they both span the column space of a the all when I had four vectors they also span the column space of a but what makes them a basis is that these guys are linearly independent there's no extra information or redundant vectors that can be represented by other vectors within the basis they're linearly independent and I'll let you go for now