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Current time:0:00Total duration:23:18

Video transcript

I've got this matrix a here it's a 2x3 matrix and just as a bit of review let's figure out its null space in its column space so the null space of a it's the set of all vectors X that are member of let's see we have three columns here so a member of r3 such that a times the vector are going to be equal to the zero vector so we could just set this up let me just just so we just need to figure out all of the X's that satisfy this in r3 so we take our matrix a 2 - 1 - 3 - 4 - 6 multiply them times some arbitrary vector in r3 here so you get X 1 X 2 X 3 and you set them equal to the zero vector it's going to be the zero vector in r2 because we have two rows here you multiply two by three matrix times a vector in r3 you're going to get a 2 by 1 vector or 2 by 1 matrix so you're gonna get the 0 vector in r3 and to solve what is essentially a system of equations you get 2 X 1 minus X 2 minus 3 X 3 is equal to 0 and so on and so forth we can just set up an Augmented matrix so we can just set up this Augmented matrix right here 2 - 1 - 3 - 4 - 6 and then augment it with what we're trying to set it equal to to solve the system and you know we're going to put up we're going to perform a bunch of row operations here to put this in reduced row echelon form and they're not going to change this the right-hand side of this augmented matrix and that's essentially the argument as to why the null space of the reduced row echelon form of a is the same thing as the null space of a but anyway that's just a bit of review so let's perform some row operations to solve this a little bit better so the first thing I might want to do is divide let's divide the first row by 2 so if I divide the first row by 2 I get a 1 minus 1/2 minus 3 1/2 and then of course 0 divided by 2 is 0 and let's just divide let's divide this row right here let's just divide it by I don't know list of I did just to simplify things let's divide it by four so I'm provide I'm doing two row operations in one step but you can do that I could have done it in two separate steps so we if we divide it by 4 this becomes minus 1 1/2 and then you get what 3 halves and then you get zero and now let's keep my first row the same I'm going to keep my first row the same it's 1 minus 1/2 minus three-halves and of course the 0 is the right-hand side and let's replace my second row with my second row plus my first row so these are just linear operations on these guys so negative 1 plus 1 is 0 1/2 plus minus 1/2 is 0 three-halves plus minus three-halves is 0 and of course 0 plus 0 is 0 so what are we left with we're left with this right here or this is another way of saying this is another way of saying that X 1 let me write it this way X 1 X 1 is you could I guess the easiest way to think about it is you're multiplying the reduced row echelon form of a now 1 minus 1/2 minus 3 halves you have a bunch of zeros here times X 1 X 2 X 3 is equal to the r20 vector this is another interpretation of this augmented matrix so this is just saying this is useless this is saying zero times ad plus zero times a plus zero times that is equal to zero so it's giving us no information but this first this first row tells us that we switch colors 1 times X 1 X 1 minus 1/2 times X 2 minus 1/2 times X 2 minus three-halves three-halves times X 3 is equal to 0 all of the vectors whose components satisfy this are in my null space so I want to write a little bit differently I could write it as X 1 is equal to 1/2 X 2 plus 3 halves X 3 or if I wanted to write my solution set in vector form I could write that my well my null space is going to be the set of all the vectors x1 x2 x3 let's satisfy these conditions that are equal to what well x2 and x3 are free variables they're associated with the non pivot entries or the non-pivot columns in our reduced row echelon form that is a pivot column that is a pivot column right there so what are what is going to I could set so let me write it this way it's going to be x2 times something plus x3 times something all right those are my two free variables and we have here X 1 is 1/2 X to 1/2 times x2 plus three-halves times x3 three-halves times x3 x2 is just going to be x 2 times 1 plus 0 times x3 x3 is going to be 0 times x2 plus 1 times x3 so our null space our null space these can be any real numbers right here right there free variables so our null space is essentially all of the linear combinations of this guy and that guy or another way to write it the null space of a the null space of a is equal to the span which is the same thing as all the linear combinations of the span of 1 half 1 0 notice these are vectors in r3 and that makes sense because the null space is going to be a set of vectors in r3 so it's a span of that and that right there and that right there so 3 halves 0 and 1 just like that and what is the column what is the column space of our original matrix a so the column space the column space of a is equal to just all of the linear the subspace created by all of the linear combinations of these guys or essentially the span of the column vectors is equal to the span of 2 minus 4 minus 1 2 minus 3 six these are all each separate vectors so it's the span of these three vectors now these guys might not be linearly independent and actually when you put this guy in reduced row-echelon form you know that the basis vectors for this are the vectors that are associated with our pivot column so we have one pivot column here it's our first column so we could say that we could use this as a basis vector and it makes sense because this guy right here is minus 2 times the sky this guy right here is what minus three-halves times that guy so these two guys can definitely be represented as linear combinations of that guy so it's equal to the span the span of just of just the vector 2 minus 4 so if you were to ask me and this is this is the basis for our column space so if you want to know the rank and this is all a bit of review the rank of a is equal to the number of vectors in our basis for our column space so it's going to be equal to 1 now everything I just did is a bit of review but with the last couple of videos we've been dealing with transposes so let's actually figure out the same ideas for the transpose of a for the transpose of a so a transpose a transpose looks like this a transpose is equal to the vector 2 minus 1 or the matrix 2 minus 1 minus 3 is the first column right there and then the second column is going to be minus 4 2 & 6 that is our transpose so let's figure out the null space and the column space of our transpose of our transpose so let's just do the let's just well let me put this in reduced row echelon form so we can get the null space let me get the null space of this guy so we could do the exact same exercise we could do the exact same X let me write it this way the null space of a transpose a transpose is a 3 by 2 matrix so it's equal to all of the vectors X that are members of r2 that are members of r2 not r3 anymore because now we're taking the transpose is null space such that a transpose times our vectors are equal to the zero vector in r3 is equal to the zero vector in r3 and we can do that the same exact way we did before we set up an Augmented matrix we could we could just put it in reduced row echelon form and set them all equal to zero so let's just do that there so if we let me just put it in reduced row echelon form so let me let me divide my first row by two let's divide the first row by two just gonna put in reduced row echelon form the first row divided by 2 is 1 minus 2 and then the second row let me divide it let me just I'll just keep it the same so minus minus 1/2 and then this last row let me divide it by 3 so it becomes minus 1 and 2 and now let me keep my first row the same 1 minus 2 and now let me replace my second row with my second row plus my first row so minus 1 plus 1 0 2 plus minus 2 is 0 you get some zeroes I'm gonna do the same thing with the third row replace it with it plus the first row once again you're going to get some zeroes so this is the reduced row echelon form of a transpose so this is the reduced row echelon form of a transpose and it's null space is the same as a transpose is null space so we could set we could say to find this null space we can find all of the solutions to this equation times the vectors x1 and x2 is equal to 0 0 & 0 these aren't vectors these are just entries right here 0 0 0 so these two lines give us no information but this first one does so we get 1 times x1 and notice this is the pivot column right here it's associated so x1 is going to be a pivot variable this is a x2 will be a free variable and just to be clear that the first column is our pivot column so if we go back to a transpose it's this first column here that is associated with the pivot column so when we talk about its column space this by itself will span the column space this is all a review what we did before we're just applying it to the transpose let's go back to our null space so this tells us that 1 times X 1 so X 1 minus 2 times X 2 minus 2 times X 2 is equal to is equal to 0 or we could say that X 1 is equal to 2 times X 2 so all of the vectors in r2 that satisfy these conditions with this entries will be in the null space of a transpose let me write it this way let me write it this way if I want to do it so the null space the null space of a is going to be the set of all the vectors let me write it here the set of all the vectors x1 x2 that are a member of r2 clearly such that such that x1 x2 is going to be equal to well our free variable is X 2 so it's X 2 times the vector times the vector so x1 has to be 2 times x2 and obviously x2 that's a 2x2 is going to be 1 times x2 so what is this going to be well this is all of the linear combinations of this vector right here of this vector right here so we could say it's equal to the span the span of our vector the span of our of the vector 2 1 now that's the null space I this was the null space of a transpose have to be very careful there now what is the column space the column space of a transpose well the column space of a transpose is the set of all vectors spanned by the columns of a so you could just say the span of this column vector in this column vector but we know when we put into reduced row echelon form only this column vector was associated with a with a pivot column so this by itself this guys the linear combination of this guy if you multiply this if you multiply him by minus two you get that guy right there so it's consistent with everything we've learned so it equals the span it equals the span of just of just this guy right here of just the vector of just the vector 2 minus 1 and minus 3 now that's just a nice neat exercise that we did you know notice that your span here it's in r3 but it's just going to be a line in r3 and we can maybe on the next video I'll do a more graphical representation of it but I did this whole exercise is to introduce you to the ideas of the null space of your transpose and the column space of your transpose think about what the column space of your transpose is it's the it's the subspace spanned by that vector I'm sorry spanned by this vector and that vector and it turns out that this guy is a multiple of that guy so we could say just by that guy but these were the rows of our original of our original matrix a so we could we could also view this as the span of the row vectors of the row vectors of our original guy right this is that column that is the basis for the column span of the R transpose matrix and of course this guy was a linear combination of that so we can also view the column span of our transpose matrix it's equivalent to the subspace spanned by these rows or we could call that the row space of a so let me write that down so the column space of a transpose and this is just general let me write this generally it doesn't just apply to this example so the column space of any of the transpose of any matrix this is called the row the row space the row space of a it's a very natural name because if a is got a bunch of rows we could call them the transpose of some vectors so that's the first row you got the second row all the way to maybe the M throw the M throw just like that these are vector transpose this they're really just rows if you imagine the space that's spanned by these vectors by the different rows that's essentially the column space of the transpose so when you transpose it each of these guys become columns so that's what the row space is now the null space the null space of our transpose let's write it like this it was all of the vectors X that satisfied this equation equals equals the zero vector right there now what happens what happens if we take the transpose of both sides of this equation so if we take the transpose of both sides of that equation well we've learned from our transpose properties this is equal to the reverse product of each of those transposes so this is going to be equal to this is a vector the vector X transpose so now it's going to become if this is a column vector before now it's going to become a row vector and then times a transpose transpose and that's going to be equal to the transpose of the zero vector or we could just write this as let's write it like this we could write this as some matrix well let me just write it like this some column vector X time what's the transpose of a transpose well that's just equal to a so you take the transpose of this of this column vector you now get a row vector you could view it as a matrix if you want it's you know if this was a member of RN this is now going to be an N by what this is an N by or 1 by n matrix if this was and a member of RN and when we and now it's essentially we kind of switch the orders and we multiply it times the transpose of the transpose we just get the matrix a and we set that equal to the transpose of the zero vector now this is interesting we now have it in terms of our original matrix a now what did the null space of our matrix a look the null-space where all of the vectors exit satisfied this equation is equal to zero so the X was on the right this is so the null spaces all the X's that satisfy this the null space of our transpose is all of the X's that satisfy this equation and this is also called so this is so let me say the set of all of the X's that such that a transpose times X is equal to 0 that is the null space of a transpose that is equal to the null null space of a transpose or we could also write this as the set of all of the X's such that the transpose of our x times a is equal to the transpose of the 0 vector and we have another name for this this is called the left null space the left null space null space of a why is it called in the left null space because now we have X on our left and just a regular null space you have X on the right but now if you take the null space of the transpose the null space of the transpose using just our transpose properties that's equivalent to this transpose vector right here actually let me write this a transpose right there this transpose vector multiplying a from the left-hand side so all of the X's that satisfy this is the left null space and it's going to be different than your null space because it was equal to notice your null space of a transpose was a span of this right here this is also the left left null space the left null space of a now what was just a regular null space of a the regular null space of a was a essentially a plane in r3 that's the null space of a the left null space of a the left null space of a is just a line in our two very different things very different things and if you go to the if you go to the row space what is the row space of a the row space of a is a line in r3 the row space of a is a line in R three well what is the column space of a a column space of a right here whereas I have it it was a well this is the only linear and literally independent vector it was essentially a line in r2 so they're all very different things and we'll study a little bit more how they're all related now there's one thing I want to relate to you we figured out that the rank of this vector right here is one right because when you put it when you put it in reduced row echelon form there was one pivot there was one pivot column and the the basis vectors are those associated with that pivot column and if you count your basis vectors that's your dimension of your space so the dimension of your column space is one and that's the same thing as your rank now what is the rank of a transpose the rank of a transpose in the example when you put it in reduced row echelon form you got one linearly independent column vector so the basis for our column space was also equal to one and in general that's always going to be the case that's always going to be the case that the rank of a which is the dimension of its column space is equal to the rank of a transpose and if you think about it it makes a lot of sense to figure out the rank of a you essentially figure out how many to figure out the rank of a you essentially just have to figure out how many pivot columns they have or another way to say it is how many pivot entries they have now to figure out when you when you figure out the row space when you figure out the column space of when you want to find the rank of your of your transpose vector you're essentially just saying and I know this is maybe getting a little bit confusing but when you want the rank of your transpose vector you're saying which how many of these columns are linearly independent or which of these are linearly independent right and that's the same question as saying how many of your rows up here are linearly independent if you want to know how many columns and your transpose are linearly independent that's equivalent to asking how many rows your original matrix are linearly independent and when you put this matrix in reduced row echelon form everything in reduced row echelon form are just row operation so just linear combinations of these things up here or you could go vice versa everything up here is just linear combinations of your matrix in reduced row echelon form so if you only have one pivot entry if you only have one pivot entry then this guy right here by himself or one pivot row that guy by himself can represent a basis can represent a basis for your row space or all of your rows can be represented by a linear combination of your pivot rows and because of that you just count that you say okay there's one in this case so the dimension of my row space is one and that's the same thing as the dimension of the columns of my column spaces transpose I know it's all of my transpose is column space I know it's getting all confusing insulating the day for me as well so that's hopefully it will convince you that the rank of our transpose is the same as the rank of our original matrix