If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:13:07

Null space 2: Calculating the null space of a matrix

Video transcript

in the last video I spoke somewhat theoretically about what a null space is and we show that it is a valid subspace but in this video let's actually calculate the null space for a matrix in this case will calculate the mate the null space of matrix a so the null space is literally just the set of all the vectors that when I multiply a times any of those vectors so let me say that the vector x1 x2 x3 x4 is a member of our null space then when I multiply this matrix times this vector I should get the zero vector I should get the vector and just to make a few points here this has exactly four columns this is a 3 by 4 matrix so I've only legitimately defined multiplication of this times a four component vector or a member of RN let me call this X and this is our vector X this is a member of r4 it has four components and then when you multiply these we need to produce the zero vector the null space is the set of all the vectors and when I multiply it times a I produce the zero vector and what am I going to get I'm gonna have one row times this and that's going to be the first entry then this row times that's the second entry then the third row so I'm going to have three zeros so my zero vector is going to be kind of the zero vector in r3 so how do we figure out the set of all of these X's that satisfy this let me just write kind of our formal notation the null space of a is the set of all vectors that are a member of you generally say RN but it's a 3 by 4 matrix so these are all the vectors that are going to be members of r4 because I'm using this particular a such that my matrix a times any of these vectors is equal to the 0 vector in this case it's going to be the 0 vector in r3 so how do we do this well this is just a straight up linear equation we can write it that way if we were to actually do perform the matrix multiplication we get 1 times x1 let me write it here let me do it in a different color 1 times x1 plus 1 times X 2 plus 1 times X 3 plus 1 times X 4 is equal to this 0 there is equal to that 0 so that times that is equal to that 0 and then this times this it should be equal to that 0 so 1 times X 1 so you get X 1 plus 2 times X 2 2 times X 2 plus 3 times X 3 plus 4 times X 4 is going to be equal to that 0 and then finally we have that times this vector should be equal to that 0 so or though you can view the dot product of that the dot product of this row vector with this column vector should be equal to that 0 so you get 4 X 1 4 X 1 plus 3 X 2 plus 2 X 1 plus X 4 I'm sorry what am i doing plus 2 X 3 right plus 2 X 3 plus X 4 is equal to 0 for X 1 plus 3 X 2 plus 2 X 3 plus X 4 is equal to 0 I know you just have to find the solution set to this and will essentially have figured out our null space now we've figured out the solution set to a system of equations like this we have four equations with or we have three equations with four unknowns and we can do that we can represent this by an Augmented matrix and then put that in reduced row echelon form so let's do that so I can represent this problem as the Augmented matrix 1 1 4 1 1 4 1 1 4 1 2 3 1 2 3 1 3 2 1 3 2 and then 1 4 1 1 4 1 and then I Ament that with the zero vector and the immediate thing you should notice is look we took the pain of multiplying this times this to equal that and we wrote this as a system of equations but then when we now we want to solve the system of equations we're kind of going back to the Augmented matrix world and what does this Augmented matrix look like well this is just our matrix a right there that's just matrix a right there and that's just the zero vector right there and to solve this and we've done this before we're just going to put this Augmented matrix into row echelon form and what you're going to find is when you put it in a row echelon form this right side is not going to change at all because no matter what you multiply or subtract by you're just doing it all times zero so you just keep ending up with zeros so as we put this into reduced row echelon form we're actually just putting matrix a into reduced echelon form so let me do that instead of just talking about it so let me start off by keeping Row one the same Row 1 is 1 1 1 1 0 and then I want to eliminate this one right here so let me just subtract let me replace Row 2 with Row 2 minus Row 1 so 1 minus 1 is 0 2 minus 1 is 1 3 minus 1 is 2 4 minus 1 is 3 0 minus 0 is 0 you can see the zeros aren't going to change and then let me replace this guy with 4 times this guy - this guy so I get so I can I don't get rid of this so 4 times 1 minus 4 is 0 4 times 1 minus 3 is 1 4 times 1 minus 2 is 2 4 times 1 minus 1 is 3 4 times 0 minus 0 is 0 now I want to get rid of if I want to put this in reduced row echelon form I want to get rid of that term and that term so let me keep my middle row the same my middle row is 0 1 2 3 middle row the same so that's 0 on the Augmented side of it although these zeros are never going to change so it's really just a little bit of a exercise just to keep writing them and then my first row let me replace it with the first row minus the second row so I can get rid of this one so 1 minus 0 is 1 1 minus 1 is 0 1 minus 2 is minus 1 1 minus 3 is minus 2 and 0 minus 0 zero and let me replace this last row with the last row minus the middle row so 0 minus 0 is 0 1 minus 1 is 0 2 minus 2 is 0 I think you see where this is going 30 minus 3 is 0 and obviously 0 minus 0 is 0 so this system of equations has been reduced to has been reduced to just by doing reduced row echelon form this problem if I just rewrite what I this right here this is this can be written as a system of equations of x1 x1 minus x3 - x4 right the 0 x2 s is equal to 0 and then this second row right here it could be written as there's no x1 so you just have an x2 + 2 x 3 + 3 x2 is equal to 0 and this obviously gives me no information whatsoever and so I can solve this I can solve this for X 1 and X 2 and what do I get I get X 1 is equal to x3 + x4 actually I made a mistake here this is x1 minus x3 minus 2 times x4 is equal to 0 so if I rewrite this I get x1 is equal to x3 + 2 X 4 and then I get x2 let me do that in green x2 is equal to minus 2 X 3 - 3 x2 and so if I wanted to write the solution set to this equation if I wanted to write in terms of this I could write X 1 X 1 X 2 X 3 X 4 is equal to what's x1 equal to it's equal to x3 times 1 plus X 4 plus X 4 times - right I just got this right here from this equation right here x1 is equal to 1 times x3 plus 2 times x4 that's just that right there now x2 is equal to x3 times minus 2 that's x3 times minus 2 plus x4 times minus 3 what am i doing I'm losing track of things this x2 right here is x2 + 2 X 3 + 3 X 4 is equal to 0 so X 2 is equal to minus 2 X 3 minus 3 X 4 right like that sorry I my brain isn't completely in the problem I'm making these silly mistakes but I think you understand this now and then what is x3 equal to well it's just equal to 1 times x3 is equal to 1 times x3 plus 0 times x4 right x3 is equal to x3 and what's x4 equal to it's equal to 0 times x3 plus 1 times x4 so all of the vectors in R 4 these are a member of r4 which satisfy the equation our original equation ax is equal to 0 can be represented as a linear combination of these two vectors of those two vectors right these are just random scalars that are member of we can pick any real number for x3 and we could pick any real number for x4 so our solution set is just a linear combination of those two vectors well what's another way of saying a linear combination of two vectors so let me write this the null space the null space of a which is just a solution set of this equation it's just all of the X's that satisfy this equation it equals all of the linear combinations of this vector and that vector well what do we call all the linear combinations of two vectors it's the span of those two vectors so it equals the span of that vector in that vector of the vector 1 minus 2 1 0 and the vector 2 minus 3 0 1 and this is our null space and before letting it go let me just point out one interesting thing right here we we represented our our system of equation like this and we put into a reduced row echelon form so this is a and this is 0 this right here is let me make sure I have some space let me put it right here that right there is the reduced row echelon form of a and so we're essentially this equation this is this is a linear equation that says trying to solve this problem the reduced row echelon form of a times our vector X is equal to 0 so this all the solutions to this are also the solutions to our original problem to our original ax is equal to 0 so this what's the solution of this all the X's that satisfy this these are the null space of the reduced row echelon form of a right so this is so here all of the X's this is the null space this problem if we find all the solution set all of the X's here this is the null space of the reduced row echelon form of our matrix a but we're saying that this problem is the same problem as this one right so we can write that the null space of a is equal to the null space of the reduced row echelon form of a and that might seem a little bit confusing hey why you even writing the cell but it's actually very useful when you're actually trying to calculate null spaces because we didn't even have to write a big Augmented matrix here we could say okay take our matrix a put it in reduced row echelon form and then figure out its null space so we would have gone straight to this point right here this is the reduced row echelon form of a and then I could have immediately solved these equations right I would have just taken the dot product of the reduced row echelon form or not the dot product the matrix vector product of the reduced row echelon form of a with this vector and I would have gotten these equations and then these equations would immediately I can just kind of rewrite them in this form and I would have gotten our result but anyway hopefully you found that reasonably useful