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let's say I have some matrix a if I'm trying to determine the null space of a I'm essentially asking I'm essentially just asking look if we set up the equation a X is equal to the zero vector the null space of a the null space of a is all the X's that satisfy this equation so there's all the X's that satisfy that satisfy that equation ax is equal to the zero vector or you can call it this system and the way you would solve it and we've done this many times this was many videos ago you would make an Augmented matrix with this so the Augmented matrix would look like that you'd have the zero vector on the right hand side and you'd perform a bunch of row operations to put the left hand side into reduced row echelon form so you would put do a bunch of operations the left hand side will go into reduced row echelon form let's call that reduced row echelon form of a and then the right-hand side is just going to stay zero because you perform the same row operations but when you perform those row operations on 0 you just get the zero vector right here and then when you unmeant when you create the system back from this right here because these two systems are equivalent you're essentially going to have your solution set look something like this you're going to have your let me write it like this your solution set is going to be equal to some scalar multiple you know let's say that of your free variables your free variables are going to be the scalar multiples and you've seen this multiple times so I'll see a pretty general it's going to be some multiple times let's say you know vector one plus some other scale scalar times vector 2 these scalars tend to be your free variables times vector 2 all the way to you know I don't know whatever C times your n vector I'm just trying to say general we haven't seen any examples that had more than two or three vectors here but this is what essentially your null space is spanned by these vectors right there you get an equation you get a solution set that looks something like that and you call you call that your null space we've done that multiple time's your null space is that so it's all the linear combinations or it's the span of these little vectors that you get here and one and two all the way to n n this is nothing new I'm just restating something that we've seen multiple multiple times we've actually did this in the previous video I just maybe never wrote it exactly like this but what about the case when you're solving the inhomogeneous equation so the inhomogeneous equation looks like this so if I want to solve ax is equal to B I would do something very similar to this I will create an Augmented matrix I have a on the left hand side and I'd put B on the right hand side and then I'll put it I'll perform a bunch of row operations to put a into reduced row echelon form so let me do that so this left hand side will be the reduced row echelon form of a and then the right hand side whatever operations I did on a I have to do on the on the entire row so I'll also be doing them to be so I'll have some new vector here maybe I'll call it maybe I'll call it the vector maybe I'll call that the vector B prime it's going to be different than B but let's just call it B Prime and so when you when you go back to your I guess when you go out of the Augmented matrix world and rewrite it as a system and you solve for and we did this in the last video you'll get your solution set your solution set that satisfies this is going to be X is going to be equal to this B prime whatever this new vector is this B prime plus something that looks exactly like this that looks exactly like that in fact I'll copy and paste it it'll look exactly like this let me see if I if it did copy let me copy and paste it edit copy and let me paste it so it'll look something that looks exactly like that and we said in the last video we said in the last video that that you can kind of given this you can kind of think as the solution set to the inhomogeneous equation is equivalent to some particular solution is equal to some particular solution let's call that X particular some particular solution plus some member of your null space so you could say plus some homogeneous solution so if you just pick particular values for a B and C all of the different multiples of the vectors that span your null space you'll get some particular homogeneous solution so what I what I implied in the last video and I didn't show it to rigorously is that any solution to the inhomogeneous system let me write it this way any solution and do it in white any that's not white any solution to the inhomogeneous system to in Homo genius genius system a X is equal to B this is a claim I made will take the form will take the form some particular solution some particular solution that was this right here maybe I should do it in green this is this right here did when you do the reduced row echelon form it becomes that B prime vector plus some homogeneous solution plus some homogeneous solution so some member of the nulls place some member of the null space now I didn't prove it to you but I implied that this is the case and what I want to do in this video is actually do a little bit of a a more rigorous proof but it's actually fairly straightforward so let's let's first of all verify that this is a solution so let's verify that this is a solution so let's just put this into our original equation so if we remember our original equation was ax is equal to B so let's verify so is let me write it as a question is that particular solution plus some homogeneous solution a solution a solution to ax is equal to B well to do that you just put that in the place of X so let's try it out so a times this guy right here times some particular solution plus Holmwood some homogeneous solution is going to be equal to a times the particular solution plus a times some member of my null space and what is this equal to that is going to be equal to B right we're saying that this is a particular solution to this equation that is going to be equal to B and that this is going to be equal to the zero vector because this is a solution to our homogeneous equation so this is going to be equal to B plus zero or it's equal to B so a times this vector right here is indeed equal to B so this is a solution this is a solution yes now the next question is does every solution to the inhomogeneous system or does any solution to the inhomogeneous system take this form so does any does any solution X 2 ax equal to B take the form X is equal to some particular solution plus a member of our null space or plus a homogeneous solution so to do that let's take let's test out what happens when we multiply the vector a times a times a times X let's just let me write it this way let's say that X X is any solution to ax is equal to B let's start off with that and let's see what happens when we take a times X minus some particular solution some particular solution to this so when we distribute the matrix vector product you get a time's our any solution minus a times our particular solution now what is this going to be equal to we're saying that this is a solution to X equal to B so this is going to be equal to B and of course any particular solution to this when you x is also going to be equal to b so it's going to be B minus B so that's going to be equal to the zero vector or another way to think about it is X the vector X minus our particular solution is a solution to a times X is equal to zero think about this if you take this in parentheses right here and you put it right there when you multiply it times a you get the zero vector we just did that you get the zero vector because when you multiply each of these guys by a you get B and then you get B minus B and so that you get zero so you can say that X minus so are any solution X minus the particular solution of X is is a member is a member is a member a member of our null space of our null space right by definition your null space is all of the X's that satisfy this equation and let's say so since it's a member of our null space we can say that it is equal to so are any solution minus our particular solution is equal to some member of our null space we could say that it's equal to a homogeneous solution there might be more than one a homogeneous solution a homogeneous solution now if we just add our particular solution to both sides of this we get that any solution remember we assumed that X is any solution to this that any solution is equal to our homogeneous solution is equal to a homogeneous solution plus plus a particular source or plus our particular solution so we've proven it both ways that this is a solution to our in homogeneous equation and that any solution to our inhomogeneous equation takes this form right here now why am I so why am I so concerned with this you know I've been kind of fixated on this inhomogeneous equation for some time but we've been talking about the notion of a transformation being one one a transformation being one-to-one that was one of the two conditions for a transformation to be invertible now to be one-to-one so let me draw a transformation here so let's say this is my domain X and this is my co domain right here Y and I have a transformation that maps from X to Y I have a transformation from the maps from X to Y in order for T to be one-to-one so alright like this 1 to 1 in order for T to be one-to-one that means for any B that you pick here for any be for any B that is a member of our codomain there's at most at most one solution to a times X and I'm assuming that a is our transformation matrix so we can write our transformation T as being equal to some matrix times our vector in our domain so this would be a X if this is X right here so T would map from that to that right there so in order for our transformation to be one-to-one that means you pick any be here there has to be at most one solution to ax is equal to B or another way to say that is that there is at most one guy that maps into that element of our codomain there might be none so there could be no solutions to this but there has to be at most one solution now we just said we just said that any solution to an in homogeneous so we just said that any we write it in blue any solution takes the form takes the form if there is a solution so if there isn't a solution that's fine we that will still satisfy one-to-one so but if there is a solution any solution is going to take the form X particular plus a member of a plus a member of your null space plus a with this guy right here is a member of the null space this this thing right here just applies to that guy right here any solution if they exist if there are no solutions fine you can still be one-to-one but if you do have a solution you can have it most one one person maps to it and any solution will take this form we just I just showed you that now in order to be one-to-one this can only be one solution the solution set can only be one solution this can only be can only be one solution we can only have one solution here right which means what does that mean that means that our that this guy right here cannot be more than one vector it just has to be one vector this is only there's only one particular solution right there but this guy right here has to be well there's for any solution set depending on how you define it there's only one particular vector there but this guy the only way that you're only going to have one solution is if your null space the only way you're going to have one solution is if your null space is trivial if it only contains the zero vector your null space will always at minimum contain the zero vector and the last video I think I you know just off the cuff that all your null space has to be empty but now your null space will also will always by definition by the fact that it is a subspace it will always contain a zero vector you can also always multiply a times zero to get the to get zero so your null space will always contain that but or don't have only one solution your null space can only have the zero vector so that this can only be zero and so that your only solution is going to be the particular solution that you found depending on how you got there but it's only going to be your particular solution so let me put this this way so if you sew one to one in order to be one to one in order to be one to one your null space of your transformation matrix has to be trivial it has to contain only only the zero vector now we've covered this many many videos ago what does it mean if your null space is only contains the trivial vector let me make this clear so if your transformation vector looks like this a 1 a 2 all the way to a n and you're multiplying it times you know x1 x2 all the way to xn and the null-space is all of the X's that satisfy this equation zero and you're going to have em zeroes right there so if your null space is trivial and we're saying that that is a condition for this for you to be one to one for your transformation to be one to one the transformation that's specified for this by this matrix if your null space is trivial what does that mean that means that the only solution to another way of writing this is a one alright this x1 times a1 plus x2 times a2 all the way to X n times a n is equal to the zero vector these are equivalent statements right here I just multiplied each of these terms times these respective column vectors these are the same thing now if you say that your null space has to be equal to 0 you're saying that the only solution to this equation right here the only scalars that satisfy this equation sorry these aren't let me actually because I wrote the scalars as vectors so this guy right here this statement right here is equivalent to x1 times a1 plus x2 times a2 plus all the way to xn times a n is equal to the zero vector where the x1 through xn 0 scalars now if we say the null space is 0 we're saying the only way that this is satisfied is if your x1 all the way to xn is equal to 0 and this means this is our definition actually of linear independence that means that a1 so the null space being 0 also means that your column vectors of a your column vectors let me write it this way it also means that a1 a2 all the way through a n are linearly independent independent now what does that mean if all of these guys are linearly independent what is what is going to be the basis for your column space remember the column space is the span the column space of a is equal to the span of a1 a2 all the way to a n what we just said if we're if we're dealing with a one to one or one of the conditions or the condition to be one to one is that your null space has to be zero if or only contain the zero vector if your null space contains a zero vector then all of your columns are linearly independent if all of these columns fan span your column space and they're linearly independent they're linearly independent then they form a basis so that means that a 1 a 2 all the way to a n are a basis are a basis are a basis or our column space and then that means if all of our column vectors here are linearly independent they obviously span our column space by definition and they're all are linearly independent they form the basis so the dimension of our basis so the dimension of our column space the dimension of our column space that's essentially the number of vectors you need to form the basis it's going to be equal to n we have n columns so it's going to be equal to n or another way to say it is that the rank the rank of your matrix is going to be equal to n so now we have a condition for something to be one-to-one something is going to be one-to-one if and only if if and only if the rank of your matrix is equal to and and you could go both ways if you assume if you assume something is one-to-one then that means that it's null space here has to only have the zero vector so it only has one solution if it's null space only has the zero vector then that means it's columns are linearly independent which means that they all are part of the basis which means that you have n basis vectors or you have a rank of n let's go the other way if you have a rank of n that means that all of these guys are linearly independent if all of these guys are linearly independent then the null space is just the zero vector and then that if the null space is just a zero vector this part of your solution disappears and then you're only left with one solution so you're one-to-one so you're one-to-one if and only if the rank the rank of your transformation matrix is equal to and