52 - The rank of T

I want to define what is the rank of a linear transformation the rank of a linear transformation and here is the definition it may look weird as at least the first time you see it why would you define it as such but then I'm gonna make a remark so the rank of a linear map T is denoted by R of T and by definition it's the dimension of the image that's the rink the dimension of the image and we denoted by R of T ok so what I'm gonna do now is write a theorem and prove it and the the key thing in that theorem is the following fact we mentioned several times and we're starting to get the feeling towards the fact which we haven't proven yet that any linear transformation can be presented by a matrix okay we know what the rank of a matrix is right it's the number of nonzero rows in an echelon form or we said that it's the same as the dimension of the row space right so we know what a rank of a matrix is what I'm going to claim now is that if T is presented by a matrix then the rank of T defined as such coincides with the rank of a which presents T ok so the two notions in fact coincide okay so let's write that let's write what we're trying to say so theorem let a some m-by-n matrix be a matrix over some field F define the linear transformation going from FN to F M by the matrix T of V equals take that given matrix and multiply it by V okay V is in the domain space in FN it's a vector with n entries so it's n by 1 so a fits to this is well defined a is M by n V is n by 1 the result is M by 1 which is a vector in FM right so this is precisely a transformation given by multiplication by a fixed matrix okay so if this is an if we're defining T this way we're not saying yet that any T is given like this that we're gonna say later but if T is given like this then we're gonna have four statements ah let's call it zero the first statement we're gonna call statement number zero because we already proved it t is linear this we argued before it did that something did find like this is a linear transformation we showed this this was example two in some and I think in the first clip where we discussed linear transformations remember that okay so first of all T is linear one the image of T the image of T as a linear map is related to the matrix a obviously because T is given by some matrix and the way is it it's related is the image of tea is or equals call of a the image is the column space of a okay we're gonna see it it's not too difficult to which is what I refer to before writing this theorem the rank of T coincides with the rank of V the rank of a we know what it means it's a number of non-zeros in the echelon form the rank of T is the dimension of the image okay and three which I'm out of space so I'll write here and luckily it's short three the kernel of T is none other than the null space of a remember what the null space of a is all the solutions the space of solutions to the homogeneous system ax equals zero remember that okay so let's prove this I'm not gonna prove zero because we've done it okay if you don't recall it's very good practice you need to take v1 and v2 show that T of V 1 plus V 2 is the same as T of V 1 plus T of V 2 that's precisely the the distributive property of matrix multiplication remember we did that everybody ok ok ok good ok let's show one one is where most of the work is him in this in this spirit so I want to show that the image of T is the same as the column space of a so proof 1 let's take the standard basis for the domain space so I'm going to take 1 0 0 dot 0 and then 0 1 0 dot zero all the way up to zero zero zero one this is the standard basis for F n do you agree so this let's give them names let's call this one e 1 e 2 all the way up to e n this is a basis the standard basis of F n of the domain space do you agree maybe we should put these squiggly brackets here good everybody good okay therefore by a theorem we had in the previous clip if we do T to them T of V 1 T of V 2 all the way up to T of V n what do we know about this set it spans the image exactly do you agree if you take a spanning set in particular a basis of the domain T of it spans the image we prove that ok but Lennon let's understand what is T of e 1 what is T of e 1 T of e 1 we know what T is T is given by a matrix a fixed matrix so T of e 1 by definition of what T is it's a times e 1 right do you agree that was the definition of T in this in this context in this theorem so it's a a let's write what a is a 1 1 all the way up to a 1 M and then a 2 1 a 2 M all the way to no in right it's an M by n matrix and then am-1 all the way up to a em in this is a I just wrote what a is explicitly gave all the entries names right x e 1 e 1 is this 10.0 do you agree what do we get what is the product of a times e 1 it's a 1 1 times 1 plus a 1 2 times 0 plus a 1 3 times 0 plus a 1 4 times 0 do you see that the first entry is just gonna be a 1 1 everybody else hits a 0 in the multiplication right this times this only a 1 1 hits the 1 and survives do you see that the second entry is this row times this column only a 2 1 hits the 1 all the rest get multiplied by 0 do you see that and likewise all the way up to a em 1 when I the last entry is this last row times the column a 1 only a.m. one meets the 1 multiplicatively and survives the rest hits 0 and die do you agree so what is that it's the first column of a do you agree okay so this is the first column of a do you agree ok and likewise likewise you can do it for all the call for all of these T of e eyes and in general T of e J is going to be equal to a times e G that's the definition and that's gonna be if you spell out the details it's gonna be the Jayce call him of a-ok if you need to do one or two more do another one duty of III and convince yourself that what you get is the third column of a okay so T of this standard basis sends you precisely to the columns of a and this statement is that the columns of a spanned the image right these are the columns of a and these span the image so the columns of a spanned the image the columns of a also spend call a so they're the same right so conclusion the image of T is the same as the column space of a good okay now let's do property part two of the theorem let's recall what part two said part two said the rank of T is a transformation which is just the dimension of the image equals the rank of a the matrix which prescribes T okay by which T is defined and that's gonna be very easy but now we are gonna rely on theorems that we proved earlier so R of T by definition is the dimension of the image of T that's how we just defined it do you agree but the image of T is the column space of a that was part one so this is the same as the dimension of the column space of a maybe I should put this here do you agree so this is by this is by the definition of the rank of a transformation and this is by Part one of the theorem which we just proved do you agree now we know that the dimension of the column space equals the dimension of the row space this is a theorem that we proved I don't even remember if he proved it or not but this is a theorem that we mentioned earlier okay and the dimension of the row space is RV is the rank of a matrix we didn't define it as such we argued that it equals but in many places its defined as such ok so again this is a theorem that we had earlier so both of these are earlier theorems okay and if you look at the ends of this equality the rank of T equals the rank of a a is a matrix t is a linear map when T is given by the matrix a good okay what about part three part three said that the kernel of T is just the null space of a ok the kernel of T is the null space of a this in fact is is almost the definition almost the definition what is the kernel of T the kernel of T is all the guys that T sends to 0 what is the null space of a its the solutions to the homogeneous system ax equals 0 okay so it's precisely those guys V such that AV equals 0 that's what it means to be in the null space do you agree so that's the same thing a V equals 0 that precisely means that T of V is 0 because T of V is defined to be AV does everybody see that so it's just the definitions so let's write it kernel of T is precisely those elements those V's in F n such that T of V is zero right that's the definition of the kernel this is the same as those V's such that AV equals zero because T of V is a V that's the transformation we're considering right and this is precisely by definition the null space of a all the V's that solve this homogeneous system of equations where a is the coefficient matrix good everybody ok so having said all of this having said all of this we're now ready to state the rank nullity theorem for transformations when they're given by matrices ok so we call so this this so maybe let's write here this completes the proof of this theorem right ok so recall here's the rank nullity theorem that we had that the previous version that we had of it when we all we knew of with systems of equations and matrices ok so let a X equal 0 B be a homogeneous system in n unknowns then n equals equals the rank of a plus the dimension of the null space of me right that was the rank nullity system rank nullity theorem good okay now let's look at the statement we had so far if if T is given by T of V equals a times V if T is a linear transformation not any linear transformation but one that is given by multiple by multiplication by a fixed matrix then this n this n is precisely the dimension of V one I I should add that here maybe if T sorry if T from V 1 to V 2 is given by T of V equals AV then so n is the diamond I'm erasing again sorry T from F n to FM I want to be precisely in the context of the previous theorem right that's the only situation in which I can multiply by a matrix a right okay good sorry everybody with me okay so if we're in the context of the previous theorem if we have a linear transformation that's given by multiplication by a fixed matrix then what is n and is the dimension of the domain space of F n right then the dimension of the domain space let's call it V 1 this is V 1 let's take a color this is v1 and this is v2 in general right the dimension of V 1 is in do you agree the rank of a the rank of a is the rank of T that was part 2 of the previous theorem the rank of T is just the rank of a do you agree by the previous theory the dimension of the null space is the same as the dimension of the kernel because the null space is the kernel by part 3 so the dimension of the kernel of T equals the dimension of the null space of a because the kernel and the null space are the same in this situation ok and I want to add one more thing R of T by definition is the dimension of the image of T that's how we defined R of T good so taking these three ingredients together with the rank nullity theorem for a itself we get the following statements statement that the dimension of V 1 equals the rank of a the rank of T the dimension of the image plus the dimension of the null space the dimension of the kernel so this is the form that the rank nullity theorem takes and when I write it like this you don't it's a fear about dimensions you don't see any rank you don't see any nullity you just see dim dim dim dim dim do you agree you can only almost sing this theorem right so that I mention of the domain space equals the dimension of the image plus the dimension of the kernel we saw it in all the examples that we had right we verified it for each example whether they were geometric examples or examples like this or examples that were completely unrelated to matrices in any form right they all satisfy this and we just proved it but only for this case right the claim is that this is true in general for any linear map T not just the linear map T given by multiplication of a matrix ok this these this line of argument shows this only when T is given by multiplication by a matrix do you agree ok so we will prove this we will prove that this is true this is valid for any linear map but that would be next time okay so is the the picture clear is what's going on here clear okay so next time I'm going to prove the first thing I'm going to do is prove this as a standalone theorem without being related to matrices in any way okay and the reason I want to spend time on it is because the proof is kind of exhibits ideas that are important okay it's not a very difficult proof not a very long proof but I still want to do it okay and then we want to continue to the argument that we still are only hinting at that this is in fact a general situation that if you start with a completely abstract T from v1 to v2 you can somehow generate a matrix a such that T can be written in this form for that matrix a okay it still may look a bit mysterious but we have some clues as to how it goes already okay but we'll get there any questions about all this okay there's you may need to read it again and connect the dots kind of there's a lot of a lot of words here that if they confuse you you need to just think of the definitions and read everything we did in this last in this current lesson again okay what is the kernel what is the null space what is the rank of a of T the image and why why these three statements are true and that gives this for free in this case good okay

Channel: Technion

Views: 13,106

Rating: 4.9607844 out of 5

Keywords: Technion, Algebra 1M, Dr. Aviv Censor, International school of engineering

Id: HaEVxx5xVII

Length: 25min 17sec (1517 seconds)

Published: Sun Nov 29 2015