51 - Properties of Ker(T) and Im(T)

let's write a theorem a theorem of five sections and the theorem is gonna satisfy this title it's going to be some properties of the kernel in the image so here it is so suppose that T from v1 to v2 is a or let t be a where is it let t be a linear transformation linear map then one the kernel of T is we know that it's part of v1 it's a subset of e1 right it's elements in v1 that satisfy something namely that T sends them to zero and V 2 but the claim is that it's not just any subset it's a sub space a sub vector space of V 1 we said it we mentioned it but we never proved it let's do it to the image is a subspace the range of v2 not just the subset a subspace 3 T is 1 to 1 1 to 1 is a function you know what a 1 to 1 function is a function is called 1 to 1 and remember that a linear map is to start with it's just the function right it's a function that satisfies two more Purdy's the linearity properties but as a function it can be a one to one function and one to one function means that if you take two different guys here in v1 t sends them to two different guys here it doesn't send two different elements in the domain to the same element in the range okay that's a one-to-one function T is one to one if and only if the kernel of T is zero contains only the zero vector okay for T is on to its range on to v2 so what is an honour to function and on to function means that every element in the range is indeed the image of some element in the domain okay so T is on to if and only if the image of T is all of v2 in fact number four is here just for for the statement to be kind of complete number four is the definition it's not it's not a theorem that's the definition of being gone to do you agree a function is on two if its image is the entire e do you agree okay so this is not really a theorem this is in fact a definition nevertheless I want it written here because you can see the the duality the nice duality and it's easy to refer to and number five is is a bit different in it maybe maybe we'll throw in number five as a separate theorem later let's let's handle these four okay so let's start with one I want to show that the kernel is a subspace how do we show that something is a subspace it has to satisfies all the property of being a vector space with respect to the same operations in the ambient space in v1 okay but we know a theorem we don't really need to check all of them it suffice us to check that it's not empty closed under linear under sums and closed under scalar multiplication that's it okay so let's do it proof of one so we want to show that the kernel is not empty right we know T of zero is zero this we proved before remember zero is always in the kernel therefore zero belongs to the kernel right because T sends it to zero and therefore the kernel is not the empty set do you agree good so this is one thing we need to show the other thing we need to show we need to take two guys in the kernel call them U and V in the kernel and show that u plus V is also in the kernel okay so if u and V belong to the kernel this is our W no this is what we're claiming is a subspace then what does it mean that they belong to the kernel it means that T of U is 0 and T of V is 0 that's what it means to belong to the kernel therefore what is T of their sum u plus V T is a linear transformation T of a sum is the sum of the t's right that's the property of being a linear transformation and since both of them are 0 its 0 plus 0 which is 0 therefore if u and V are in the kernel T of U plus V is also 0 and therefore U plus V is also in the kernel do you agree likewise if V is in the kernel then by definition T of V is zero therefore T of alpha v right by being a linear transformation means that you can pull out the alpha that's the second property of being a linear transformation right and this is alpha times 0 which is 0 therefore alpha V is in the kernel as well good so these three bullets show precisely the necessary conditions and the sufficient conditions for being a subspace do you agree good ok proof of two proof of two homework same idea very very similar try to do it on your own ok good let's prove number three proof 3 so let's recall what 3 set let's go back to the theorem itself remember that we don't need to prove 4 because 4 is just the definition a function is on 2 if and only if the image is the is the entire range but we do need to prove 3 so 3 3 is a statement and if and only if statement therefore we need to prove both directions and we're going to prove them separately so the first direction is if T is a one-to-one function then the kernel is trivial only zeros in the kernel second direction if the kernel is trivial then T is one to one question no no three is not a definition three requires proof no a function is called one-to-one if you take two different guys they they get sent to two if X is different than Y then f of X is different than F of Y that's the definition of being one-to-one it doesn't have a priori anything to do with the kernel right you in calculus you saw the definition of our one-to-one function you never mentioned the kernel right these are two different notions okay they're not different it turns out that they're well they are different the kernel is 0 is the same as being one-to-one but that requires proof okay good here it's precisely the definition that's how we defined an onto function okay okay and if you haven't seen these definitions so the definition of what being one-to-one is I'm gonna write in a minute so you're gonna see it the definition of being on two I'm gonna tell you what it is it's precisely this the definition of being on two is the fact that the image is everything okay okay so let's prove three so first of all this direction this direction assume the T is on - ah the T is 1 to 1 what it means by definition that if you take U and V that have the T sends to the same image if T of U equals T of V that can only happen that implies that U was in fact V that's what it means to be one-to-one okay a different way of saying there so assume T is one-to-one by definition this is what it means and this is equivalent equivalent to sing a different way of saying the same thing is that if you is different than V then T of U is different than T of V this and this are the same thing this implies this is the same as not this implies not this that's what I wrote here the reason I chose this one is that this is what I'm gonna work with okay so that's the definition of being one-to-one the word kernel is not there do you agree okay and it's not gonna be there if it's not a linear transformation okay it's only related with we're gonna use the linearity okay so so this is what it means to be one-to-one by definition let's prove that the kernel is trivial so how are we going to show that the kernel is trivial what does it mean that the kernel is trivial it means that the only guy in the kernel is zero there's nobody else there okay so a way of proving this take somebody in the kernel call it V I'm gonna show that V is zero and then I'm done I took some general vector in the kernel and showed that it's well it's just the zero vector okay so take V in the kernel T of V is zero why because it's in the kernel but we also know that T of zero is zero that's a general fact about linear transformations that be proved in the very beginning remember so T of V and T of zero are equal do you agree and that implies T of e equals T of zero were assuming one two oneness that implies that V is zero so here we used one two one where do we use the linearity here right here this we proved previously and this is true for a linear transformation not every function takes zero to zero right not every function sends zero to zero zero two zero a linear one does okay a linear transformation does okay so good do you see why this proves the claim it's very simple very straightforward as long as you're clear on on the roles being played by each guy here okay what what are we trying to prove what do we know why did we show it everybody good I don't understand what you what do you mean by technically I claim that this is a complete proof my claim is that the kernel is zero there's nobody else in the kernel only zero I'm proving it I'm taking somebody general in the kernel and showing that that somebody is none other than zero now you're gonna say take somebody else in the kernel why why do I need to take somebody else no I didn't I took V and kernel t any V you want not specific take any V in the kernel boom it has to be zero that's it only zero is in the kernel good everybody okay but that's just one direction now we need to show the other direction so now we're assuming that the kernel is zero we need to show that it's a one-to-one that okay so assume that the kernel is zero okay how do we prove that it's a one-to-one map I want to show this this is what it means to be one-to-one okay so take take U and V such that T of U equals T of Z and we need to show that u equals to V right if we show that if T of U equals C of V that implies that u equals V we're done so one-to-one map okay so take T of U equals T of V and let's look at consider T of U minus V so what is U minus V what is subtraction this is T of U plus the inverse of V the additive inverse that's how we define u minus V right that's the same thing good now we're using linearity again what is T of a sum the sum of the t's so this is T of U plus T of minus V what is T of minus V in the same little lemma the same little write the same place where we showed that T of 0 is 0 we showed that T of minus V is minus T of V ok obvious because this is like a minus 1 here you can pull it out by linearity so this is T of U minus T of V but we're assuming that T of U and T are V are equal we took a u of in a V that satisfy this so this is just 0 right so T of U minus V is zero do you agree what does that tell us about u minus V it's in the kernel we just showed that teen of it is zero the T sends it to 0 that means that u minus V is in the kernel do you see that ok so I'm gonna write it and on the next board in a second but we're assuming that the kernel is zero so it means that u minus V is zero okay so this implies that u minus V belongs to the kernel because T of it is zero and in turn this implies that u minus V is zero because we're assuming in this direction that the kernel only contains 0 and therefore u equals V good okay so nothing difficult going on here everything is very straightforward you just have to understand what you're assuming what you're trying to prove and how to express it ok sometimes that are those are not trivial things even even if right we didn't use any deep theorems here or the Steinitz exchange lemma or I don't know what very very straightforward you just have to understand the the players who vs. who ok ok so this concludes the proof of that four part theorem now we know that the kernel and the image are in fact subspaces and we know that the kernel is trivial if and only if the map is one-to-one okay what I want to do is add another theorem another theorem maybe we'll write it down here I even don't need to erase this so another theorem let T from V 1 to V 2 be a linear map then oh not then I need another statement if u 1 u 2 dot dot dot up to u n span V 1 so I'm taking a spanning set for V 1 a bunch of vectors in V 1 such that the entire space is all linear combinations of these guys that's a spanning set right then the set T of u 1 T of u 2 dot dot dot T of you in spans V 2 okay so if you think a spanning set send the guys in the spanning set to v2 you get a spanning set for the range okay for the image oh mistake not V to fix this they don't spam v2 there's no chance for them to span v2 unless unless it's on two right they can only spin what's obtained by t they span the image of t which is a subspace of V too good okay let's prove this so how do I prove it how do I prove something like this what's the logic behind it so not much I'm gonna take an element in the image an element in the image and show that this general element in the image is only near combination of these guys that's it right that's all I need to show do you agree so take take somebody in the image what does somebody in the image look like it's T of something right so take T of V this is a general a general element in the image I could say take some W in the image but that's not capturing the fact that it's in the image if it's in the image its T of something that's what it needs to be the image right so take T of V in the image I want to show that T of V is a linear combination of these guys right what do I know I know that V is a linear combination of these guys because this is a spanning set right so V equals some alpha 1 u 1 plus alpha 2 u 2 plus alpha in VN because V is somebody in V 1 you in thank you beats me why I converted to use ear they should have been all these anyway so V is a linear combination of these guys because this is a spanning set for v1 therefore what is T of V T of V is T of this thing T of alpha 1 u 1 plus alpha 2 u 2 plus alpha nu in right by linearity T of a sum is the sum of the t's so this equals T of alpha 1 u 1 plus T of alpha 2 u 2 plus dot dot T of alpha n u n this is the additive part of being linear right now again by linearity again by linearity T of a scalar times the vector the scalars can hop out so this equals alpha 1 T of u 1 plus alpha 2 T of u 2 plus dot dot dot alpha in T of U n do you agree and that's it I'm done I showed that I took somebody in the image it's a linear combination of these guys done ok the the step from here to here which I did in two steps is very very standard we're gonna do it in many many many examples and proofs and all the time when you have T of a linear combination its the linear combination of the T's of the of the original ones for any linear transformation ok so this step is you should you should really feel why why it's a very this is the essence being a linear transformation in fact good he isn't v1z is an element of v1 that therefore it's a linear combination of a spanning set of v1 therefore T of V now everything lives in in v2 once UT something it's in the range it's in v2 okay Tia V is and v2 it's tier of this linear combination being a linear map I can take this step which I took in steps because it's the first time I'm doing it and now I show that T of V is a linear combination of the T's of the U is good okay so let's do an example this was this um we've been rather abstract and theoretical this clip let's do an example it's filling the example there it is so example let's take T going from m2 of our what is m2 of art right two by two matrices to r2 of X these are polynomials of degree less than or equal to two with variable X and coefficients in our okay I have to tell you what T does this is just saying what the domain and the range are now I have to tell you what the rule is right that's how you prescribe a function okay so T of a two by two matrix call it ABCD is gonna be a polynomial of degree less than or equal to two and the polynomial is going to be a plus B that's the free coefficient plus C plus 2 D X plus two C plus four d x squared good we need to check that it's a linear map okay and you need to check that it's a linear map a clue a way to see it but we didn't prove this yet is that every component of the image is a linear combination of the components of the element in the domain okay do you see that so because we said it we already know that it's a linear map but we didn't prove that yet and you need to verify manually show that if you take two guys T of one plus T of the other is T of the sum okay show that if you multiply by alpha you get alpha times T okay so verify that T is linear do it do it very important to do it okay let's see what the kernel of T is so what is the kernel of T the kernel of T is all the guys that get sent to zero zero is the zero polynomial so the kernel of T it's a subspace of the domain it's all the guys of the form ABCD that satisfy that t of them is zero when would this polynomial be the zero polynomial if a plus B has to be zero C plus two Dini has to be 0 and 2 C plus 4 D has to be zero but that's already incorporated in the second requirement do you agree so this is the kernel does everybody agree good ok we can rewrite this it's it's very specific right from here it follows that B is minus a so it's going to be matrices of the form a B has to be minus a and then D is minus C over 2 so if here we have C and there's no relation between a and C there's no requirement of that sort right but D has to be minus C over 2 do you agree so this is the kernel all the matrices of this specific form are precisely the ones in the kernel okay and we can even write this we can even find easily find a basis for that right it's all the elements of the form a times a times this matrix plus C times matrix do you agree so this is in fact the basis for this kernel and I just wrote it as a spin but there it's obviously spanned by this in order to say that it's a basis we need to say something else right that these two are linearly independent and they obviously are because one is not a multiple of the other okay so so let's write that these are linearly independent right and therefore this is a basis these two this set of two elements is a basis for the kernel not for v1 v1 the basis for V 1 we can take the the standard basis right 1 0 0 0 0 1 0 0 etc but this is a basis for the kernel for only for the guys that get sent to 0 and what's the dimension of it so the dimension of the kernel is 2 right because there are 2 basis elements so these are linearly independent and thus a basis good ok so this is the kernel we deciphered what the kernel is and we know its dimension okay let's try to understand what the images do we get all polynomials of degree 3 if not which ones do we get what the images what are the polynomials in r2 of X that are T of some two by two matrix by this formula by this rule okay so in order to do that what I want to use is the last theorem that we had that said take a spanning set of the domain we called it u 1 to u in duty to it that gives you a spanning set of the range right that was the previous theorem so in order so let's write let's find the image of this T in this example we're in this example so here's a basis for the domain m2 of art I'm just gonna take the standard basis that I know well 1 0 0 0 0 1 0 0 we called it a 1 e to remember no we called it a 1 1 e 1 to e 2 1 e 2 to remember that notation 0 0 1 0 and 0 0 0 1 there you have a basis for M 2 of art ok if we see these guys if we do T of this T of this T of this and T of this we're gonna get a spanning set for the image that was the previous theorem let's glance at it for a minute if you take a spanning set for V 1 and a basis is of course a spanning set duty to them you get something that spans the image and that's how we're gonna decipher what the image is ok so T of these guys let's find out what T of these guys is so T of these this is not formal language T of these right T is not a verb okay but I'm considering T as a verb here ok let's T these guys T of them using T as a as everything in fact so T of these will give a spanning set for the image that was the previous theorem do you agree so let's find what T of these are the formulas on the previous board okay so let's glance at it for a minute I want to do T of 1 0 0 0 so I get a 1 here and 0 is everywhere everywhere else so T of the first element is just the polynomial 1 T of 0 1 0 0 I get a 1 here and 0 everywhere everywhere else so again I get just the polynomial 1 right T of 0 0 1 0 0 0 1 and 2 right it's going to be X plus 2 x squared do you see that and T of 0 0 0 1 is gonna be 2 X plus 4 x squared good so let's write that so we get the spanning set that we get for the image is 1 1 X plus 2 x squared + 2 X + 4 x squared this is a spanning set for the image do you agree okay by the previous theorem okay now is it a basis no obviously they're linearly dependent one of them just appears here twice and this one is just um double this one right so we can obviously throw these two away they contribute nothing there are linear combinations of the others and then we're left with 1 and x + 2 x squared which are linearly independent obviously third one is not a multiple of the other okay so good everybody followed these arguments I'm not going to write all the details I'm just gonna write 1 + X + 2 x squared is a basis for the image any polynomial in the image is going to be a linear combination of these two guys and obviously this is not all the polynomials right there's no restriction on the free coefficient but there is a big restriction on the relation between X and x squared do you see that okay what is the dimension of the image to the dimension of the image is to the dimension of the kernel was 2 they sum up to 4 right which is the dimension of the domain space okay then I mention of the range space by the way is 3 in this case right polynomials of degree less than or equal to 2 or of deme of is is a three dimensional vector space right but the dimension of the image is to the dimension of the kernel was 2 they sum up to the dimension of the domain space for good everybody ok so we'll stop this one here everything we did in this in this lesson was kind of accumulate very basic facts very basic ingredients that will enable us to do two major things we want to do at least at least in the near future one is show that every linear transformation is can be presented by a matrix and the other is to show that rank nullity theorem in the context of of of linear transformations namely that the dimension of the image plus the dimension of the kernel equals the dimension of V 1 that's a theorem we're heading for ok so to be continued

Channel: Technion

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

Id: r13KE8iTjyc

Length: 40min 7sec (2407 seconds)

Published: Sun Nov 29 2015