 You try to first check which properties do not depend on the set under question, but rather on the operations themselves right. If I do that, I would see that this commutativity presumably has nothing to do with what? The set I have chosen, it is about the operation right, so this commutativity I may not need to check. Same with the associativity here, same with the associativity here and the distributivity here right, what else can I say? So this closure, this identity and this inverse have to be checked for this addition right, where this closure has to be checked okay. Of course this has to be also checked right sorry, yeah so if you are convinced that it is closure, so depends on how we are proposing to check, what is the test criteria we are putting together. So if you are putting together a clever test criteria, then we might actually do away with two checks, with one check we can do this. So we are trying to design that test, test for when it is a subspace. Then that you are inheriting from a parent space, when it is a subspace that is what we are trying to check. So we want an efficient test for that. So you have to figure out that this one in the field acting on this must also give you back this, oh so this should be sorry W right, yeah so this should be W. So now we have to propose such a scheme, but before that let us also turn our attention to the following. I am going to claim something like this, again this requires sort of a proof okay, what is the proof? See the statement of this is profound. On the right hand side what do we have? In the language of vector spaces and all, what is this? This is the additive inverse of a vector V okay, for all V and V. This is the additive inverse of a vector, is it not and on the left hand side I am giving you a recipe for obtaining the additive inverse. What am I saying? I am saying go ahead, look for the multiplicative identity in the field, take its additive inverse. So the additive inverse of the multiplicative identity of the field acting on this vector by means of the scalar multiplication that I have defined leads to the additive inverse of the element itself, yeah again those ones and zeros should not be read as something trivial. Can we prove this? How do we prove this then? So what is our starting point? 0 dot V is equal to sorry right, so 1 plus minus 1 dot V right and then because this I know what I have already seen this right, what this object is going to be right, so this is nothing but the 0 of the vector space no right, that is what we have just proved while back and what is this? Distributivity 1 times V plus minus 1 acting on V, now this is of course just V plus minus 1 acting on V, let me add equals to both sides so that they still remain equal. So I am just going to add minus V which is the additive inverse of V to both sides and I am going to have minus V plus V and I am going to combine them through associativity so that I have this, this of course reduces to the identity, the additive identity in the vector space which is 0 of V plus minus 1 on V which means that the additive identity is nothing but the additive inverse sorry the additive inverse is nothing but the additive inverse on the field of the multiplicative identity acting on the vector yeah, so you know these tidbits here and there we need to convince ourselves that these are true and now with this being said we are in a position to propose a condition that has to be checked in order for something to be a subspace and here is the proposition okay just a one line check what is the consequence if this is true we will try and convince you that this is indeed true but let us first see what is the consequence of this if this is true, the consequence is you pick out two arbitrary elements without loss of generality from the set W and pass it through just this one check the closure take any one of the objects take the other object scaled by alpha, so this is a scalar multiplication that is carried out with W1 and see where if this passes the test so you do not assign some special properties to your W1 and W2 just choose any arbitrary W1 and W2 any arbitrary alpha and if you can show that despite that arbitrariness this object let us call it W hat still resides inside W and you do not need anything outside of W let us say inherited from the parent space V then that W is rightfully a subspace in itself in other words it is going to be also a vector space in its own right that is this object is also a vector space okay, so how do I try and convince you of this I am just going to check out against these meaningful properties because you see the ones I have ticked out do not need to be checked as I said because they are inherent in the operations themselves okay, so how do we check for the closure, so if I choose alpha is equal to 1 if I choose alpha is equal to 1 then is not the closure under vector addition given because what I am doing is just picking out W1, W2 arbitrarily and adding them and I am claiming that each of them must belong to W so that means by choosing alpha is equal to 1 I am assuring myself that this operation of addition must be closed and this must be swept over all possible alphas so I have just chosen one alpha one legitimate alpha which is just one so that means if I have verified that this is true for all alpha all W1 all W2 then alpha is equal to 1 is just one such alpha so that has also been that base has been covered in my check so that means this test is holistic in the sense that it is already checked for the closure if something has passed this test it has already passed this closure test that is all I need to convince myself that once something passes that test it passes all the remaining tests remaining meaningful tests I mean these I do not care about because they come about just as a consequence of the operation themselves right. What about the identity? What about the identity in view of what we have just proved choose any W call it W2 call that same W as W1 and choose alpha to be minus 1 so for this you choose W1 is equal to W2 is equal to sum W right and alpha is equal to minus 1 in view of what I have just proved a while back when you take minus 1 times this W it is nothing but the additive inverse so the additive inverse gets added with the number itself it must result in the 0 the additive identity and therefore if you have swept over all possible alpha as all possible W's W1 and W2 you have already covered the case when 0 must be a part of the set W because otherwise it would not have been closed right so therefore this check also verifies this what about the inverse once you know that the additive identity is part of the set choose W1 is equal to some W W2 is equal to the 0 I know that the 0 belongs now because of this second point that is already been checked so therefore W2 is equal to 0 must also be a part of it and choose alpha is equal to minus 1. So therefore for every object W that you might pick for you so therefore for every object W that you might pick from the set W big W any little W that you pick from big W 0 of course also comes from big W because of our previous point here and with alpha is equal to minus 1 you have the point that minus of W that is the additive inverse must also be a part of this W you have checked for the existence of the additive inverse in W itself in other words the addition property is all check out yeah what do we need to check now the closure how do we do that well is it really difficult because we are already scanning over all possible alphas. So choose W1 is any W here W2 as 0 and alpha sweep over all possible alpha in F that is what you are required to do under this check because you are going to check for all possible alpha that is already specified as part of your prescription. So you are anyway going to do it as part of this check so that means if you have performed this check this also happens to reduce to a special case of that check. So this closure property is there what about this well the one has to be part of the field right and again the closure as your friend said a while back your friend said that this property is a consequence of the closure. So nothing to be checked here right because it is inherited already one is see the field is already inherited here so that one must be there because of the multiplicative identity in the field that must exist in a field therefore this one must be there and the original parent spaces operations are still legitimate here so this is also right. So therefore in other words at least even if I am not written a complete formal proof of this you can now convince yourself that given that I pick out a big W which is a subset of this big V and inherits these same operations under the same field if I ask you the question as to whether the newly constructed structure also has the properties of a vector space in other words if it is a subspace of the original vector space it just suffices that you check this against arbitrary objects a couple of arbitrary objects in W and an arbitrary scalar yeah so maybe I will take a couple of examples now and show you how it is done so that you get a feel for how this how this check works and how elegant it is okay. So I am going to first claim that certain objects are subspaces so the moment I say something is a subspace there must be something that is an original a bigger vector space that is sitting over there whose subspace I am claiming this new object to be okay. So let us say because we understand matrices is better let us just take the examples of some matrices is there any questions please ask yeah yeah no it is not circular in order to prove this I did not require this I mean sorry in order to prove this I required that in order to prove this I did not require this oh you mean but that is independent of that I had already proved you know sure I mean probably his confusion came from there so I thought maybe you are taking the same sorry yeah no no but but you see you are asking for this to be part for all possible alphas so for alpha is equal to minus one this has to be a part that is what the check is at that point you are saying that whether this check covers this base now this check allows me to put alpha is equal to minus one nothing prevents me I may not know that whether individually it belongs or not but the overall object definitely belongs I do not yet know I do not yet know later on it will it will come to later on it will come to effect that because of the closure but actually because of the closure property you can say that if you take two objects and you take their sum then they must belong there right so if you take two objects and their sum and on the right hand side you have another object and you take its inverse on both sides so any sum the constituent components cannot help but belong to that vector space otherwise it would not be closed because even when you are subtracting so called subtraction you are basically doing this additive inverse and adding it right so you are adding objects in any case so that closure will guarantee that it has to exist so it is all connected in this step you are using this step but in this step you are not using any property other than the fact that which is why I proved it just before this that you have to know that minus one acting on that is the additive inverse of the multiplicative identity acting on any vector results in the additive inverse of the vector but that is independent of any of these properties that is something that came from the parent spaces property itself so that is the only thing I have used in proving this I have not used this in proving this I have just used this in proving this so it is not circular right so I will just take the example of matrices and I will cook up two such objects specifically so consider a belonging to general field by the way not real or complex okay m cross n matrices over any general field okay now the first thing you have to understand is fm is a vector space over f and fn is a vector space over f do you agree with these again I am not going to prove this just take this as an exercise to convince yourself that when you have n tuples or a finite number of tuples of objects from a certain field yeah and you stack them up as a column of numbers in fact you can stack them up as rows of numbers also it matters not so long as you know how that vector addition in Euclidean spaces works here of course when I say over the field that means it respects the fields addition and multiplication operation so if it is the module if it is the finite field for example you will have the modular operations so when you take an object from the field multiplied with an object that is an n tuple of the objects from that field and you are multiplying them in the usual scalar manner every number gets multiplied a scale you have to take the modular multiplication so it somehow embeds that operation of the field to that n tuple okay so I am not explaining all of that in detail of course but I hope that I expect you to understand that in that common sense if I want to be very precise I must also mention that but you know I have not defined that plus n dot assuming that you understand what the plus n dot is okay now these are vector spaces I am going to define the following objects one I am going to call this as image of a is equal to the following set which is x belongs to of course it is n okay I am going to define a second object this is called the kernel of a so where is this y coming from by the way what do you think this is a subspace of if at all it is a subspace f m right okay so this apparently comes from f m so this comes from well of course it is there in the definition itself so you might as well guess this is m is f m now if I want to prove that these two are subspaces given that I believe I have left it as an exercise for you to prove but given that I believe that these are indeed vector spaces what I am required to show is that this image is a subspace of f m f to the m m tuples of members in the field and kernel is a subspace of n tuples in the field right how do we show this based on what we have just proposed over there and hopefully given you a an idea about why this proposition works okay so what I have to do is I have to pluck out two arbitrary objects I have to pluck out two arbitrary objects in this right so for one pick y 1 is equal to a x 1 for some x 1 in f to the n and y 2 is equal to a x 2 for some x 2 in f n if I am saying that y 1 and y 2 must be in the image of a then they must be representable in the form a x 1 and a x 2 there must be some x 1 and x 2 such that y 1 and y 2 are representable in this manner now consider y 1 plus alpha y 2 right what is this equal to this is nothing but a x 1 plus alpha a x 2 and because of the usual properties of matrix multiplications we can say that this is nothing but x 1 plus alpha x 2 but what is this x 1 plus alpha x 2 can I not write this as a x hat where x hat belongs to an n tuple of members in the field but that by definition implies that y 1 plus alpha y 2 is representable in the same form that would imply that it is a part of the image of a this means y 1 plus alpha y 2 must belong to image of a and I am done I picked out arbitrary y 1 and y 2 members from the image of the set a sorry this set that is the image of a right and then I showed you that y 1 plus alpha y 2 cannot help but also belong to the image of a which means that the image must be a subspace we know it is already something that is sitting inside this m tuple and m tuple this m tuple is already a vector space so all that I needed to check was this by that proposition so that is the power of that proposition anytime you are asked to check whether something or show that something is a vector the subspace and you know it is sitting inside some bigger vector space you only have to know how to massage it into this form the same thing we can make it work for the kernel let us quickly do that in a minute or so and you will have more exercises by tonight on these sort of topics so the next thing is say x 1, x 2 belong to kernel of a implies a x 1 is equal to 0 and a x 2 is equal to 0 right this is even simpler consider x 1 plus alpha x 2 where alpha is arbitrary then it implies that a acting on x 1 plus alpha x 2 is equal to because of the distributivity of matrix multiplication with a vector this is a x 1 plus alpha a x 2 of course individually there are 0s so this is also 0 which means where x 1 plus alpha x 2 must also belong to the kernel of a and the kernel of a that set is sitting inside an n tuple that is f n therefore it must be a subspace of f n so the image of a is a subspace of f m the kernel of a is a subspace of f n these are only two special examples of subspaces in the next lecture we shall look at more general examples such as things called span of a set of vectors and we will see that even that turns out to be a subspace and we will hopefully deal with a lot more other abstract objects thank you.