Definition of V** (double dual) and an amazing miracle Dual Space Definition: • Dual Space Dual Spaces Playlist: • Dual Spaces Subscribe to my channel: / drpeyam
In infinite dimensions, the vector space can be seen as a subspace of the bidual. In this video, you showed that there was an isomorphism between a vector space and its bidual (in finite dimensions). Is there any chance you could talk about the more analytical aspects of all this? For example, could you show that this isomorphism is also an isometry? As always, this has been a great video. I love this whole series. Thanks for making an approachable series into such an abstract topic! I really appreciate it.
3:40 That reference to Zelda A Link To The Past it seems you like Zelda in general since it's not the first time you speak about Zelda :) Btw great video as always keep the good work 👍
Interesting observation: If you think about it, when you have two spaces v and w , and a transformation T from v to w then T transpose will be from w* to v* and T transpose transpose will be from v** to w**. But since T transpose transpose is just T it means T sort of do the same to v and v**.🤔
Hello! Thank you for your wonderful explanation! (The video indeed helped me a lot!) Still I have a question, I hope you would answer for it whenever possible, so that I can improve my knowledge about the concept! Here's my question: the function psi get x(a vector in V) as an input, however when you're proving the linearity of psi, you inserted f(which is obviously a function in V*), so that it became psi(x+cy)(f). I'm bit confused with the concept; shouldn't an input always be a vector, since psi is a mapping from V to V**? How can it possible for us to input f? What does it actually mean? Anyway, I appreciate that I can get such a nice lecture for free, thank you again, and I would be waiting for your reply!
Yes so psi takes x as an input and spits out something in V** so psi(x) is an element of V** hence a super function from V* to R. Hence the input to psi(x) is f
Dr. Peyam there is an integral that Breaking my mind it is the indefinite Integral of { 1/((1+x^n)^((n+1)/n))} there has been nothing I could find to help me solve this😔
Hi! Awesome playlist! I wanted to ask something on the relation between V and V**. Maybe in math there is no reason to distinguish 'being' and 'behaving' , but when we say V=V** do we mean that they are the same or that they behave the same? I think ( and please tell me if it makes any sense) that thanks to that SUPER isomorphism of which you spoke we can see the vectors of V as linear function (belonging to V**) which are the evaluation of a function in V* on the vector of interest. Thus my doubt is: the element of V** are functions (as element of (V*)*) or vectors (as element of V=V**)? Or maybe there is no need to ask such a thing since what matters is the behaviour in those cases? Thaks in advance from an italian student ^^
oh sorry i accidentally wrote in bold, whenever there is a bold line there are two stars , one on its right and one on its left ( so the relation of wich i refere is between V and V** and so on)
Best explanation on double duals I have seen. Also is the notation of L(V, F) the same thing as Hom(V, F)? If so, then why is there two notations? Anyways, great video! What can I say, you can always count on Dr Peyam to have the best explanations!!
I think that the dual of an infinite dimensional vector space is always strictly larger than the space. Therefore, the double dual of V is isomorphic to V if and only if V is finite dimensional.
@@drpeyam Looking up the definition on wikipedia, it seems like a reflexive space is one where the continuous dual is isomorphic to the original space, not the entire dual. I am pretty sure the entire dual space of a vector space has to have a higher dimension when said vector space is infinite.
@@drpeyam I am guessing you are using the map g goes to (f goes to integral of fg) as an isomorphism of L2 into L2 dual? In this case, every such map, f goes to integral of fg is bounded, and hence continuous, by Cauchy Schwarz. Wikipedia says that it is the continuous dual of L2 that is L2, not the regular dual space.
will newman is correct here. A vector space V is isomorphic to V** if and only if V is finite-dimensional, and this can be seen by noting that the dual space of an infinite-dimensional vector space has a strictly larger dimension than the space itself (so the double dual will also have strictly larger dimension). And yes, according to the Wikipedia article on reflexive vector spaces, the definition of a reflexive vector space is one that is isomorphic to the continuous dual of its continuous dual. If you want to go a different route, you can think about modules instead of vector spaces. There is always a natural map (the one you described in this video) from any module to its double dual. For vector spaces (regardless of dimension), this map is always one-to-one. The map may not be one-to-one for modules, so we call a module M *torsionless* if the natural map M → M** is one-to-one. A module M is called *reflexive* if the natural map is an isomorphism. So the concept of "reflexive" is more interesting in module theory than for abstract vector space ;)
There’s a very elegant isomorphism from V to V** with a neat formula. To find an isomorphism from V to V* you have to go through a basis, which is uglier
@@drpeyam I'm not sure if this is the case, but my interpretation of what you're saying is that (categoricaly) there is a natural isomorphism between V and V**.
i have some questions : 1) when we say that two vector spaces are isomorphic , we mean that they are isomorphic according to a linear transformation ?, like that two vector space V and W can be isomorphic if we talked abt a specific LT f and not isomorphic for another LT g . 2) when u proved that V and V** are isomorphic u proved that according to the HAT "^" linear trasformation , is there any other LT that also isomorphic ??
Isomorphic means there is a LT between V and V**. There is no such thing as isomorphic with a specific LT, unless maybe you say it’s reflexive if hat is an isomorphism.
@@drpeyam thank you for your answer , i have another question : V and V** are only isomorphic for the hat linear function ? thanks again . i mean if there is ONLY ONE (OBLY ONE) LT that is bijective between two vector spaces V and W , then we say that those two are isomorphic , we only need ONE bijective linear transformation to say that two vector spaces are isomorphic ???
Thanks for the video, I learned why V^{**} and V are isomorphic! I think you can shorten the proof of $psi being 1-1: We will show that Ker($psi) = 0. If $psi(x) =0, f(x) = 0 for all f in V*. In particular, f(x)=0 even if function f is the identity function. Hence x=0. Thanks!
your description of V** is by far better than in books!!! Thank you so much
This explains why, in quantum mechanics, we sometimes write = f(x)
That is projecting f along \ket{x}
In infinite dimensions, the vector space can be seen as a subspace of the bidual. In this video, you showed that there was an isomorphism between a vector space and its bidual (in finite dimensions). Is there any chance you could talk about the more analytical aspects of all this? For example, could you show that this isomorphism is also an isometry?
As always, this has been a great video. I love this whole series. Thanks for making an approachable series into such an abstract topic! I really appreciate it.
3:40 That reference to Zelda A Link To The Past it seems you like Zelda in general since it's not the first time you speak about Zelda :)
Btw great video as always keep the good work 👍
My favorite video game ❤️
Interesting observation: If you think about it, when you have two spaces v and w , and a transformation T from v to w then T transpose will be from w* to v* and T transpose transpose will be from v** to w**. But since T transpose transpose is just T it means T sort of do the same to v and v**.🤔
There’s a video on that, actually!
hhhhhh i was thinking of the same thing but didnt reach the isomotphic part , nice thats very cool , i was like woooow
Hello! Thank you for your wonderful explanation! (The video indeed helped me a lot!) Still I have a question, I hope you would answer for it whenever possible, so that I can improve my knowledge about the concept! Here's my question: the function psi get x(a vector in V) as an input, however when you're proving the linearity of psi, you inserted f(which is obviously a function in V*), so that it became psi(x+cy)(f). I'm bit confused with the concept; shouldn't an input always be a vector, since psi is a mapping from V to V**? How can it possible for us to input f? What does it actually mean? Anyway, I appreciate that I can get such a nice lecture for free, thank you again, and I would be waiting for your reply!
Yes so psi takes x as an input and spits out something in V** so psi(x) is an element of V** hence a super function from V* to R. Hence the input to psi(x) is f
Amazing video!!!!
Lamont Cranston would be proud! Who knows what transforms lurk in the hearts of math? The shadow knows! The weed of mathematics bears basis fruit!
Dr. Peyam there is an integral that Breaking my mind it is the indefinite Integral of { 1/((1+x^n)^((n+1)/n))} there has been nothing I could find to help me solve this😔
Hi! Awesome playlist! I wanted to ask something on the relation between V and V**. Maybe in math there is no reason to distinguish 'being' and 'behaving' , but when we say V=V** do we mean that they are the same or that they behave the same? I think ( and please tell me if it makes any sense) that thanks to that SUPER isomorphism of which you spoke we can see the vectors of V as linear function (belonging to V**) which are the evaluation of a function in V* on the vector of interest. Thus my doubt is: the element of V** are functions (as element of (V*)*) or vectors (as element of V=V**)? Or maybe there is no need to ask such a thing since what matters is the behaviour in those cases? Thaks in advance from an italian student ^^
oh sorry i accidentally wrote in bold, whenever there is a bold line there are two stars , one on its right and one on its left ( so the relation of wich i refere is between V and V** and so on)
Cool. Thank you very much.
Best explanation on double duals I have seen. Also is the notation of L(V, F) the same thing as Hom(V, F)? If so, then why
is there two notations? Anyways, great video!
What can I say, you can always count on Dr Peyam to have the best explanations!!
I think so, but Hom(V,F) is more general
I think that the dual of an infinite dimensional vector space is always strictly larger than the space. Therefore, the double dual of V is isomorphic to V if and only if V is finite dimensional.
I don’t think so, there are some spaces called reflexive spaces, where the double dual is isomorphic to the original space
@@drpeyam Looking up the definition on wikipedia, it seems like a reflexive space is one where the continuous dual is isomorphic to the original space, not the entire dual. I am pretty sure the entire dual space of a vector space has to have a higher dimension when said vector space is infinite.
Not true, though! The dual of L^2 is isomorphic to L^2, which is isomorphic to the double dual of L^2
@@drpeyam I am guessing you are using the map g goes to (f goes to integral of fg) as an isomorphism of L2 into L2 dual? In this case, every such map, f goes to integral of fg is bounded, and hence continuous, by Cauchy Schwarz. Wikipedia says that it is the continuous dual of L2 that is L2, not the regular dual space.
will newman is correct here. A vector space V is isomorphic to V** if and only if V is finite-dimensional, and this can be seen by noting that the dual space of an infinite-dimensional vector space has a strictly larger dimension than the space itself (so the double dual will also have strictly larger dimension).
And yes, according to the Wikipedia article on reflexive vector spaces, the definition of a reflexive vector space is one that is isomorphic to the continuous dual of its continuous dual.
If you want to go a different route, you can think about modules instead of vector spaces. There is always a natural map (the one you described in this video) from any module to its double dual. For vector spaces (regardless of dimension), this map is always one-to-one. The map may not be one-to-one for modules, so we call a module M *torsionless* if the natural map M → M** is one-to-one. A module M is called *reflexive* if the natural map is an isomorphism. So the concept of "reflexive" is more interesting in module theory than for abstract vector space ;)
8:00 reflexive case
I do not understand in what sense the map from V to V** (double dual) is _more_ isomorphic than the map fro V to V* (single dual).
There’s a very elegant isomorphism from V to V** with a neat formula. To find an isomorphism from V to V* you have to go through a basis, which is uglier
@@drpeyam I'm not sure if this is the case, but my interpretation of what you're saying is that (categoricaly) there is a natural isomorphism between V and V**.
i have some questions :
1) when we say that two vector spaces are isomorphic , we mean that they are isomorphic according to a linear transformation ?, like that two vector space V and W can be isomorphic if we talked abt a specific LT f and not isomorphic for another LT g .
2) when u proved that V and V** are isomorphic u proved that according to the HAT "^" linear trasformation , is there any other LT that also isomorphic ??
Isomorphic means there is a LT between V and V**. There is no such thing as isomorphic with a specific LT, unless maybe you say it’s reflexive if hat is an isomorphism.
@@drpeyam thank you for your answer ,
i have another question : V and V** are only isomorphic for the hat linear function ?
thanks again .
i mean if there is ONLY ONE (OBLY ONE) LT that is bijective between two vector spaces V and W , then we say that those two are isomorphic , we only need ONE bijective linear transformation to say that two vector spaces are isomorphic ???
thank u!
Thanks for the video, I learned why V^{**} and V are isomorphic!
I think you can shorten the proof of $psi being 1-1:
We will show that Ker($psi) = 0. If $psi(x) =0, f(x) = 0 for all f in V*. In particular, f(x)=0 even if function f is the identity function.
Hence x=0. Thanks!
The identity function isn’t in V*, because we need the output of f to be a scalar!
@@drpeyam Whoops, my logic only exists when V=F. Thanks.
If no basis is needed, maybe these would be good for finding lengths of curves and areas of surfaces.
LarryRiedel can you please explain why?
@@waterfirecards5128 I wish I could. Maybe you can get something from the Wikipedia page on Metric Tensors en.wikipedia.org/wiki/Metric_tensor
god bless
Very nice D peyam
Ahah!