Hello comments readers! For clarity, it seems this series is already done, not a work in progress. There is a playlist link in the description to the other 14 (unlisted) videos. Hope you have a free afternoon.
When I was much younger it was difficult for me to understand that the evaluation of polynomials at a single place was a linear functional. But when you evaluate at a single place all the monoms become constant and only the coefficients are variable. You get a linear functional. That was a very important step for me. Later I learned that this was true also with other base functions, not only monoms. And it is also true with defined integrals and linear functions on matrices to the real numbers . Linear vector spaces are very powerful and the heard of mathematics.
Thank you so much for this comment. My mathematical training insofar has been in the context of physics, leaving some fairly large gaps in my mathematical foundations; such comments help subside a great deal of uneasiness
I came here because I am starting to get into Functional Analysis, and Operator Algebras. The video begins with an analogy about Legend of Zelda. Yes, I am in the correct place. I love your videos. They really bring the topics into perspective.
This answered my questions perfectly. I have been using O'Neill's "Elementary Differential Geometry" to teach myself, and he just referred to "dual space" in his definition of one-forms. After checking my linear algebra text (no dice) and prowling the internet's math forums for an hour of wasted time, I came here. Perfect explanation for what I needed without being overly complicated (unlike Stack Exchange). Thank you!
Thank you! I've been struggling to have an intuitive explanation of a dual space and you just gave it to me, with examples and everything. And the dimensionality and isomorphic explanation was great!
Regarding your last comment about the inequality of dimensions of an infinite dimensional vector space and its dual: The precise statement is that the dimension of the dual space of an infinite dimensional vector space is the cardinality of the field to the power of the dimension of V. This is called the Erdös-Kaplansky theorem. Since a field has at least two elements this is at least 2^(dim V), which is definitely bigger than dim V.
I'm currently learning Linear Algebra 1 in Germany and that's one topic which was explained to me very abstractly. You're intuitive explanations and good examples helped me a lot to get a deeper understanding of the topic, thank you very much! Subscribed :)
Is the 'scalar space' of a vector space reduced only to complex and real numbers? In 8:11 you said that a transformation wasn't a lineal fuctional because it didn't map to R nor C. But could it? Could and operation be defined so we could make elements from R2 scalars(I suppose it's possible since you could make them look like complex numbers)? But I'd like something more general. I've had this question in the back of my mind since I started to learn about vector spaces. In some exercises we were told to see if a transformation was linear or not, so we had to see if f(x+y) = f(x) + f(y), and if f(kx) = kf(x), for all scalars k and all vectors x y. But taking the first 'axiom'(?) of linearity, if we let x = y, then f(2x) = f(x+x) = f(x) + f(x) = 2f(x). This tells us that sometimes the second axiom doesn't have to be proven. Then I discovered it is valid also for negative and rational numbers, but it wasn't for irrational numbers. My guess is that it is valid only for non enumerable sets. Of course I'm only talking about numbers, but, as I said before, can we consider some scalar set that weren't numbers? I don't know if I worded it right, I'm just very curious! Thanks! By the way, I love your videos.
@Santiago Sanz I also was wondering about R2 acting as a field during the video, and I also thought about defining addition and multiplication as for the complex numbers. But you wrote that you wanted something more general. In what way would you like R2 as a field to be more general? I do not quite follow how the rest of your comment pertains to making elements of R2 act as scalars in a more general way. Thank you.
Haha, I liked how you were speaking in English about relationship between functional analysis and then you pronounce "Fourier" as french! Very good video, doc!
Thank you for the video, Professor! I found it highly informative and useful! :D Just for clarification - for the linear functional given at 2:55, what would f(0) look like and would it equal to zero?
@@drpeyam Yep, logically I see that, however I can't escape the thought that a degree-zero polynomial can just be any constant, and in this case, subbing 1 leaves the value unchanged because there is no x to speak of. What am I missing?
Very nice video dr. Peyam! I love you channel so much! In my functional analysis course we had some examples on infinite dimensional spaces like the dual space of L^p is L^q for 1
In your non-example 6, do you depend on the fact that R^2 is not a field? Since R^2 obviously is a vector space wouldn’t it be sufficient to define a multiplication and a division for R^2 being a field? Borrowing the multiplication and division operators from the complex numbers over the reals seem to produce a field for me. Is there anything else restricting R^2 from being the underlying field of the vector space V?
R^2 is not a field, unfortunately. Because in a field, you must have that if x and y are nonzero, then xy is nonzero, but with your multiplication, you get (1,0) times (0,1) is (0,0), even though both are nonzero
One of the tricky things about abstract algebra is that we often put structures together to form another structure. Often, the *way* we are putting structures together is the most important thing. As such, we often have a "natural" way to define the algebra within that new structure. For example, given any two rings R and S, we can form a new ring RxS by taking elements to be pairs (r,s) and defining addition and multiplication to both be done component wise. i.e., (r1,s1)+(r2,s2) = (r1+r2, s1+s2) and (r1,s1)*(r2,s2) = (r1*r2, s1*s2). We call this the "direct product" of R and S. A priori, you could be able to get different ring structures on the set of pairs (r,s). For example, if we look at the ring of real numbers, then on the set of pairs of real numbers, we could define addition component-wise and then do multiplication differently so it matches up with complex number multiplication. But this is not always guaranteed to be possible if you take two rings as above. As such, whenever we use the notation RxS for two rings R and S, there is a guaranteed ring structure on it (and it is natural in a meaningful sense), so we actually _mean_ to use that ring structure. R^2 is just shorthand for RxR, so when we think of R^2 as a ring, it has component-wise multiplication. So R^2 is not a field for the reason Dr. Peyam said. So, the thing to remember, is that when we're defining a structure by combining structures, there's usually a specific way in which we're doing it, and our notation is implied to mean that we are using *that* specific way.
Danke dir für deine Videos! Wir besprechen derzeit Dual Spaces und Basis in Linearer Algebra. Ich hasse und liebe das Fach irgendwie Ich find es immer wieder interessant, was für Verbindungen es in den Lehrtexten gegen Ende von Kapiteln gibt (bspw. Dualbasen mit Spalten und Zeilenvektoren von einer Matrix) aber irgendwie immer wieder nervig und oder anstrengend, einige "alte" Sachen, zum Beispiel bezüglich der Matrixdarstellung von Homomorphismen, hin zu nehmen weil man was vergessen hat, oder wieder in einem anderen Kontext anzuwenden... Ich freu mich einfach auf die angewandten Fächer die noch kommen im Studium :D
And that's the cool thing, there is no canonical isomorphism from V to V*. If you have a metric (a dot product), then there is one, but else you can pretty much map them in whatever way you want (with the restriction that the map is an isomorphism, of course!)
@@drpeyam if what you say is true, then what is the trace of this matrix [1,2]? Where is the diagonal? It is not 1 or 2 or the sum because they are the coordinates, and they surely change when one changes the basis; while the trace is an invariant. You should check your definitions....
corey Sure you can! The formula tr(A) = sum i = 1 to n of a_ii is still valid, except instead of n you replace it with the smaller of the number of rows and columns
@@drpeyam But that definition is only valid for a square matrix. Assuming that A is an nxn matrix. mathworld.wolfram.com/MatrixTrace.html I cannot find a definition for the trace of an nxm matrix
I am confused at ruclips.net/video/OGO3HGlOQO4/видео.html; the trace of a matrix is only defined when it is a squared matrix, while in the video you "computed" the trace of a 2x3 matrix. ¿How is that possible?
@@drpeyam the trace is an invariant of any linear operator and coincides with the sum of the eigenvalues. If your definition is correct, then suppose the matrix [1,2] (not diagonalizable), this matrix has two columns, therefore, your definition of trace doesn't make sense as there's no elements A21 or A22. The trace of a matrix is only defined for endomorphisms. In my years of academic and research, I have seen such a definition of trace. Admiting an error es more noble than creating confusion, specially in science ;-)
It’s definitely not an error. Your definition is valid for square matrices, but not for non square ones. My definition is an extension of the definition for non square matrices
Hello comments readers! For clarity, it seems this series is already done, not a work in progress. There is a playlist link in the description to the other 14 (unlisted) videos. Hope you have a free afternoon.
I want to thank you and the entire mathematics community. We are so lucky we have people like you teaching this for free
Thank you!!!
Much cooler than MySpace
We have double dual spaces V**.
But do we have n-dual spaces?
@@YanickSP I think he made a video on that already?
LMAO, MySpace 😂😂😂
Finally, someone who can explain linear functionals. Thank you for providing so many examples to make it all very clear. I truly appreciate you!
When I was much younger it was difficult for me to understand that the evaluation of polynomials at a single place was a linear functional. But when you evaluate at a single place all the monoms become constant and only the coefficients are variable. You get a linear functional. That was a very important step for me. Later I learned that this was true also with other base functions, not only monoms. And it is also true with defined integrals and linear functions on matrices to the real numbers . Linear vector spaces are very powerful and the heard of mathematics.
Thank you so much for this comment. My mathematical training insofar has been in the context of physics, leaving some fairly large gaps in my mathematical foundations; such comments help subside a great deal of uneasiness
It is very pleasant to learn from someone who seems as excited to be teaching as I am to be listening.
I came here because I am starting to get into Functional Analysis, and Operator Algebras. The video begins with an analogy about Legend of Zelda. Yes, I am in the correct place. I love your videos. They really bring the topics into perspective.
This answered my questions perfectly. I have been using O'Neill's "Elementary Differential Geometry" to teach myself, and he just referred to "dual space" in his definition of one-forms. After checking my linear algebra text (no dice) and prowling the internet's math forums for an hour of wasted time, I came here. Perfect explanation for what I needed without being overly complicated (unlike Stack Exchange). Thank you!
Thank you! I've been struggling to have an intuitive explanation of a dual space and you just gave it to me, with examples and everything. And the dimensionality and isomorphic explanation was great!
Immediately won me over with the Zelda reference. Good stuff professor. Good stuff.
:)
love the level of enthusiasm, and the content is informative too.
Today, I found this video and WOW, i finally getting understanding of duality after 30 years!!!!
That is amazing, I’m so happy to hear that! 😊
Thank you very much for this series. I'm always glad when I find out you have videos on the topic I am trying to learn. Keep it up.
Thank you VERY much! You have opened the door to differential forms and Exterior Calculus for me.
Those examples assist me a lot in understanding what a functional means. Thanks a lot!!
Regarding your last comment about the inequality of dimensions of an infinite dimensional vector space and its dual: The precise statement is that the dimension of the dual space of an infinite dimensional vector space is the cardinality of the field to the power of the dimension of V. This is called the Erdös-Kaplansky theorem. Since a field has at least two elements this is at least 2^(dim V), which is definitely bigger than dim V.
Whoa!!!
This was published on the 19th of June? By gawd, that's a week away from my exam, and you might've just saved my grade!
Thank you!
Yay!!! There are more videos on my playlist if you’re interested!
@@drpeyam Yes, thank you! I'll study them all
Your videos are a lifesaver.
thanks very much. Best course for dual space ever!
You are a great man! And I thank you so much for your Videos, cause otherwise I wouldn't have a chance in my exams!
Great Job, keep up the work.
The cake is the awesome teaching quality. The ace of the cake is the awesome sense of humor of Dr πm
Awwwwww ❤️
I'm looking forward to getting an intuition about transposes, and maybe a little about tensors and contravariant components.
yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeees. 2 days until the exam LA1 and I finaly found a good series on in! :):):) TYSM
Dual space is really interesting when it comes down to optimization problems (looking for min/max...)
I'm currently learning Linear Algebra 1 in Germany and that's one topic which was explained to me very abstractly. You're intuitive explanations and good examples helped me a lot to get a deeper understanding of the topic, thank you very much! Subscribed :)
Beautiful. Simply Beautiful
Sorry if this is a stupid question, but at 8:03 you say it's a non-example because the 'output' is in R^2, but we're mapping to R^2?
That is why f is not inside the dual space I think. Inside the dual space are functions and they have to have the mentioned properties.
so i can add Dr. Peyam after 3blue 1brown in the list of best mathematics related youtubers.
Thats what I was going to comment as well
at 4:30 how can you define the trace of a non square matrix?
Sum a_ii
I clicked on this this video when it had 314 views, must be a video by Dr πm.
Very helpful explanation Thank you 🙏🏻
We got Kaka explaining dual spaces b4 GTA6 😀🙂👌
GOOD EXPLAIN, THANK YOU. and it's interesting to watch someone writen on the left hand :)
Is the 'scalar space' of a vector space reduced only to complex and real numbers? In 8:11 you said that a transformation wasn't a lineal fuctional because it didn't map to R nor C. But could it? Could and operation be defined so we could make elements from R2 scalars(I suppose it's possible since you could make them look like complex numbers)? But I'd like something more general.
I've had this question in the back of my mind since I started to learn about vector spaces. In some exercises we were told to see if a transformation was linear or not, so we had to see if f(x+y) = f(x) + f(y), and if f(kx) = kf(x), for all scalars k and all vectors x y.
But taking the first 'axiom'(?) of linearity, if we let x = y, then f(2x) = f(x+x) = f(x) + f(x) = 2f(x). This tells us that sometimes the second axiom doesn't have to be proven. Then I discovered it is valid also for negative and rational numbers, but it wasn't for irrational numbers. My guess is that it is valid only for non enumerable sets. Of course I'm only talking about numbers, but, as I said before, can we consider some scalar set that weren't numbers?
I don't know if I worded it right, I'm just very curious! Thanks!
By the way, I love your videos.
It works for any field! And your argument is correct, it’s valid up to rational numbers, but for real numbers it’s a problem
@@drpeyam Thanks for answering! Could you give me an example of a "weird" scalar set?
Ratios of polynomials, like (t^2 - 1)/(t^2 + 1)
@Santiago Sanz I also was wondering about R2 acting as a field during the video, and I also thought about defining addition and multiplication as for the complex numbers. But you wrote that you wanted something more general. In what way would you like R2 as a field to be more general? I do not quite follow how the rest of your comment pertains to making elements of R2 act as scalars in a more general way. Thank you.
@@RalphDratman More general as in, if I have a vector space (V,k), what characteristics do those sets have for my ideas to verify
Very useful this examples. thx a lot for this content!
Looking forward to this series, really interesting!
I like the energy!
Awesome content👍👍👍👍
Haha, I liked how you were speaking in English about relationship between functional analysis and then you pronounce "Fourier" as french!
Very good video, doc!
I love Dr peyam 😍
Linear algebra wohooo 😘
Thank you for the video, Professor! I found it highly informative and useful! :D
Just for clarification - for the linear functional given at 2:55, what would f(0) look like and would it equal to zero?
For any linear transformation we have f(0) = 0
@@drpeyam Yep, logically I see that, however I can't escape the thought that a degree-zero polynomial can just be any constant, and in this case, subbing 1 leaves the value unchanged because there is no x to speak of.
What am I missing?
@@ivanskopin7723Oh but here the zero vector is the 0 polynomial! Constant polynomials would be p = c
@@drpeyam Aaaaah, I see! Thank you very much! :D
at 4:27 how did you take diagonal and its sum. Because it is non- square matrix
Sum of a_ii
Very helpful !
And then a duel-space is where you challenge your opponent to a math duel.
Cries in Galois
For me dual spaces are somewhat the equivalent to Transposition of Matricies. Like „transposing a Vectorspace“
Why dim of linear transformation is product of dimensions at 10:01 ?
There’s a video on that in my linear transformations playlist!
@@drpeyam thank you for Fast answer, will také a look
He got me with dual spaces and the like by explaining it with Legend of Zelda!
Very nice video dr. Peyam! I love you channel so much!
In my functional analysis course we had some examples on infinite dimensional spaces like the dual space of L^p is L^q for 1
I wanted to do that actually, but the proof is too long 😣
Yes, that's correct! It is a long and technical proof.
In your non-example 6, do you depend on the fact that R^2 is not a field?
Since R^2 obviously is a vector space wouldn’t it be sufficient to define a multiplication and a division for R^2 being a field? Borrowing the multiplication and division operators from the complex numbers over the reals seem to produce a field for me. Is there anything else restricting R^2 from being the underlying field of the vector space V?
R^2 is not a field, unfortunately. Because in a field, you must have that if x and y are nonzero, then xy is nonzero, but with your multiplication, you get (1,0) times (0,1) is (0,0), even though both are nonzero
Dr Peyam Thank you!
Dr Peyam Giving this a bit more thought, wouldn’t the multiplication give (0, 1) since 1*i = i?
But then you’re just replacing R^2 with C, which is a field.
One of the tricky things about abstract algebra is that we often put structures together to form another structure. Often, the *way* we are putting structures together is the most important thing. As such, we often have a "natural" way to define the algebra within that new structure.
For example, given any two rings R and S, we can form a new ring RxS by taking elements to be pairs (r,s) and defining addition and multiplication to both be done component wise. i.e., (r1,s1)+(r2,s2) = (r1+r2, s1+s2) and (r1,s1)*(r2,s2) = (r1*r2, s1*s2). We call this the "direct product" of R and S.
A priori, you could be able to get different ring structures on the set of pairs (r,s). For example, if we look at the ring of real numbers, then on the set of pairs of real numbers, we could define addition component-wise and then do multiplication differently so it matches up with complex number multiplication. But this is not always guaranteed to be possible if you take two rings as above.
As such, whenever we use the notation RxS for two rings R and S, there is a guaranteed ring structure on it (and it is natural in a meaningful sense), so we actually _mean_ to use that ring structure. R^2 is just shorthand for RxR, so when we think of R^2 as a ring, it has component-wise multiplication. So R^2 is not a field for the reason Dr. Peyam said.
So, the thing to remember, is that when we're defining a structure by combining structures, there's usually a specific way in which we're doing it, and our notation is implied to mean that we are using *that* specific way.
Danke dir für deine Videos! Wir besprechen derzeit Dual Spaces und Basis in Linearer Algebra. Ich hasse und liebe das Fach irgendwie
Ich find es immer wieder interessant, was für Verbindungen es in den Lehrtexten gegen Ende von Kapiteln gibt (bspw. Dualbasen mit Spalten und Zeilenvektoren von einer Matrix) aber irgendwie immer wieder nervig und oder anstrengend, einige "alte" Sachen, zum Beispiel bezüglich der Matrixdarstellung von Homomorphismen, hin zu nehmen weil man was vergessen hat, oder wieder in einem anderen Kontext anzuwenden...
Ich freu mich einfach auf die angewandten Fächer die noch kommen im Studium :D
Dankeschön :)
If i have a V=R above Q, then if I create a linear funcional, the range of this transformation need to be on Q ?
Yes it’s whatever field you’re studying
In Dr. Peyams math book: "...by Miracle 10.2.1 we see that..." :D
Thank you sir
Helped a ton!
Cette fois ci Dr Peyam m`a appris qqchose ! Bravo
Toujours, toujours :)
That's a Miracle! If V are finite dimensional, V and V* are isomorphic. We
And that's the cool thing, there is no canonical isomorphism from V to V*. If you have a metric (a dot product), then there is one, but else you can pretty much map them in whatever way you want (with the restriction that the map is an isomorphism, of course!)
Examples why V* is useful please
Check out the playlist
good explanation. thank you :)
Thank you very much
But what is the principal difference beetween vector space and dual space?
At least I understand the Zelda part :D
Thankyou sir
No, you are number one
Awwwww
Sir what is the difference between function and functional
Góod question
Functions send stuff into some set, functionals send stuff into a field.
@@_ilsegugio_ Thanks
Soooo exciting!!
I love the Zelda reference 🥰
Me tooo ☺️
I'm saving your videos for my linear algebra 2 final exam after two weeks
It will do a lot of help thanks doc.piAm
Dual space!!! That's great !!!
Dr. Peyam, that isomorphism is cool, but imho ur the true miracle.
Awwwwwwwwww ❤️❤️❤️
Frate tu sei un grande sei er numero uno ao
You can take the trace of a non square matrix?
Of course
@@drpeyam if what you say is true, then what is the trace of this matrix [1,2]? Where is the diagonal? It is not 1 or 2 or the sum because they are the coordinates, and they surely change when one changes the basis; while the trace is an invariant. You should check your definitions....
It’s 1, since it’s defined to be the sum of a_ii
We have double dual spaces V**.
But do we have n-dual spaces?
Check out the video on V***** on my playlist
Hey! Can you solve the DE f'(x)=f(x+a)?
Are you high when recording?
No, why?
You can't take the trace of a non-square matrix! so at 4:30 , the transformation is not an element of a dual space. It doesn't make sense.
It still is!
But I thought you can't take the trace of non-square matrices?
Please give me the definition for the trace of a non-square matrix. I'm extremely interested in this. Please, cite your source as well.
corey Sure you can! The formula tr(A) = sum i = 1 to n of a_ii is still valid, except instead of n you replace it with the smaller of the number of rows and columns
@@drpeyam But that definition is only valid for a square matrix. Assuming that A is an nxn matrix.
mathworld.wolfram.com/MatrixTrace.html
I cannot find a definition for the trace of an nxm matrix
Thank you!!!!!!!
I am confused at ruclips.net/video/OGO3HGlOQO4/видео.html; the trace of a matrix is only defined when it is a squared matrix, while in the video you "computed" the trace of a 2x3 matrix. ¿How is that possible?
The trace is defined for any matrix actually, it’s the sum of aii where i goes from 1 to the number of columns
@@drpeyam the trace is an invariant of any linear operator and coincides with the sum of the eigenvalues. If your definition is correct, then suppose the matrix [1,2] (not diagonalizable), this matrix has two columns, therefore, your definition of trace doesn't make sense as there's no elements A21 or A22. The trace of a matrix is only defined for endomorphisms. In my years of academic and research, I have seen such a definition of trace. Admiting an error es more noble than creating confusion, specially in science ;-)
It’s definitely not an error. Your definition is valid for square matrices, but not for non square ones. My definition is an extension of the definition for non square matrices
But then what’s the use? What could you prove with such an “extension”.
From a general relativity standpoint, this is the starting of adventures in Tensor land :P
thanks
Cool. Thanks.
"press f to press t"
did you play with your keyboard again?
AndDiracisHisProphet
He did. : )
Heh, I somehow always had this association of a vector space and dual spaces being like the overworld and the subspace in super mario bros. 2
How familiar are you with tensor algebra?
Not at all 😣
@@drpeyam There's a really instructive series from eigenchris about tensor algebra and then tensor calculus. It was a life-changing event to me.
Cool! Keep on going! Love your channel!
Unlisted but amazing!
This guy is awesome
What about the dual space of a dual space and, just for fun, the dual space of a dual space of a dual space . . . of a dual space?
There’s a video on that! Check out the playlist
In French we call the linear functions with R as the arrival space "forme" ( form if you translate litteraly )
Yo doc teach 121B pls
Haha, thank you!!! I’m teaching 112A in the fall, not sure about the other quarters. Youssefpour is supposed to teach 121B in the fall, he’s excellent
Is he really?!? Aw man I was hoping to have you for at least one class before I finish my undergrad at UCI :(
This is what my dissertation was based on XD
Vaahh !!
F
How it is comfortable to have two right hands!...
red shirt. ARGH! i want to see you in a crimson sweater
Haha, argh as in you like it?
@@drpeyam how could i not ?!
Red is the color of emperor Caesar. It’s a VERY good sign for his students :))