What I don't get is the part "as you vary t, you get all the points along the straight line through v1 and v2". Well there will be a straight line, but obviously the vectors will be moving drawing a plane, not a line... Any help explaining this?
Yup! I believe this is caused by a misunderstanding of what a vector is. A vector, which is often drawn as an arrow with a beginning and an end, does not consist of all the points between the base of the arrow and the tip. The vector is just the tip of the arrow, and is therefore only a single point. So as t varies, we are only looking at how the tip of the arrowhead moves, and that does trace out a straight line. I agree this is a misleading practice, and you are certainly not the first to ask such a question, but we nevertheless do this for illustrative purposes.
tv1 + (1-t)v2 = tv1 + v2 - tv2 = t(v1-v2) + v2. v1 - v2 is the line vector connecting v1 and v2. so v2 plus any combination of that will give the whole line of vectors passing through v1 and v2.
many thanks. But as an engineer I would first ask myself: What are some real-world applications of such a space? Can anyone please provide me with a few inspiration and to help me gain a sense? Why should one be first interested in defining such a fancy constrained type of vector space?!
I hope it's alright with you, but I'll answer your question with another one. Suppose you allowed the definition to be some constant, such as 2. Take the vectors (1,0) and (0,1) in the plane (R^2). What do you get if you take all combinations a(1,0)+b(0,1), where a and b are any real numbers satisfying a+b=2? Now what if it was -1 instead of 2?
a good lecture but lacking some points that should be there because these are not obvious first for around 3:40 tv1 + (1-t)v2 = tv1 + v2 - tv2 = t(v1-v2) + v2. v1 - v2 is the line vector connecting v1 and v2. so v2 plus any combination of that will give the whole line of vectors passing through v1 and v2. and second for around 6:36 say we want affine span of those 3 vectors v1,v2,v3 then it will be,let's call lambda l for simplicity, l1v1 + l2v2 +l3v3 such that summation l =1. now to explain why the lines that connect v1,v2 or v2,v3 or v1,v3 are in the span,you can take for eg. l1=0 and l2 and l3,then sum of l2 and l3 is 1 for v2 and v3 and that will give the line which is in the affine span of our original v1,v2,v3. then also why the affine span of points on these lines,which will again be lines,is in our original affine span? because say you take the affine span of these points say x1,x2 and give them coefficients t and 1-t. then say x1= l1v1 + l2v2 +0v3 and x2 = 0v1 + l2v2 +l3v3 multiply x1 by t and x2 by 1-t and add them and you get (t l1 ) v1 + (t l2 + (1-t)l2) v2 + ((1-t) l3 ) v3 .now it can be verified that sum of these coefficients of v1 v2 and v3 is 1. just add them and use the fact that l1 +l2 =1 and l2+l3=1. this shows an affine combination of v1 v2 v3 exists that contains the affine combination of points on those lines.
Thanks for the comment. It's great to see your thinking process here. In fact, both perspectives are totally valid! It seems to me that your explanation seems more in line with your way of thinking about things. I think it's great to compare the two viewpoints to clarify in case other people have similar thoughts. As for the affine span of v1 and v2, the formula I wrote gives you the interpretation of "connecting the two points v1 and v2 by a straight line." In your version of the formula, which is t(v1-v2) + v2, the interpretation is "starting at v2, go forwards and backwards along the v1-v2 direction." I would not say that one is more obvious than the other, but would say that they are two ways of looking at the same thing. As for the comment about 3 vectors, my purpose here was to give an intuition, rather than a proof, so I am happy to see you thinking about this and supplying a proof. This is exactly what one should be doing when reading papers/books, watching lectures, or just learning!
In this video, I describe affine subspaces of R^n (or more generally vector spaces). An affine space can be defined more abstractly without viewing it as living inside of R^n (or a vector space). In this case, an affine space consists of points, rather than vectors. This is a subtle distinction! For example, imagine if you could draw an infinitely long straight line on an infinitely large paper. Then this line is an affine space. It consists only of points, not of vectors. But, if you draw a dot somewhere on the page, then your line is an affine subspace of R^2, where the dot is the zero vector. Now, every point on your line acquires a directionality with respect to the reference point you drew (the dot). Does this help clarify things for you? Or perhaps I misunderstood your question?
@@ArthurParzygnat Nice!! To recreate all R3 with affine combinarmos, do i need 2 planes like X on each other ? Otherwise can i say, i need 6 points not colinear ?
I realized that needs only four. Because the intersection of the 2 planes is a line and have 2 point in common to make this line. So need only 2 more not colinear to create 2 different planes
@@arthurlbn If you take affine combinations, then yes, you will get all of R^3. One way to see this is to pick a random point in the empty region away from these planes. Can you find a straight line through this point such that this line intersects two other points on these planes?
This is called earning a subscriber. Damn, you explain heck lot better than my own PDF-reading professor!
This is the kind of teaching we need, Thank You Arthur. Hoping to learn a lot from you
this floating head and arms knows what he's talking about
😂😂😂
This dude is writing from right to the left so well!!!
Video is inverted
As you can see he is writing with left hand in videos
@@shubhankarjaiswal2219 you're saying left-handed people can't write well? wow that's very prejudicial. tsk tsk tsk
Great contribution to TH-cam! Came here to get a better understanding of the geometrical implications for affine transformations in neural networks.
Incredibly clear explanation!
Thank you very much! Excelent explanation. I was stuck with this, and you cleared my mind. Hope you're doing well.
God bless you.
Fantastic lecture!
Thank you for the helpful and short lecture, hope to see more!
Neatly explained,
Thank you!
Beautiful explanation! Do make more videoes
Très utile. Very helpful, thanks!
The most amazing part of this video is how he's writing everything backwards without the slightest hesitation.
You know he’s using a mirror … right?
how can lambda still in the R space when it in the line or when it in the plane?
Good day! veli good content sir.
perfect alignment 6:09
What I don't get is the part "as you vary t, you get all the points along the straight line through v1 and v2". Well there will be a straight line, but obviously the vectors will be moving drawing a plane, not a line... Any help explaining this?
Yup! I believe this is caused by a misunderstanding of what a vector is. A vector, which is often drawn as an arrow with a beginning and an end, does not consist of all the points between the base of the arrow and the tip. The vector is just the tip of the arrow, and is therefore only a single point. So as t varies, we are only looking at how the tip of the arrowhead moves, and that does trace out a straight line. I agree this is a misleading practice, and you are certainly not the first to ask such a question, but we nevertheless do this for illustrative purposes.
tv1 + (1-t)v2 = tv1 + v2 - tv2 = t(v1-v2) + v2. v1 - v2 is the line vector connecting v1 and v2. so v2 plus any combination of that will give the whole line of vectors passing through v1 and v2.
Well done sir
Thank you so much!!!!
many thanks.
But as an engineer I would first ask myself:
What are some real-world applications of such a space?
Can anyone please provide me with a few inspiration and to help me gain a sense?
Why should one be first interested in defining such a fancy constrained type of vector space?!
Do they have to be 1 or the sum has to be a constant
I hope it's alright with you, but I'll answer your question with another one. Suppose you allowed the definition to be some constant, such as 2. Take the vectors (1,0) and (0,1) in the plane (R^2). What do you get if you take all combinations a(1,0)+b(0,1), where a and b are any real numbers satisfying a+b=2? Now what if it was -1 instead of 2?
Do you write everything backwards?
a good lecture but lacking some points that should be there because these are not obvious
first
for around 3:40 tv1 + (1-t)v2 = tv1 + v2 - tv2 = t(v1-v2) + v2. v1 - v2 is the line vector connecting v1 and v2. so v2 plus any combination of that will give the whole line of vectors passing through v1 and v2.
and second
for around 6:36 say we want affine span of those 3 vectors v1,v2,v3
then it will be,let's call lambda l for simplicity, l1v1 + l2v2 +l3v3 such that summation l =1.
now to explain why the lines that connect v1,v2 or v2,v3 or v1,v3 are in the span,you can take for eg. l1=0 and l2 and l3,then sum of l2 and l3 is 1 for v2 and v3 and that will give the line which is in the affine span of our original v1,v2,v3. then also why the affine span of points on these lines,which will again be lines,is in our original affine span? because say you take the affine span of these points say x1,x2 and give them coefficients t and 1-t.
then say x1= l1v1 + l2v2 +0v3 and x2 = 0v1 + l2v2 +l3v3 multiply x1 by t and x2 by 1-t and add them and you get (t l1 ) v1 + (t l2 + (1-t)l2) v2 + ((1-t) l3 ) v3 .now it can be verified that sum of these coefficients of v1 v2 and v3 is 1. just add them and use the fact that l1 +l2 =1 and l2+l3=1. this shows an affine combination of v1 v2 v3 exists that contains the affine combination of points on those lines.
Thanks for the comment. It's great to see your thinking process here. In fact, both perspectives are totally valid! It seems to me that your explanation seems more in line with your way of thinking about things. I think it's great to compare the two viewpoints to clarify in case other people have similar thoughts.
As for the affine span of v1 and v2, the formula I wrote gives you the interpretation of "connecting the two points v1 and v2 by a straight line." In your version of the formula, which is t(v1-v2) + v2, the interpretation is "starting at v2, go forwards and backwards along the v1-v2 direction." I would not say that one is more obvious than the other, but would say that they are two ways of looking at the same thing.
As for the comment about 3 vectors, my purpose here was to give an intuition, rather than a proof, so I am happy to see you thinking about this and supplying a proof. This is exactly what one should be doing when reading papers/books, watching lectures, or just learning!
@@ArthurParzygnat alright.
You say vectors but you use points in the space. The direction of the vectors makes any difference ? In the affine combination
In this video, I describe affine subspaces of R^n (or more generally vector spaces). An affine space can be defined more abstractly without viewing it as living inside of R^n (or a vector space). In this case, an affine space consists of points, rather than vectors. This is a subtle distinction! For example, imagine if you could draw an infinitely long straight line on an infinitely large paper. Then this line is an affine space. It consists only of points, not of vectors. But, if you draw a dot somewhere on the page, then your line is an affine subspace of R^2, where the dot is the zero vector. Now, every point on your line acquires a directionality with respect to the reference point you drew (the dot). Does this help clarify things for you? Or perhaps I misunderstood your question?
@@ArthurParzygnat Nice!! To recreate all R3 with affine combinarmos, do i need 2 planes like X on each other ? Otherwise can i say, i need 6 points not colinear ?
I realized that needs only four. Because the intersection of the 2 planes is a line and have 2 point in common to make this line. So need only 2 more not colinear to create 2 different planes
But, this 2 planes is enough to create all R3?
@@arthurlbn If you take affine combinations, then yes, you will get all of R^3. One way to see this is to pick a random point in the empty region away from these planes. Can you find a straight line through this point such that this line intersects two other points on these planes?
thanks a lot
what is the book name of this lecture
I based these off of Lay's Linear Algebra and its Applications, but I suspect there are some differences in the presentation.
Thank you!
You're welcome!
Thank You
I love you arthur
MVP
How tf does this work? Is he writing backwards ???
The video is simply mirrored!
you can do the same thing in pot player for example quite easy
The content is fine, but the writings is small, out of focus, and has almost the worst color because it is matching your flesh.
how the frickedy frick is he writing like that