Linear Algebra Example Problems - Subspace Example #1
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- เผยแพร่เมื่อ 14 เม.ย. 2015
- adampanagos.org
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We work with a subset of vectors from the vector space R3. We show that this subset of vectors is a subspace of the vector space via a useful theorem that says the following:
Given a vector space V, the span of any set of vectors from V is a subspace of V.
Since we're able to write the given subset of vectors as the span of vectors from R3, the set of vectors in this problem is indeed a subspace of R3.
Another way to work this problem is to show that the set of vectors satisfies the following 3 properties: 1) Contains the zero vector, 2) Is closed under addition, and 3) Is closed under scalar multiplication.
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almost 2022 and this video is still saving lives... thank you kind sir.
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Nice video and explained well. The speed of your explanation is perfect. Hope you'll upload some more videos about linear algebra.
+Joost Driebergen Make sure you check out the rest of the videos on the "Linear Algebra Example Problems" playlist, I have about 50 there now:
th-cam.com/play/PLdciPPorsHuk3Hp7QPPAtTkpW0o1UXQB6.html
very heplful video. The way of explanation is really nice sir.
I'm taking college trig right now and was curious to see an example of linear algebra since my major requires calculus 4/differential equations and I might take linear algebra.
that is so easy and quick oh god i love you
yes it really helps a lot....thanx sir
Wonderful explanation
With me majoring in civil engineering, I got curious to see what linear algebra looks like due me sooner or later having to work up to this point and boy oh boy i completely did not understand squat!! haha can't wait to actually comprehend these mystical theorems... xD It will be a long ass adventure due to me being in trig at the moment. Hopefully i don't back out and stick with civil engineering till the end, I know it's worth it... :)
Just stick with it; the topic in this video is definitely one of the more abstract and difficult topics covered in a "typical" class. Many of the beginning topics are much more computational and easier to understand. Good luck!
Shouldn't it be -1 0 1 instead of 1 0 -1?
+Zineage Yes! Great catch. Sorry for the sign error. I added an annotation to the video at the appropriate spot to note this. Thanks for watching.
Thank you man
What a delicious easy to understand, amazingly explained linear algebra subspace example problem, mmm! mmm! my brain ate this up real good :D
Thanks for the kind words! If you found the video useful make sure to check out my website (adampanagos.org) where I have a ton of other resources available and it's easier to watch my videos in a more organized fashion. Thanks, Adam.
Hey i signed up for a member request on your website and it says its waiting approval, my email is isaacenaupa@gmail.com please accept me
I just sent you an email, thanks!
I understand that U is spanned by the set of vectors we created, and that both vectors belong in R3, however, isn't it necessary to confirm that the vectors are linearly independent of each other in order to define them as a subspace of R3?
No, I don't think that's necessary. Per the theorem in the video, any time we have vectors v that are elements of V, then their span is a subspace of V. Now, if the v are not independent, then set they span will be "smaller" than we think it is (e.g. if we have v1 and v2 that are dependent, it might look like span(v1,v2) is 2 dimensional when it's only 1-D), but the span is still a subspace. Hope that helps,
Adam
Thank you, nice explanation!:)
Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam
Why are the vectors of r{1,0,-1} given the opposite way in the end?
My question too,I think it's just a mistake
Isn't span the all linear combinations in a vec space. I though what you refered to as U is set of basis vectors.
Thnk yu sir it helps a lot
Glad to hear that. Hope it helped,
Adam
So simply showing that the set of vectors in U spans some dimension that’s lower than three dimensions is enough to prove U is a subspace of R3? Is this equivalent to showing that U is closed under addition and scaler multiplication?
That's only true because we started with vector space R3, and then chose a subset of it. So, it inherits closure under addition and scaler multiplication from R3. All that's needed is to check what we did. Make sure to review the theorem at the end of the video that states all of this formally.
shoudnt there be at least 3 vectors to span 3 dimensional space?or am i mixing up something ?
In general, yes. But we're not spanning a 3-dimensional space here. The set U only has 2 dimensions. It's a subspace of R3. Every vector in U can be written as a linear combination of 2 vectors. So, by definition it's only 2-dimensional. Hope that helps,
Adam
thanks :)
Thanks a mil
You're welcome, thanks for watching.
Hey excellent video, just quick question, you say your two vectors [2,3,1] and [1,0,-1] are elements of R3, is this always the case?
Also if you had a third vectors say v3 which was [4,6,2] (a multiple of vector 1) would you include that in the span?
+MrICEDBlack Yes, [2, 3, 1] and [1, 0, 1] are always elements of R3 by definition (i.e. they are length 3 vectors and they contain real-valued numbers as their entries).
No, [4,6,2] wouldn't be included in the span notation above. By definition, the span is the collection of all linear combinations of a set of independent vectors, so we wouldn't include a linearly dependent vector in the set of vectors. Actually, the vector [4, 6,2] is already part of span{[2, 3, 1]; [1, 0, -1]} since this notation means all linear combinations of the two vectors. So, the linear combination 2*[2, 3,1] + 0*[1,0-1] = [4,6,2]+[0,0,0] = [4,6,2] is part of this set.
Thank you very much, cleared up a lot of things in my head! Subscribed and liked! Keep up the good work!
thanks
good stuff
Thanks!
This is a really fucking solid video lol
Nice
Wait those two spanning vectors aren't linearly independent because there's only two of them in R3? So what you're saying is impossible?
S = {v_1, v_2 | v in R3} cannot exist right?
burnsy96 No, these vectors are definitely linearly independent. The only way for two vectors (regardless of length) to be dependent is if one vector is a multiple of the other vector. Obviously, the vectors [2 3 1] is not some multiple of [1 0 -1] (or vice versa) so these are two linearly independent vectors.
I think you may thinking of the case where you have more vectors then dimensions? For example, if you had 5 vectors of length 4, then they would have to be linearly dependent as the number of linearly independent vectors of length 4 is at most 4.
Does that help?
Adam Panagos makes a little more sense I guess. So the subspace U spans the vector space R2 and is also a subspace of R3? Would it be also correct to say U is a valid subspace of R4, R5, R6 and so on?
burnsy96 No, we're working exclusively with R3 here and a subspace of it. The set of vectors U is a collection of all linear combinations of the vectors [2 3 1] and [1 0 -1]. Since each of these vectors has length 3, then U is a subspace of R3. Don't let the fact that we have 2 vectors mix you up. The important thing is that every element of U is a vector of length 3. For example, the vector 2*[2 3 1] + 4*[1 0 -1] = [4 6 2]+[4 0 -4] = [8 6 -2] is an element of U since it is a linear combination of the vectors [2 3 1] and [1 0 -1]. Also, clearly [8 6 -2] is an element of R3 since it has 3 coordinates. This is just one of an infinite number of vectors in U since I could have chosen any linear combination. This vector is NOT an element of R2 or R4 or R5 or any other space since all of these other spaces have different numbers of dimensions.
Maybe that helps?
Adam Panagos ah yes, I've also studied more on subspace since my original comment and that and your help have made me more confident in subspaces.
Excellent video and help good Sir, I'll subscribe to you for anymore updates.
Thanks again
Great, glad to hear that you've got it figured out.
KING!!!!!!!!!!
the add is annoying. Yes, you tube video is helping me a lot.
did you mix up the -1 and 1 in the second vector?
Yes, good catch. I used to have an annotation that popped up to note the sign error, but TH-cam removed annotations. Thanks for the heads-up.
here zero vector is not belonging from U.how can we say its a subspace?
shivani dixit The zero vector is an element of U. Since r and s can take on any real value, when r=0 and s=0 then we have the all zero vector. Hope that helps.
helpfull
+Zaid Almymoni Thanks for watching!
what I tried to learn in a week of lectures is broken down into 5 mins....
Glad I could help, thanks for watching!
is this 2 Dimension?
+Joe Liban The vectors we're working with have 3 coordinates so they are elements of R3. However, the subset of vectors we're working with is U = span([2; 3; 1], [1; 0; -1]). Since every element of U can be written as a linear combination of these two vectors the subspace U is indeed a 2-dimensional vector space. It's a two-dimensional plane in R3. Hope that helps.
Reading your answer to Joe Liban's question, can we say that U spans R2?
I kind of see what you're getting at, but we can't really say that.
When we say that some set of vectors SPANS another space, then we mean that every vector in the space can be written as some linear combination of the set of vectors.
Since the vectors in R2 are all 2-dimensional quantities, any set that would span R2 should only have 2 elements in them.
Since U is a collection of 3-D vectors it's doesn't technically span R2. Now, U is a 2-dimensional quantity, in that it consists of a plane of points. But, it's still a 2-D plane of points within a 3-D space.
Hope that helps,
Adam
Alright, I got it, thankyou
Your 2 look more like a z...
I know, I'm sorry. Just they way I write. When I write "z's" I "cross" them so my z's actually look quite a bit different than my 2's but I hear ya!
لو سمحت ممكن الترجمه بالعربي
If there's something I can do to allow a more useful translation just let me know. I think I have things setup correctly but maybe there's a different setting I can use? Let me know, thanks,
Adam
your 2 looks like a z
Yes, sorry. Just the way I write. I do "cross" my z's though so they do look a little different.