Linear Algebra Example Problems - Subspace Example #2
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- เผยแพร่เมื่อ 29 ส.ค. 2024
- 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 NOT a subspace of the vector space.
In general, given a subset of a vector space, one must show that all of the following are true:
1) Contains the zero vector, 2) Is closed under addition, and 3) Is closed under scalar multiplication.
If any of these fail, the subset is not a subspace. It turns out that for the particular example worked in this problem none of these properties are true.
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I found this example extremely helpful. Seeing the failures really helped fill in some blanks for me. Thanks for the help!
+jrp95 Awesome, glad to hear this helped!
Thank you so much for being able to explain something so simple that my Ph.D. Professor couldn't convey in 2 hours!!!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (540+ videos) you might find helpful. Thanks, Adam
I like these simple exercises, they help me understand the concept of subspaces. thank you!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
just want to say thank you so much for all these helpful videos!
ive been watching your videos for my linear algebra class, and they have been very helpful
+Qi Wu Great, glad to hear that! I still have about 15 or more to make for this "Linear Algebra Example Problems" playlist. Hopefully I can get them done in the next few weeks. Thanks for watching!
Wow, this video cleared up so many misconceptions I had. My final is tomorrow and this is the first time understanding what "closed under addition means."
Awesome, glad I could help. I hope the final went well!
Amazing videos that I keep coming back to remember the basic ideas and the applications of all the theorems. Thank you, again! Hope you come to Portugal, you will love it (I´d recommend Lisbon or Porto!!)
This saved me thank you!
+spkoumarianos Glad to have helped, thanks for watching!
Your 2's are suspiciously similar to Z. I was confused for a moment with what you were saying and what was being presented.
agree
I know, it's a common complaint unfortunately! I actually "cross" my z's when I write them, so when side-by-side a 2 and a z look quite different. However, if you've never seen my z's then 2's do look a lot like a z. Wish I'd fixed this handwriting problem 40+ years ago. =(
@@AdamPanagos I'll keep that in mind going forward. Changing your handwriting isn't too difficult. By no means do you need to do so, but I actively work on my handwriting. I'm a pen enthusiast and everything in regards to my writing has to be aesthetically pleasing. I think it comes from being a fan of 90's graffiti and my brother (former graffiti artist and writes like an architect). I print out the cursive worksheets they give to elementary students and work on them lol Thank you for your work though! It was helpful and I'll be back for more
thanks, your tutorials came in clutchhhh
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
thank you a lot! it reaIIy helped me to understand the concept
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (525+ videos) you might find helpful. Thanks, Adam
Thank you ... adam it was very helpful video.. i did not understand anything my sir taught me....but u did... thanks a lot😊
Glad to help, thanks for watching!
Adam Panagos my pleasure adam..😊
thanks adam😊
This helped alot!! thanks for taking the time to explain the other axioms!
YOU ARE GODDAM LEGEND
Thanks you, hope I was able to help! Make sure to checkout my website adampanagos.org for more material you might find helpful. Best,
Adam
Thank you!! for the help I was finally able to understand vectors spaces and subspace
Ah, this makes sense now! Thanks!!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (540+ videos) you might find helpful. Thanks, Adam
What if you wanted to check if the set of vectors was a subspace of R2 or R6? Would you just do the exact same thing?
No, that's not really possible. If we're working with vectors from R3, then all of them have 3 dimensions. As such, they aren't even part of R2 since R2 is the collection of vectors that have 2 dimensions. It's a complete "mismatch" so asking the question doesn't really make sense. Hope that helps.
I think he means in terms of checking the properties. Not using vectors from R3 to prove R2 in which case yes.
Yes, in that case the steps would be very similar.
Thank you so much! the text book just says closed under addition and scalar multiplication but it doesn't really show how we checked these!
I could not thank you enough
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
Awesome Video, really helped me out understand this concept!!!
Great, glad to hear that. Thanks for watching!
Thank you so much
Glad to help
beautiful
YOU SIR ARE AWESOME!
how about if you get a form that looks like this : {(x, y) ∈ R
2
| x ≥ 2y}. How would you tackle a problem like this
Is it necessary to check all the conditions even if the first condition becomes wrong like here this vector does not contain zero vector?? Plz answer
To be a subspace, all the listed conditions must hold. So, if you check just the first one and it fails, you know it's not a subspace. No need to check the others if all you're trying to determine is if it's a subspace or not.
thanks for this video!! Just one question, for scalar multiplication, would 2rc-sc and rc+sc meet the requirements? Why or why not?
+Adil Ahmed They meet the requirements; r and s can be any real numbers and as they are multiplied with a real number c, they still remain real numbers. Hope that helps
Thanks... but I just finished my exam and I never have to learn about Lin Alg again lol
continue making linear algebra videos please!!!
Thank sir 😌
This video was really helpful, thank you!
Thanks!
thanks
I cannot understand it properly
Hey but wouldn't U = Span { (2,1,1) , (-1,01) } therefore it would be a subspace of R^3 ?
I was following your instructions from example 1
Oh no sorry I mixed up the 2 in the second entry for a r.
Very helpful video
Thanks
+MyThundermuffin Glad you got it figured out, thanks for the nice feedback.
The third question, how about if I put c=1 and put it into 2c? Won't it become 2.
Yes, but the form of the vector has to hold for ALL possible values of c. We need to be closed under scalar multiplication for all values of c. Holding true for just c = 1 isn't sufficient. Hope that helps,
Adam
ok thanks
Ooo that's a 2?? I thought it was a capital Z. Face palm on my part.
No worries, you can just blame it on my poor handwriting!
Lovely
The subspace does not have to have a zero vector instead of that condition, the subspace has to be nonempty which is clearly the case of [0 2 0] given in the example!
I think you may be a little confused about the definition of a subspace. Let V be a vector space and H be a subset of V. There are 3 properties that must be satisfied for H to be a subspace of V. The three properties are:
1) The zero vector of V is in H
2) H is closed under vector addition
3) H is closed under scalar multiplication
The above definition is the one I'm using here. See en.wikipedia.org/wiki/Linear_subspace#Properties_of_subspaces for more details.
Some authors use the following equivalent definition:
1) H is non-empty
2) H is closed under vector addition
3) H is closed under scalar multiplication
These are equivalent definitions of a subspace.
Note that while [0 2 0] is in U and thus it is not empty; it is still not closed under scalar multiplication. If c = 0, then 0*[0 2 0] = [0 0 0] and this is not an element of the U. Thus U is not closed under scalar multiplication.
Hope that helps.
Adam Panagos Really thanks for your quick clarification. I was confused because the book I have never mentioned the zero, but nonempty condition. Thank u!
@@AdamPanagos oooh now its clear thank you
Thank you sooooo much
+Tengisbold Khurelbaatar You're welcome, glad to have helped!
your channel is awesome
subscribed :D
+Brix Monton Thanks for subscribing, glad I could help.
Let G be the set of all vectors of the form [a+2b a+1 a] a,b is in R.Is G a subspace of R^3.
please help me Thank you
There are several properties that a subspace must satisfy. One is that it is closed under addition. It's easy to construct a simple example to show this isn't true for this case.
For example, if a = b = 0, then [0; 1; 0] is an element of G.
Or if, a = b = 1, then [3; 2; 1] is an element of G.
However, if we add these two vectors, we have [0; 1; 0] + [3; 2; 1] = [3; 3; 1]. If the last element of this vector is "a", then we must have a = 1. But the second element must be a+1, but 1 + 1 = 2, not 3. So, there isn't a value for a and b that we can find to yield [3; 3; 1].
So, having added two vectors in the G, we obtain a vector that is NOT in the form G. Since it's not closed under vector addition, then it cannot be a subspace of R^3.
Hope that helps!
Adam
+Adam Panagos thank you very much
how the heck are you writing so fast
I used Doceri, an app for my iPad that let's me pre-record all the writing. Then, I can "play back" the writing while I record the audio. Much easier to get all the writing done first!
I am from India