This makes sense out of Lay's statement, a few subsections later, that the imposition of a coordinate system makes a space of n vectors "behave like R^n". You must really think of vectors in a coordinate-free way to understand the difference between R^n and a space of n independent vectors
1)In that case the associated scalar Field's range will decide the range of vector space . 2) Please also explain how is Vector space is applicable to the relativistic Quantum Mechanics.
Why would you use that book? Take Hubbard's, or Susan Colley's or Marsden's Book. Unless you are studying Biology or something like that, then Stewart's Book is fine.
I have a question. I want to preface this with saying I think that this series of lectures is very helpful and I only am highlighting mistakes in order to help someone who might be confused by them. At aprox 15:30 in the video you say that p(x)=a sub 0 is called "the" zero polynomial. That makes it sound as if it should function as the identity element, but clearly it is not. So I am wondering what is the significance of the thing the you refer to as "the zero polynomial" ? Or did you mean to say that the "zero vector" is the"zero polynomial" where a sub 0 =0 ? Also at the 10:30 mark you state that matrix multiplication is not associative, which is not correct.
at 15:30 there are lots of mistakes. She never actually said what it means for polynomials to be a vector space, like what P_n is. Also the polynomials of degree 0 are the constant polynomials, not the zero polynomial.
Also, P_n denotes the set of polynomials of degree at most n, not the set of all polynomials. While I'm sure a lot of people find these videos helpful, there are moments where I don't think she actually knows what she's talking about.
I tried different values for the Practise exercise... and for u = [ - 2 and 7] and v = [3 and -1], u+v is a Vector space. Please elaborate. I'm confused
If you mean the practice example at 11:31 the thing is that even if for some u and v vectors they follow the rule and u + v still remains inside V with just one case of vectors u and v that when sumed dont belong to V then the axiom is broken. By just finding a case that doesnt follow the axiom(normaly called proof by counter example)you prove that the axiom isnt true, as for it to be true it has to be valid in ALL cases(no exceptions), else you cant generalize that property . Hope that helped.
Since vectors are essentially matrices, they can only be multiplied if the dimensions work out. If A is 2x3 and B is 3x4, then we can multiply AB but not BA.
Blown away by how much easier this material is to grasp from watching your videos as compared to my college. Thanks for the uploads
Ma'am, please never stop making educational videos. You have an gifted quality for teaching. Lots of Respect from Bangladesh.
This is the first actual sensible tutorial i've found on vector spaces. thank you so much for uploading these.
so sad I discovered this channel two days before my first midterm, definitely gonna use these videos later.
was on the verge of despair when you said I know you can do it
lots of respect... I was afraid of vector spaces, as my concepts were not clear, but now, I'm enjoying it. thank you very much ....
This is the only playlist on youtube about the topic. Thank you a lot!
*This channel is underrated. You should have more Subs.*
Thanks
You are highly appreciated Mrs/miss Kimberly Brehm
Thank You Soooooooo Much for the help of "Sky Wolves" in this regard ...❤
You taught me an entire semester worth of maths just by these short videos. Kudos.
I am watching your discrete mathematics course as well. Thanks for uploading this.
finally iam understanding it i dont know how i can thank u
THANK YOU SO MUCH
it is an universial thing, when teachers in colleges explain their courses worse than youtube tutorials. Thank you!
Well I’m a college professor 👩🏻🏫
Wow so much clarity,love you.
hats off mam! Thanks for explaining it in such a simple way :)
You are incredible thank you!!!!
Thank you so much ma’am I’ll check back after my midterm 🥰😭
YOU SAVED MY LIFE
Happy to help!
The subspaces of R^2 are the zero vector, lines through the origin, and the whole plane R^2 itself.
Best Linear Algebra on the net
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Thank you for helping me, God bless you
This is really helpful! Thanks a lot!!
you've saved my linear algebra mark so much
Happy to help!
Thank you very much.. It helped me a lot.. And saved me a lot of work and time.. ❤❤
I wish my professor can teach like you so i dont need spend hours of time searching a small concept to understand it :(
thank you so much !!!!!!!!!!!!!!!!!!!!!!!!!
I understand so much better
This makes sense out of Lay's statement, a few subsections later, that the imposition of a coordinate system makes a space of n vectors "behave like R^n". You must really think of vectors in a coordinate-free way to understand the difference between R^n and a space of n independent vectors
Thank you so much Ma'am.
helped a lot mam can't thank enough
6:54 you only have me😂❤️
1)In that case the associated scalar Field's range will decide the range of vector space . 2) Please also explain how is Vector space is applicable to the relativistic Quantum Mechanics.
Kimberly you are
Thank you so much!
V nice explanation 👍
thank you. it helped alot
This video is very helpful.
Thank you no, thank you very much
Mam how can i acess the video 4.4.1&4.4.2 those videos arent here ...its 4.3.2 and 4.5.1 can you please help me with it
THE SET OF ALL VECTORS IN A SPHERE OF FINITE RADIUS IS A REAL VECTOR SPACE TRUE OR FALSE?
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thanks
Can you please consider going over multivariable Calculus 1 and 2 (Stewarts Calculus chapters 11-17). It would be greatly appreciated.
Why would you use that book? Take Hubbard's, or Susan Colley's or Marsden's Book. Unless you are studying Biology or something like that, then Stewart's Book is fine.
@@el_witcher Gilbert stranger is better.
I have a question. I want to preface this with saying I think that this series of lectures is very helpful and I only am highlighting mistakes in order to help someone who might be confused by them.
At aprox 15:30 in the video you say that p(x)=a sub 0 is called "the" zero polynomial. That makes it sound as if it should function as the identity element, but clearly it is not. So I am wondering what is the significance of the thing the you refer to as "the zero polynomial" ? Or did you mean to say that the "zero vector" is the"zero polynomial" where a sub 0 =0 ?
Also at the 10:30 mark you state that matrix multiplication is not associative, which is not correct.
at 15:30 there are lots of mistakes. She never actually said what it means for polynomials to be a vector space, like what P_n is. Also the polynomials of degree 0 are the constant polynomials, not the zero polynomial.
Also, P_n denotes the set of polynomials of degree at most n, not the set of all polynomials. While I'm sure a lot of people find these videos helpful, there are moments where I don't think she actually knows what she's talking about.
I tried different values for the Practise exercise... and for u = [ - 2 and 7] and v = [3 and -1], u+v is a Vector space.
Please elaborate. I'm confused
If you mean the practice example at 11:31 the thing is that even if for some u and v vectors they follow the rule and u + v still remains inside V with just one case of vectors u and v that when sumed dont belong to V then the axiom is broken. By just finding a case that doesnt follow the axiom(normaly called proof by counter example)you prove that the axiom isnt true, as for it to be true it has to be valid in ALL cases(no exceptions), else you cant generalize that property . Hope that helped.
@@ronicave8522 well explained!
why is there a vector hat on V for axiom 6? where it states is c(u) is in V
RANK OF A MATRIX IS THE SAME AS THE DIMENSION OF ITS RANGE SPACE TRUE OR FALSE
can u pick any value for u in the last example?
for u+v should be in V is u also have to be in V or it can be any vector in all 4 spaces
which book are you following maam?
Why there is no multiplicative axiom on vector in vector space?
Since vectors are essentially matrices, they can only be multiplied if the dimensions work out. If A is 2x3 and B is 3x4, then we can multiply AB but not BA.
@@SawFinMath Thank you for sharing your knowledge and making our life easier
where can I get the book? or what is the title of the book?
Linear Algebra and Its Applications by David Lay 6E
@@SawFinMath thanks
P(t) + cutie 😂 16:18
isn't zero a real number in 10:57?
so wouldn't that mean that the axiom 6 holds?
Yes but x and y are both MORE than 0, not more than or equal too, so [0 0] is not in V since 0 is not more than 0,