Linear Algebra Example Problems - Subspace Example #5

[@lyndseyhubbard5973]

I wish I could like this 1000 times. I was sick and missed some classes and it has been tough doing this solo. I appreciate your help! This made things a lot better.

[@AdamPanagos]

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

[@lesliesantiago765]

I'm a student of mathematics and this teacher just saved my entire life! Thank you.

[@AdamPanagos]

Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (540+ videos) you might find helpful. Thanks much, Adam

[@lyndseyhubbard5973]

SAME

[@DemarcusRichardson]

You're really great at this! Thank you!

[@AdamPanagos]

+Demarcus Richardson (MarcDBeats) Glad you liked the video, thanks!

[@MegaAlindo]

Brilliantly explained, thanks a lot

[@AdamPanagos]

You're welcome, I'm glad you found the video useful. Make sure to checkout my website adampanagos.org for the rest of the linear algebra videos in this series. Thanks. Adam

[@steiner6085]

Great videos! Extraordinary

[@AdamPanagos]

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

[@dinithisahanika7823]

Love from sri 🇱🇰 lanka ❤️

[@AdamPanagos]

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, Adam

[@ChefFarisMom3]

Can you please explain how to prove even and odd polynomials to be the subspecies of P_n.

[@abdullahmajed7554]

Thanks a lot from saudi arabia ❤️

[@AdamPanagos]

You're welcome, thanks for watching! Always happy to see people from across the globe enjoying the material. Best, Adam

[@bossman4112]

what if you had the 2 x^2 terms cancel out by them being negative and positive it wouldn't be closed then right?

[@yashmishra12]

Constant a2 can be a real number and in your case, it would be equal to 0, which still stands :)

[@rajp5307]

Nice explanation..sir!😊😊😊

[@NothingMoreThanMyAss]

I have a question though, when you're finding an arbitrary constant to see whether the polynomial is closed under multiplication, what if you chose C to be x? Then that would make the 2nd degree polynomial a 3rd degree which is no longer the same as the original degree which means it's no longer closed under multiplication. What's up with that?

[@mansthulani7682]

You can't choose C to be x, C is a constant and x is a variable o_O

[@dennercassio]

God damn lol

[@eevibessite]

@@mansthulani7682 kzbin.info/www/bejne/ZnLLm2uJgaman8k

[@pandulathennakoon3826]

Thanks man, much appreciated!

[@AdamPanagos]

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

[@prashantranjan6487]

i would like to point out one mistake , when you put a2=0 in a 2 degree eq , the equation will no longer remain a 2 deg eq, regardless you can put a1=a0=0 and x=0 to obtain the 0 vector

[@AdamPanagos]

I think you've misunderstood the video. This is working with P2, the set of all polynomials of degree less than or equal to 2. So, P2 also includes constants and first order polynomials.

[@HCH-mo5ds]

Hi prof, but wasnt n defined as >=3? 2 does not satisfy, so I think P2 is not in P3 because for it to be defined as "of order 3 at most" as you mentioned Pn should be defined as n

[@bossman4112]

would this still hold for n>= 2 or no I am assuming yes but just want to confirm

[@hammadhassan9409]

I am a little bit confuse and need a help. What if i take u=a1+ a2(x)^3 and v=a1+(-a2(x)^2) ,both these polynomial have degree of 2 ..but their sum has degree of 1 which is not belong to polynomial of degree 2..which shows that it is not a subspace of P2.

[@AdamPanagos]

Play close attention to how Pn is defined. Pn consists of all polynomials of at MOST degree n. So, if we're talking about P2, this includes all constants, all first-order polynomials, and all second-order polynomials. So, the example you gave does indeed belong to P2 since the final sum has degree 1 which still satisfies the " at most degree 2" contraints. Does that help?

[@VikashKumar-bs4lq]

Did u not given then video of basis and dimension

[@nhlanhlamlanzi6950]

great,can you now help me how to show that a certain vector belongs to some set of vectors given.

[@AdamPanagos]

If you're given vectors x1, x2, ..., xn and want to know if the vector y can be written as a linear combination of them, then we just need to find scalar coefficients ai such y = a1*x1 + a2*x2 + ... + an*xn. This can be done seeing if there's a solution to the system of linear equations [x1 x2 ... xn][a1 a2...an]^T = y

[@nhlanhlamlanzi6950]

A big thanks to you,you are doing an excellent job,please we need more help like these

[@musratshaheen2946]

Thanks alot

[@AdamPanagos]

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.

[@neetusharda4952]

adding anything with 0 will obviously result in the number it was added to, so why do we even consider it, you did not show any graph too for this question, in the previous questions I was able to understand the 0 property just because there were graphs shown and we were able to interpret from the graphs that whether 0 is there or not, but I am not able to understand the 0 adding thing.

[@AdamPanagos]

What you said is not necessarily true. If there is a zero element in the set of objects we're considering, then adding zero to an object doesn't change anything. But there doesn't have to be a zero element. As a simple example, if I'm working with the set {1, 2, 3, 4, 5}, there is no zero element of this set. So, establishing some set of mathematical objects does or doesn't have a zero element is an important part of being a subspace. Hope that helps. Adam

[@parthgupta2742]

what is the constant in p(x) not equal to 0? because then it will not contain 0 vector

[@eduardoschiavon5652]

If it doesn't contain the zero vector, it's not a subspace

[@jashanpritgujjar0113]

@@eduardoschiavon5652 Thank you so much sir. I have the same doubt.

[@cameronsantiago3155]

Why is p2 a subset of p3?

[@AdamPanagos]

Because of all the reasons we checked in the video.

[@HCH-mo5ds]

@@AdamPanagos Hi prof, but wasnt n defined as >=3? 2 does not satisfy, so I think P2 is not in P3 because for it to be defined as "or order 3 at most" Pn should be defined as n

[@Sertsiyaset]

thank you so much

[@AdamPanagos]

+JeasWB You're welcome, thanks for watching!