Examples of SubSpaces and Non SubSpaces of Polynomial Space

[@Amantheparadise]

Very interesting ❤

[@gigglemug247]

It cleared all my doubts ✌️

[@DrMathaholic]

Great.. Happy to hear that :)

[@Rahul.G.Paikaray27]

Clear sir 👍👍👍

[@DrMathaholic]

Great👍😊

[@hafizur12349]

Thank you sir

[@DrMathaholic]

Welcome :)

[@thenewdimension9832]

Thankyou so much sir u saved me 💐💐💐💐

[@DrMathaholic]

Welcome :)

[@Lokesh-vo9uz]

Sir what is the dimension of polynomial function with 3x^2 + 1, is it 3 or 2 ?

[@DrMathaholic]

@Lokesh-vo9uz dimension is defined for a subspace..not for a single polynomial.. Can you state the complete question?

[@Lokesh-vo9uz]

@@DrMathaholic Sir this is question { What is the dimension of the vector space V consists of all polynomials are in the form of p(x) = ax^2+c } this is my general doubt like do we need to consider the vector of 'x' term in dimension or not, bcz in polynomial functions dont have 'x' term. If we consider that zero vector of 'x' term in subspace then dimension is 2+1=3 {x^2, x, constant} If we dont consider it then dimension is 2 right bcz subspace contains vectors of {x^2, constant} So what is dimension for this subspace ?

[@DrMathaholic]

@Lokesh-vo9uz here you are taking all polynomials of the form ax^2+c that is you are taking spanning of the set {x^2,1}. So dimension is 2. No, the x will not come here...

[@Lokesh-vo9uz]

@@DrMathaholic Got it sir, thank u

[@Shiksha291]

Sir if degree of n greater than equal 1 h to tb ye subspace hoga ya nhi... please reply

[@DrMathaholic]

Nai Hoga.. F(X)= X AND G(X)= -X. Both are of degree 1 but addition is not of degree greater equal 1

[@Shiksha291]

Okay sir

[@DrMathaholic]

@@Shiksha291 👍

[@SiniDhurua]

Sir in polynomial of degree 3 there is missing of zero vector but sir we can express 0 as 3 degree polynomial so why there is missing plz help me sir

[@DrMathaholic]

@SiniDhurua 0 polynomial has no degree... Degree for a polynomial is always unique..

[@SiniDhurua]

​@@DrMathaholic thank you❤

[@SiniDhurua]

​@@DrMathaholicThank you❤ sir

[@mathematicalsociety9928]

sir agr hm x ki power n positive lyn to wo W ko belong krta hy phr wo subspace ho gi

[@DrMathaholic]

Sorry, I didn't get your question.

[@mathematicalsociety9928]

sir i am saying that if we let positive x power 4 instead of negitive it will become 2x power 4 and it belongs to W

[@DrMathaholic]

@@mathematicalsociety9928 yes correct... But then, I am giving counterexample. So I have one specific example.. That is , addition need not always belong to W.

[@thetechblogger5385]

Sir I didn't get one question. The question is ' Let V=P2(R). Then W={a0x^2+a1x+a2 belongs to P2 : a0+a1=0} is the subspace of V or not?

[@DrMathaholic]

It's a subspace.. Try to prove it..if you get stuck then write down that step, I will check..

[@thetechblogger5385]

@@DrMathaholic Let u,v belongs to P2 and alpha belongs to R. u= a0x^2+a1x+a2 v= b0x^2+b1x+b2 For this to be a vector space alpha.u +v =(alpha.a0+bo)x^2+(alpha.a1+b1)x + (alpha.a2+b2) What to do after this? Did I assume everything correct?

[@DrMathaholic]

@@thetechblogger5385 a0+a1=0 and b0+b1=0 so in u+v coefficient of x^2+ coefficient of x = a0+b0+a1+b1=0 . So the condition is satisfied. So u+v is in W.

[@thetechblogger5385]

@@DrMathaholic Yes Sir u+v is in W but how to prove then scalar multiplication? i.e. alpha.u belongs to W if aplha belongs to R

[@DrMathaholic]

@@thetechblogger5385 alpha u= alpha*a0+ alpha*a1= alpha*(a0+a1)=alpha*0=0. So alpha*u is in w

[@Learnwithme.07]

Is polynomial at most degree 2 a subspace if p’(x) = x?

[@DrMathaholic]

You mean S={ p(x) | deg p(x)

[@Learnwithme.07]

@@DrMathaholic and their addition also then gives 2x Thanks a lot for your reply!!!

[@DrMathaholic]

@@Learnwithme.07 yes correct.. Welcome..

[@navjotzsingh]

Sir why 3 degree polynomial is not a vector space .. If it is ..then how? Plz help me with this doubt

[@navjotzsingh]

Sir plz tell me

[@DrMathaholic]

Degree = 3 can't form a subspace.. Take p(x)= x^3 and q(x)= -x^3+ x Then p +q = x which is a polynomial of degree 1 and not of degree 3.

[@navjotzsingh]

@@DrMathaholic sir ..we can do it same for p4 . then why p4 is a vector space Plz tell..

[@navjotzsingh]

@@DrMathaholic plz reply sir

[@DrMathaholic]

@@navjotzsingh P4 is what? If you are taking polynomials of degree less equal 4 then yes, it's a subspace. But if you are taking equal to 4 then it's not a subspace

[@joicet5618]

sir can you prove a polynomial with degree >=n

[@DrMathaholic]

please see at 6:45 . There you will see that taking x^{n=1} and - x^{n+1} +1 and adding will give a poly of degree less than n. so its not a subspace

[@thetechblogger5385]

Let V=P2(R). Then which of the following subsets of V are subspace of V. i. S1={p belongs to V: p'(1)=0} ii. S2= {p belongs to V: p(-1)=1} Sir I am confused on how to start this ques.

[@DrMathaholic]

Have you seen 4 and 5th question in the video? Same idea.. (i) is the subspace (ii) is not as 0 polynomial does not satisfy the given condition..so (ii) is not a subspace..

[@thetechblogger5385]

@@DrMathaholic Yes Sir ! Is this solution correct as well: ii.Let p1,p2 belongs to W this implies p1(-1)=p2 (-1)=1 p=alpha.p1+p2 p(-1)=alpha.p1(-1)+p2(-1) p(-1)= alpha +1 p(-1) =1 for only alpha=0 and not for all values of alpha belongs to R. Therefore it is not a subspace. Is this correct as well?

[@DrMathaholic]

@@thetechblogger5385 yes...correct

[@thetechblogger5385]

@@DrMathaholic Thank you Sir You are helping me a lot!

[@DrMathaholic]

@@thetechblogger5385 welcome..

[@lightdevil7143]

What If U = { f € P/f has rational coefficient} is subspace or not???

[@DrMathaholic]

No.. Take f from U and pi a real number which is not rational.. Then pi*f does not belong to U as coefficients are no more rationals...

[@lightdevil7143]

@@DrMathaholic ok thank you so much sit it's really mean lots ❤️❤️

[@DrMathaholic]

@@lightdevil7143 👍😊

[@lightdevil7143]

@@DrMathaholic Is S={ (x,y,x)| |x| = |y| =|z| } space of a vector space ? this is not V.S right because When we do -a(x,y,x) is not in S . Is this right

[@DrMathaholic]

@@lightdevil7143 not because u=(1,-1,1) and v=(1,1,-1) are in S since modulus of each element is 1 but u+v=(2,0,0) and here |2| not equal to |0|

[@thetechblogger5385]

Sir please suggest a way to prepare theorems? Sir can you prepare a lec on Span of Subset?

[@DrMathaholic]

Hi, Yes, I want to make but occupied with college work..hopefully soon. Regarding theorems, just try to understand the meaning and make sure step by step proof is clear.. Any queries then you can ask here.

[@thetechblogger5385]

@@DrMathaholic Ok Sir Thank you so much Sir! :)

[@DrMathaholic]

@@thetechblogger5385 welcome

[@DrMathaholic]

Have you seen this lecture on span?? kzbin.info/www/bejne/mabMZqyud7yjobM

[@thetechblogger5385]

@@DrMathaholic Sir can you please explain the proof of the theorem "If S is a non-empty subset of a vector space V, then [S] is the smallest subspace of V containing S."

[@abhiclassesdu515]

❤❤

[@ojosworld2648]

Sir can you solve this set as a subspace of P i.e {p€P / degree of p=4 } This set is not subspace can you give us example for this plz.

[@DrMathaholic]

X^4 and -x^4+x. Addition is not a poly of deg 4

[@ojosworld2648]

Thankyou sir

[@DrMathaholic]

@@ojosworld2648 welcome

[@ojosworld2648]

Plz.give examples for these also 1.{p€P / degree of p≤3 } Subspace ,yes 2.{p€P / degree of p≥5} Subspace ,no 3.{p€P / degree of p≤4 and p'(0)=0} Don't know? 4.{p€P / p(1)=0} Subspace yes Help me with all the questions through examples.

[@ojosworld2648]

@@DrMathaholic ??

[@ankitchowdhury8575]

How to find dimension of a subspace such that p(x)=p(2x) is satisfied

[@DrMathaholic]

take p(x) = ax+b and use the condition, we get a=0. take p(x) = ax^2+x+c and use the given condition , we get a=0 and b=0. Similarly we always get all coefficients 0 except constant. So p(x) = c. so I think dimension is 1. Do you have the answer key?

[@ankitchowdhury8575]

@@DrMathaholic yes sir I got it. I do not have the answer key but your explanation is right. Dim will be 1. Thank you sir ❤️

[@DrMathaholic]

@@ankitchowdhury8575 great..