Tricks to determine whether a given subset of R^n is a subspace or not.

[@aidanlee8567]

best video on subspaces, you deserve a shoutout

[@DrMathaholic]

Thank you 😊

[@musicattheendofmovies3702]

you are the best person i have seen explain this concept so easy and precise i love it .

[@DrMathaholic]

Thank you.. glad to hear that🙏😊

[@_flexion]

Really well explained!! Thank you Sir

[@DrMathaholic]

Thank you and welcome 🙂

[@deborah380]

Thank you for doing examples with nonzero constants and powers other than 1!

[@DrMathaholic]

Welcome:)

[@jimmyhunnicutt1923]

you may have just saved my grade1 this is the first time I'm getting it, thank you!

[@DrMathaholic]

Welcome.. all the best 👍

[@georgehinneh6277]

I love ❤ the hints you gave , it will be very helpful in mcq's

[@DrMathaholic]

😊👍

[@lipansan]

Thank you very much! Best explaning ever!

[@DrMathaholic]

Thank you 😊

[@LOLjerel]

My professor is making us do both separately but this looks way easier lol Thank you!!

[@mitra.1]

Sir what about these 2. Will these form a subspace? 1. {(x, y, z) : x + y = z} 2. {(x, y, z) : x = y^2

[@DrMathaholic]

1 is a subspace. 2 is not as it contains degree 2 term. For ex: (1,1) and (1,-1) satisfies the equation but their addition (2,0) does not satisfy the given equation. So addition fails..

[@mrbleetoe]

You are the best, thank you!

[@DrMathaholic]

Thank you for your kind words 😊😊

[@whiteshadow5881]

W video, love it, saved it, i regret judging it because it used a chalkboard and came here last

[@DrMathaholic]

Glad to hear that.. All the best😊

[@gauravdateer9931]

Sir your explanation is tooo good 💯🔥

[@DrMathaholic]

Thank you 😊

[@BhagyashreeBhoomshetty]

Thank you sir❤ Well explained 🙏

[@DrMathaholic]

@@BhagyashreeBhoomshetty thank you ..welcome 😊

[@barakamtawa6571]

thanks dr mathaholic am understand you very very good thanks🙏🙏

[@DrMathaholic]

Very happy to hear that:)

[@saimamanzoor2781]

Very very helpful vedio...thnkuu soo much

[@DrMathaholic]

Welcome 😊

[@jayavaradhanp8405]

Sir make more videos on real analysis it will be helpful for net preparation.

[@DrMathaholic]

Yes sure.. I will definitely try..

[@Icywings-v1b]

Very much thankful

[@DrMathaholic]

Welcome 😊

[@mathematicsbya.g]

Thanks for the trick ❤

[@DrMathaholic]

Welcome :)

[@craftmedia3448]

Sir I am confused on this question Show that the following subspace is corresponding to vector space R³ The set of plane passing through the origin {(x,y,z) | ax+by+cz=0 where a,b,c is a scalar}

[@craftmedia3448]

Wether it is subspace or not Sir

[@DrMathaholic]

Try to show that 3 conditions are satisfied. 1. This set contains zero vector (0,0,0) bcoz when u replace x,y,z by 0,0,0 it satisfies the equation ax+by+cz=0. 2. Suppose (x1,y1,z1) and (x2,y2,z2) satisfies the equation ac+by+cz=0. Can you show that (x1+x2,y1+y2,z1+z2) also satisfies the equation? 3. Suppose (x1,y1,z1) satisfies the equation. Can you show that alpha(x1,y1,z1) also satisfies the equation? Where alpha is any real number.

[@DrMathaholic]

Let me know if u get stuck.

[@DrMathaholic]

@craftmedia3448 it's a subspace..

[@craftmedia3448]

@@DrMathaholic ok Sir I try

[@KWW9999]

You are a legend

[@DrMathaholic]

🙏😊

[@sturkyturky9792]

Hello, was wondering if (x,y,z) in R^3, (4x - 9y)^2 = z^2 is a subspace, thank you

[@DrMathaholic]

Nope. As degree is not 1 so not a subspace. Take (1,0,4) and (0,1,9), these 2 points satisfy given equation but their addition (1,1,13) does not satisfy the given equation.

[@sturkyturky9792]

@@DrMathaholic Shouldn't it be (0,-1,9) satisfies the equation? because if you use (0,1,9) you get -9^2 = 9^2, and then by using (0,-1,9) the addition is (1,-1,13) which satisfies: (4(1)-9(-1))^2 = 13^2 which becomes 4 + 9 = 13, 13 = 13?

[@theresahfosuah3869]

@@sturkyturky9792since such an equation doesn’t satisfy all real numbers regarding closure under addition, it’s not a subspace

[@mohitjain6460]

Very nice explained sir....🎉🎉🎉

[@DrMathaholic]

Thanks and welcome...keep watching 😊

[@LightYagami-pw4wu]

thanks sir

[@103anushkasingh3]

Very helpful video thanks sir!!😊

[@DrMathaholic]

Glad to hear that.. welcome 😊

[@top1malayalam145]

Thank you so much sir

[@DrMathaholic]

Welcome :)

[@sultanalajmi6089]

S={(x,y)| x=2y and 2x=y} is this subset in the subspace?

[@DrMathaholic]

@@sultanalajmi6089 yes, it's a subspace

[@priyankas91]

Sir please explain this example:S={(x,y,z)/x≥y≥z},it is subspace or not

[@DrMathaholic]

No, Take x= (3,2,1) and scalar c= -1. Then x vector is in the set S but c*x is not in S. So scalar multiplication fails.

[@mitra.1]

Sir what about this one (x, y, z) : x = y = z . Will this form a subspace?

[@DrMathaholic]

Yes, it forms a subspace.

[@ecmrn]

so helpful!

[@DrMathaholic]

Thank you 😊

[@isurudilshan7172]

x^2 + y^2 + z^2 = 0. Sir can this be a subspace ? Seems like the only vector in this set is (0,0,0). So isn't it enough to form a subspace. The problem is confusing me. Btw your video is help me a lot..

[@DrMathaholic]

This is a singleton set.. {0} is always a subspace. We call it a trivial subspace..

[@isurudilshan7172]

@@DrMathaholic thank u sir..♥️

[@fayadhpm]

Thank you sir

[@DrMathaholic]

Welcome:)

[@tusharpersai5790]

in that sin x example it is not a vector space since it does not satisfy the vector addition property, so there is no point of checking whether it is a subspace or not? am i right sir

[@DrMathaholic]

Very right 👍👏

[@mustakeemsheikh1854]

W={(x,y,z)}€R3/(y2=0)} so sir wheather this is sub space or not please tell And you're explanation is awesome ❤

[@DrMathaholic]

As y is a real number so y^2=0 implies that y=0. So ultimately the set is {(x,0,z) | x,z are real numbers}. This set is nothing but xz plane and hence its a subspace.

[@DrMathaholic]

Thank you 😊

[@khadkabaniya4545]

Sir what about x^2+y^2-z^2=0. Will it form a subspace??

[@DrMathaholic]

No, (1,0,1) and (0,1,1) satisfy this equation but their addition is (1,1,2) which does not satisfy the given equation.. So addition property fails..u, v satisfies but u+v do not

[@shwetashukla4402]

Thanku Sir🙏

[@DrMathaholic]

Welcome :)

[@rajilakshmi3566]

{(a,b,c)}/a=b+c} please explain how it is subspace of R3

[@DrMathaholic]

Let (x,y,z) nd (a,b,c) belongs to given set. We have a=b+c and x=y+z. For addition, (x+a,y+b,z+c) we gave, x+a=y+z+b+c=y+b+z+c so (x+a,y+b,z+c) belongs to the set. Can you try for scalar multiplication?

[@AbayR-yw4zz]

Can we do the same method when there are x y z and w

[@DrMathaholic]

@@AbayR-yw4zz yes...

[@gamerom9324]

Sir can you tell which playlist we have to follow for 1 st chapter and its basics for ode and MVC plz

[@DrMathaholic]

Hi Here is the Playlist for unit I. Enjoy:) kzbin.info/aero/PLwaXU7G6UrbefJGT7WXm-ONSzcNoR4foC

[@gamerom9324]

@@DrMathaholic tysm sir😁

[@DrMathaholic]

@@gamerom9324 :)

[@gameparty3704]

a savior

[@tanvikadahiya3751]

Please explain sir.... W = {( x, 2y, 3z) : x,y,z belongs to R} Will W be a subspace of R³

[@DrMathaholic]

Yes... because W is x(1,0,0)+y(0,2,0)+z(0,0,3)= span{(1,0,0), (0,2,0), (0,0,3) }...

[@DrMathaholic]

From above one can see that W=R^3. So yes, W is a space..

[@tanvikadahiya3751]

Thank you sir

[@DrMathaholic]

@@tanvikadahiya3751 welcome..

[@jayavaradhanp8405]

How to check solution set of differential equation form a vector space or not

[@DrMathaholic]

Check this video.. I have given the solution kzbin.info/www/bejne/g2jIf3qDeZKMe8U

[@fabnavc4691]

Thank u sir

[@DrMathaholic]

Welcome :)

[@malaykhare1006]

sir if there is a set C2(complex) over R2, then it will be a vector space right?

[@DrMathaholic]

R2 is not a field. So c2 over R2 do not have VS structure. C2 over R is VS

[@malaykhare1006]

@@DrMathaholic okay sir thank you

[@ArnavDeshmukh-f7e]

W ={ (x1,x2,x3)/x1=x2} is it a subspace of W?

[@DrMathaholic]

Yes, it's a subspace..

[@olivierdebruijn8704]

hank you for the very nice explanation. Will this form a subspace: W1={(x,y)∈R^2 |xy≥0}⊂R^2?

[@DrMathaholic]

Welcome.. No, it won't. Take (0,1) and (-1,0) both are in W1 but when u add them, its (-1,1) which is not in W1 as product of 1 and -1 is not >=0

[@olivierdebruijn8704]

@@DrMathaholic thank you for your very fast respone! I have another question, is it correct that W3 ={(x,y,z)∈R3 | ax+by+cz=d} is a subspace in R3 for all values of a,b,c and d=0?

[@DrMathaholic]

@olivierdebruijn8704 welcome .yes , when d is 0, its plane passing through origin. Hence it do forms a subspace..

[@KeketsiFrancisSebapo]

how about this ,,, V =  a 1 b c  : a, b ∈ R  under the usual matrix operations.

[@DrMathaholic]

No. Because this set does not contain the zero vector, which is zero matrix.

[@kukuyiehalem6391]

"Show that W = {(x, y, z)|x = y and 2y = z} is a subspace of IR³." Sir is this a subspace?

[@DrMathaholic]

Yes.. all terms of Linear and there is also no product of variables and there is no constant. So yes it's a subspace. Other way, W={ (x,x,2x) / x belongs to R}. Since y=x and z=2y=2x . So W is span of (1,1,2). So W is a subspace.

[@ektakesharwani8115]

Sir your explanation is very nice please provide some important questions of group theory related uppsc exam

[@DrMathaholic]

Thank you.. Well, I dont have any such collection but I am sure you will get it online... All the very best...

[@rishirajbehera7183]

x = 5y does it forms a subspace?

[@DrMathaholic]

Yes...

[@rishirajbehera7183]

Sir there's a question which I'm not able to understand. It says, "Explain why no list of six polynomials is linearly independent in P4?" @@DrMathaholic

[@DrMathaholic]

@@rishirajbehera7183 see if this video helps:: kzbin.info/www/bejne/p5y1m4Z6f8tgm5Y

[@DrMathaholic]

@rishirajbehera7183 If V is a space of dimension n then you take any subset with more than n elements, it will always be linearly dependent. Here P4 has dim 5, so any set with more than 5 elements will always be dependent. This is due to- A set B is a basis if and only if it is maximal linearly independent set.

[@rishirajbehera7183]

Thank you so much sir :) @@DrMathaholic

[@ankitdadarwal192]

If degree is same on both side then that is subspace

[@DrMathaholic]

I didn't get the question.. Set S={ p(x) | deg(p(x)) = what?? } What is the set S?

[@sruthisaravanan82sruthisar15]

{(x,y,z) belongs to V: x+y+z=0} sir this is Subspace or Not

[@sruthisaravanan82sruthisar15]

Sir plz reply

[@DrMathaholic]

Yes..it's a subspace..its a plane passing through origin

[@lordbryann]

the vector satisfies the equation, ggggg thanks

[@ujjwalmahajan7581]

Hello sir I am having my LA ESE on Monday, tell me some important concepts and questions so that I can pass the exam...please sir....I am kindaa scared :{ I am from COEP 1st sem

[@DrMathaholic]

Be relax.. see the videos n also read the same topic from text book. After that see solved examples from text book and then try to solve problems from tutorial sheet. If you get stuck then ask me here. After this, if u get time then go to exercise problems from textbook

[@ujjwalmahajan7581]

@@DrMathaholic ok sir thank you for the guidance

[@DrMathaholic]

@@ujjwalmahajan7581 all the best..

[@ujjwalmahajan7581]

@@DrMathaholic :)

[@BirhanuDejen-qy3xd]

You are uniqe go on and tank you

[@DrMathaholic]

Thank you and welcome 🙂

[@srinathshrestha3899]

s = {{x,y,z,} belongs to R^3 : x+y=0 or y-z=0}

[@DrMathaholic]

Yes, it forms a subspace. We have, x=-y and y=z. Thus, S={( x,-x,-x) } I.e. s is a 1 dimension subspace of R^3

[@srinathshrestha3899]

@@DrMathaholic sir ! This questions was in our mid sem exam , and the it's not a subspace because (1,-1,0) and (0,1,1) are counter example

[@DrMathaholic]

@Srinath Shrestha oh, it's "or" in between.. I thought its "and"

[@DrMathaholic]

@@srinathshrestha3899 Under the "or" condition, it's not a subspace..whereas under "and" it's a subspace..

[@DrMathaholic]

So whenever there is an or condition between more than 1 equation then it's never a subspace..

[@ArvindKumar-cq8km]

Noce

[@REKHASHARMA14]

Thank you so much sir

[@DrMathaholic]

@@REKHASHARMA14 welcome :)

[@Anjanbiswas003]

Thank you sir

[@DrMathaholic]

Welcome 😀

[@singh40506]

Thank you sir

[@DrMathaholic]

welcome :)