you are the best person i have seen explain this concept so easy and precise i love it .
@DrMathaholic
Жыл бұрын
Thank you.. glad to hear that🙏😊
@_flexion
2 жыл бұрын
Really well explained!! Thank you Sir
@DrMathaholic
2 жыл бұрын
Thank you and welcome 🙂
@jimmyhunnicutt1923
6 ай бұрын
you may have just saved my grade1 this is the first time I'm getting it, thank you!
@DrMathaholic
6 ай бұрын
Welcome.. all the best 👍
@deborah380
4 ай бұрын
Thank you for doing examples with nonzero constants and powers other than 1!
@DrMathaholic
4 ай бұрын
Welcome:)
@georgehinneh6277
8 ай бұрын
I love ❤ the hints you gave , it will be very helpful in mcq's
@DrMathaholic
8 ай бұрын
😊👍
@lipansan
8 ай бұрын
Thank you very much! Best explaning ever!
@DrMathaholic
8 ай бұрын
Thank you 😊
@LOLjerel
6 ай бұрын
My professor is making us do both separately but this looks way easier lol Thank you!!
@gauravdateer9931
Жыл бұрын
Sir your explanation is tooo good 💯🔥
@DrMathaholic
Жыл бұрын
Thank you 😊
@whiteshadow5881
Жыл бұрын
W video, love it, saved it, i regret judging it because it used a chalkboard and came here last
@DrMathaholic
Жыл бұрын
Glad to hear that.. All the best😊
@mrbleetoe
Жыл бұрын
You are the best, thank you!
@DrMathaholic
Жыл бұрын
Thank you for your kind words 😊😊
@mitra.1
Жыл бұрын
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
Жыл бұрын
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..
@BhagyashreeBhoomshetty
3 ай бұрын
Thank you sir❤ Well explained 🙏
@DrMathaholic
3 ай бұрын
@@BhagyashreeBhoomshetty thank you ..welcome 😊
@barakamtawa6571
9 ай бұрын
thanks dr mathaholic am understand you very very good thanks🙏🙏
@DrMathaholic
9 ай бұрын
Very happy to hear that:)
@sturkyturky9792
6 ай бұрын
Hello, was wondering if (x,y,z) in R^3, (4x - 9y)^2 = z^2 is a subspace, thank you
@DrMathaholic
6 ай бұрын
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
6 ай бұрын
@@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
3 ай бұрын
@@sturkyturky9792since such an equation doesn’t satisfy all real numbers regarding closure under addition, it’s not a subspace
@jayavaradhanp8405
2 жыл бұрын
Sir make more videos on real analysis it will be helpful for net preparation.
@DrMathaholic
2 жыл бұрын
Yes sure.. I will definitely try..
@saimamanzoor2781
Жыл бұрын
Very very helpful vedio...thnkuu soo much
@DrMathaholic
Жыл бұрын
Welcome 😊
@craftmedia3448
10 ай бұрын
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
10 ай бұрын
Wether it is subspace or not Sir
@DrMathaholic
10 ай бұрын
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
10 ай бұрын
Let me know if u get stuck.
@DrMathaholic
10 ай бұрын
@craftmedia3448 it's a subspace..
@craftmedia3448
10 ай бұрын
@@DrMathaholic ok Sir I try
@103anushkasingh3
Жыл бұрын
Very helpful video thanks sir!!😊
@DrMathaholic
Жыл бұрын
Glad to hear that.. welcome 😊
@isurudilshan7172
Жыл бұрын
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
Жыл бұрын
This is a singleton set.. {0} is always a subspace. We call it a trivial subspace..
@isurudilshan7172
Жыл бұрын
@@DrMathaholic thank u sir..♥️
@mathematicsbya.g
8 ай бұрын
Thanks for the trick ❤
@DrMathaholic
8 ай бұрын
Welcome :)
@REKHASHARMA14
2 ай бұрын
Thank you so much sir
@DrMathaholic
2 ай бұрын
@@REKHASHARMA14 welcome :)
@mitra.1
Жыл бұрын
Sir what about this one (x, y, z) : x = y = z . Will this form a subspace?
@DrMathaholic
Жыл бұрын
Yes, it forms a subspace.
@sultanalajmi6089
21 күн бұрын
S={(x,y)| x=2y and 2x=y} is this subset in the subspace?
@DrMathaholic
21 күн бұрын
@@sultanalajmi6089 yes, it's a subspace
@tusharpersai5790
Жыл бұрын
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
Жыл бұрын
Very right 👍👏
@priyankas91
3 ай бұрын
Sir please explain this example:S={(x,y,z)/x≥y≥z},it is subspace or not
@DrMathaholic
3 ай бұрын
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.
@AbayR-yw4zz
Күн бұрын
Can we do the same method when there are x y z and w
@DrMathaholic
20 сағат бұрын
@@AbayR-yw4zz yes...
@gamerom9324
2 жыл бұрын
Sir can you tell which playlist we have to follow for 1 st chapter and its basics for ode and MVC plz
@DrMathaholic
2 жыл бұрын
Hi Here is the Playlist for unit I. Enjoy:) kzitem.info/door/PLwaXU7G6UrbefJGT7WXm-ONSzcNoR4foC
@gamerom9324
2 жыл бұрын
@@DrMathaholic tysm sir😁
@DrMathaholic
2 жыл бұрын
@@gamerom9324 :)
@KWW9999
Жыл бұрын
You are a legend
@DrMathaholic
Жыл бұрын
🙏😊
@singh40506
Жыл бұрын
Thank you sir
@DrMathaholic
Жыл бұрын
welcome :)
@LightYagami-pw4wu
20 күн бұрын
thanks sir
@malaykhare1006
2 жыл бұрын
sir if there is a set C2(complex) over R2, then it will be a vector space right?
@DrMathaholic
2 жыл бұрын
R2 is not a field. So c2 over R2 do not have VS structure. C2 over R is VS
@malaykhare1006
2 жыл бұрын
@@DrMathaholic okay sir thank you
@khadkabaniya4545
Жыл бұрын
Sir what about x^2+y^2-z^2=0. Will it form a subspace??
@DrMathaholic
Жыл бұрын
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
@aishwariyajayan5100
Жыл бұрын
Very much thankful
@DrMathaholic
Жыл бұрын
Welcome 😊
@jayavaradhanp8405
2 жыл бұрын
How to check solution set of differential equation form a vector space or not
@DrMathaholic
2 жыл бұрын
Check this video.. I have given the solution kzitem.info/news/bejne/rm2btHuDfV-KeJg
@mustakeemsheikh1854
Жыл бұрын
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
Жыл бұрын
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
Жыл бұрын
Thank you 😊
@rajilakshmi3566
Жыл бұрын
{(a,b,c)}/a=b+c} please explain how it is subspace of R3
@DrMathaholic
Жыл бұрын
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?
@ecmrn
11 ай бұрын
so helpful!
@DrMathaholic
11 ай бұрын
Thank you 😊
@kukuyiehalem6391
4 ай бұрын
"Show that W = {(x, y, z)|x = y and 2y = z} is a subspace of IR³." Sir is this a subspace?
@DrMathaholic
4 ай бұрын
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.
@olivierdebruijn8704
Жыл бұрын
hank you for the very nice explanation. Will this form a subspace: W1={(x,y)∈R^2 |xy≥0}⊂R^2?
@DrMathaholic
Жыл бұрын
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
Жыл бұрын
@@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
Жыл бұрын
@olivierdebruijn8704 welcome .yes , when d is 0, its plane passing through origin. Hence it do forms a subspace..
@ArnavDeshmukh-f7e
Жыл бұрын
W ={ (x1,x2,x3)/x1=x2} is it a subspace of W?
@DrMathaholic
Жыл бұрын
Yes, it's a subspace..
@ektakesharwani8115
2 жыл бұрын
Sir your explanation is very nice please provide some important questions of group theory related uppsc exam
@DrMathaholic
2 жыл бұрын
Thank you.. Well, I dont have any such collection but I am sure you will get it online... All the very best...
@tanvikadahiya3751
2 жыл бұрын
Please explain sir.... W = {( x, 2y, 3z) : x,y,z belongs to R} Will W be a subspace of R³
@DrMathaholic
2 жыл бұрын
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
2 жыл бұрын
From above one can see that W=R^3. So yes, W is a space..
@tanvikadahiya3751
2 жыл бұрын
Thank you sir
@DrMathaholic
2 жыл бұрын
@@tanvikadahiya3751 welcome..
@gameparty3704
11 ай бұрын
a savior
@rishirajbehera7183
7 ай бұрын
x = 5y does it forms a subspace?
@DrMathaholic
7 ай бұрын
Yes...
@rishirajbehera7183
7 ай бұрын
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
7 ай бұрын
@@rishirajbehera7183 see if this video helps:: kzitem.info/news/bejne/0qGI0Id6g5hemGk
@DrMathaholic
7 ай бұрын
@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
7 ай бұрын
Thank you so much sir :) @@DrMathaholic
@KeketsiFrancisSebapo
4 ай бұрын
how about this ,,, V = a 1 b c : a, b ∈ R under the usual matrix operations.
@DrMathaholic
4 ай бұрын
No. Because this set does not contain the zero vector, which is zero matrix.
@ujjwalmahajan7581
2 жыл бұрын
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
2 жыл бұрын
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
2 жыл бұрын
@@DrMathaholic ok sir thank you for the guidance
@DrMathaholic
2 жыл бұрын
@@ujjwalmahajan7581 all the best..
@ujjwalmahajan7581
2 жыл бұрын
@@DrMathaholic :)
@shwetashukla4402
2 жыл бұрын
Thanku Sir🙏
@DrMathaholic
2 жыл бұрын
Welcome :)
@sruthisaravanan82sruthisar15
Жыл бұрын
{(x,y,z) belongs to V: x+y+z=0} sir this is Subspace or Not
@sruthisaravanan82sruthisar15
Жыл бұрын
Sir plz reply
@DrMathaholic
Жыл бұрын
Yes..it's a subspace..its a plane passing through origin
@fabnavc4691
2 жыл бұрын
Thank u sir
@DrMathaholic
2 жыл бұрын
Welcome :)
@ankitdadarwal192
Жыл бұрын
If degree is same on both side then that is subspace
@DrMathaholic
Жыл бұрын
I didn't get the question.. Set S={ p(x) | deg(p(x)) = what?? } What is the set S?
@lordbryann
Жыл бұрын
the vector satisfies the equation, ggggg thanks
@srinathshrestha3899
Жыл бұрын
s = {{x,y,z,} belongs to R^3 : x+y=0 or y-z=0}
@DrMathaholic
Жыл бұрын
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
Жыл бұрын
@@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
Жыл бұрын
@Srinath Shrestha oh, it's "or" in between.. I thought its "and"
@DrMathaholic
Жыл бұрын
@@srinathshrestha3899 Under the "or" condition, it's not a subspace..whereas under "and" it's a subspace..
@DrMathaholic
Жыл бұрын
So whenever there is an or condition between more than 1 equation then it's never a subspace..
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