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.
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..
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.
@@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?
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}
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.
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..
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?
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
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.
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
@@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?
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
@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.
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
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
![INSANE Hack to Find Span of Any Vectors [Passing Linear Algebra]](http://i.ytimg.com/vi/9R6zQfd6vlQ/mqdefault.jpg)
best video on subspaces, you deserve a shoutout
Thank you 😊
you are the best person i have seen explain this concept so easy and precise i love it .
Thank you.. glad to hear that🙏😊
Really well explained!! Thank you Sir
Thank you and welcome 🙂
Thank you for doing examples with nonzero constants and powers other than 1!
Welcome:)
you may have just saved my grade1 this is the first time I'm getting it, thank you!
Welcome.. all the best 👍
Thank you sir❤
Well explained 🙏
@@BhagyashreeBhoomshetty thank you ..welcome 😊
My professor is making us do both separately but this looks way easier lol Thank you!!
I love ❤ the hints you gave , it will be very helpful in mcq's
😊👍
Thank you very much! Best explaning ever!
Thank you 😊
W video, love it, saved it, i regret judging it because it used a chalkboard and came here last
Glad to hear that..
All the best😊
Sir your explanation is tooo good 💯🔥
Thank you 😊
Thank you so much sir
@@REKHASHARMA14 welcome :)
tomorrow is my exam and I'm watching this now, nice explanation sir👍👍
All the best.. :)
You are the best, thank you!
Thank you for your kind words 😊😊
thanks dr mathaholic am understand you very very good thanks🙏🙏
Very happy to hear that:)
Thank you sir
welcome :)
Sir make more videos on real analysis it will be helpful for net preparation.
Yes sure..
I will definitely try..
Very very helpful vedio...thnkuu soo much
Welcome 😊
Thanks for the trick ❤
Welcome :)
Very helpful video thanks sir!!😊
Glad to hear that.. welcome 😊
"Show that W = {(x, y, z)|x = y and 2y = z} is a subspace of IR³." Sir is this a subspace?
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.
Sir what about these 2. Will these form a subspace?
1. {(x, y, z) : x + y = z}
2. {(x, y, z) : x = y^2
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..
Very much thankful
Welcome 😊
You are a legend
🙏😊
Sir please explain this example:S={(x,y,z)/x≥y≥z},it is subspace or not
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.
so helpful!
Thank you 😊
Hello, was wondering if (x,y,z) in R^3, (4x - 9y)^2 = z^2 is a subspace, thank you
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.
@@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?
@@sturkyturky9792since such an equation doesn’t satisfy all real numbers regarding closure under addition, it’s not a subspace
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}
Wether it is subspace or not Sir
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.
Let me know if u get stuck.
@craftmedia3448 it's a subspace..
@@DrMathaholic ok Sir
I try
the vector satisfies the equation, ggggg thanks
Thanku Sir🙏
Welcome :)
a savior
how about this ,,,
V =
a 1
b c
: a, b ∈ R
under the usual matrix operations.
No.
Because this set does not contain the zero vector, which is zero matrix.
Thank u sir
Welcome :)
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..
This is a singleton set.. {0} is always a subspace. We call it a trivial subspace..
@@DrMathaholic thank u sir..♥️
{(a,b,c)}/a=b+c} please explain how it is subspace of R3
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?
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
Very right 👍👏
W={(x,y,z)}€R3/(y2=0)} so sir wheather this is sub space or not please tell
And you're explanation is awesome ❤
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.
Thank you 😊
How to check solution set of differential equation form a vector space or not
Check this video..
I have given the solution
th-cam.com/video/M7eNGMD-YEc/w-d-xo.html
Sir your explanation is very nice please provide some important questions of group theory related uppsc exam
Thank you.. Well, I dont have any such collection but I am sure you will get it online... All the very best...
Sir what about this one (x, y, z) : x = y = z . Will this form a subspace?
Yes, it forms a subspace.
Sir can you tell which playlist we have to follow for 1 st chapter and its basics for ode and MVC plz
Hi
Here is the Playlist for unit I.
Enjoy:)
th-cam.com/play/PLwaXU7G6UrbefJGT7WXm-ONSzcNoR4foC.html
@@DrMathaholic tysm sir😁
@@gamerom9324 :)
Sir what about x^2+y^2-z^2=0. Will it form a subspace??
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
Noce
Please explain sir....
W = {( x, 2y, 3z) : x,y,z belongs to R}
Will W be a subspace of R³
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) }...
From above one can see that W=R^3.
So yes, W is a space..
Thank you sir
@@tanvikadahiya3751 welcome..
hank you for the very nice explanation. Will this form a subspace: W1={(x,y)∈R^2 |xy≥0}⊂R^2?
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
@@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?
@olivierdebruijn8704 welcome
.yes , when d is 0, its plane passing through origin. Hence it do forms a subspace..
sir if there is a set C2(complex) over R2, then it will be a vector space right?
R2 is not a field. So c2 over R2 do not have VS structure.
C2 over R is VS
@@DrMathaholic okay sir thank you
If degree is same on both side then that is subspace
I didn't get the question..
Set S={ p(x) | deg(p(x)) = what?? }
What is the set S?
x = 5y does it forms a subspace?
Yes...
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
@@rishirajbehera7183 see if this video helps::
th-cam.com/video/qkRjSDJf-e4/w-d-xo.html
@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.
Thank you so much sir :) @@DrMathaholic
You are uniqe go on and tank you
Thank you and welcome 🙂
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
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
@@DrMathaholic ok sir thank you for the guidance
@@ujjwalmahajan7581 all the best..
@@DrMathaholic :)
W
={ (x1,x2,x3)/x1=x2} is it a subspace of W?
Yes, it's a subspace..
{(x,y,z) belongs to V: x+y+z=0} sir this is Subspace or Not
Sir plz reply
Yes..it's a subspace..its a plane passing through origin
s = {{x,y,z,} belongs to R^3 : x+y=0 or y-z=0}
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
@@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
@Srinath Shrestha oh, it's "or" in between.. I thought its "and"
@@srinathshrestha3899 Under the "or" condition, it's not a subspace..whereas under "and" it's a subspace..
So whenever there is an or condition between more than 1 equation then it's never a subspace..
Thank you so much sir
Welcome :)
Thank you sir
Welcome 😀
Thank you sir
Welcome:)