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Determine if S is a subspace of V 
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#1
Feb1314, 07:00 PM

P: 8

My answers aren't all correct and I am not sure why..
Problem: Determine whether the given set S is a subspace of the vector space V. A. V is the vector space of all realvalued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=f(b). B. V=R^n, and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix. C. V=C^1(R), and S is the subset of V consisting of those functions satisfying f′(0)≥0. D. V=P_4, and S is the subset of P_4 consisting of all polynomials of the form p(x)=a(x^3)+(bx). E. V=R^n×n, and S is the subset of all symmetric matrices. F. V=R^4, and S is the set of vectors of the form (0,x2,3,x4). G. V=R^n×n, and S is the subset of all matrices with det(A)=0. I chose A,B,C, E, G. A: sine an cosine are real valued functions. in the case of sine,on the interval [0,2pi] f(0)=f(2pi). B:Just looking at the dimensions A has n columns which means( i think) n unknowns therefore falls in R^n C: if f(t) = t then f'(t)=1 which is greater than 0 at t=0. E:a symmetric matrix is a square matrix G:A must be square in order to take a determinate. I did not choose D and F: D: because P(x) is of degree 3 and V=P_4 F: does not contain the 0 vector. Any insight would be greatly appreciated. 


#2
Feb1314, 09:25 PM

P: 455




#3
Feb1314, 09:48 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,488




#4
Feb1314, 09:50 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,488

Determine if S is a subspace of V



#5
Feb1314, 10:14 PM

P: 8




#6
Feb1314, 10:27 PM

P: 8

Im supposed to pick the ones that are sub spaces. The ones under "I chose" are the ones a chose and the ones under" I did not choose", i did not choose. I hope that clears it up. I'm supposed to look to see if scalar multiplication and vector addition hold. 


#7
Feb1414, 01:05 AM

P: 455

In particular, you need to check that (1) ##u+v## is in ##S## whenever ##u## and ##v## are in ##S## and (2) ##\alpha u## is in ##S## for all real numbers ##\alpha## whenever ##u## is in ##S##. For instance, in (A), if ##f## and ##g## are in ##S##, then ##f(a)=f(b)## and ##g(a)=g(b)##. So (1) ##(f+g)(a)=f(a)+g(a)=f(b)+g(b)=(f+g)(b)##, so ##f+g## is in ##S##, and (2) for all real ##\alpha##, ##(\alpha f)(a)=\alpha f(a)=\alpha f(b)=(\alpha f)(b)##, so ##\alpha f## is in ##S##. So ##S## is a subspace of ##V## in (A). Based on what you wrote in the original post, I would strongly encourage you to, in each case, take a minute (or more) to figure out exactly what ##V## is, understand what it's members "look like", understand the vector addition and scalar multiplication, and verify that it is a vector space. Then look at the defining aspect of ##S## and truly understand what it means for a member of ##V## to also belong to ##S##. Then go ahead and check (1) and (2). Don't take this the wrong way, but almost nothing you said in the first post is really pertinent to this problem. I'm not trying to put you down. I just don't want you to waste time trying to salvage that work. Also, check to make sure that you copied (F) down correctly. 


#8
Feb1414, 09:41 AM

P: 8




#9
Feb1414, 09:48 AM

P: 8

And in response to:
check to make sure that you copied (F) down correctly. It's copy pasted. But checked anyway and it's correct. 


#10
Feb1414, 11:46 AM

P: 455

Also, there is a onestep check; for all ##u## and ##v## in ##S## and all real ##\alpha## and ##\beta##, ##\alpha u+\beta v## is also in ##S##. But I've never really found that to be very useful in practice. It's usually just as easy, if not easier, to check the conditions separately. The onestep check is more for when you completely understand the concept and just want to get through the problem as quickly as possible. 


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