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 one year ago
W := {(x_1; x_2; x_3) in V l (x_2)=(x_1)(x_3)}
Is W a subspace of R^3 (subspace test)
 one year ago
W := {(x_1; x_2; x_3) in V l (x_2)=(x_1)(x_3)} Is W a subspace of R^3 (subspace test)

This Question is Closed

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1there are certain criteria for "subspace"

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do you recall what they are?

TedG
 one year ago
Best ResponseYou've already chosen the best response.1I know show 0 vector is in set, closure under addition and closure under multiplication, i have shown the first two but for the last i am unsure on whether i have done it right so just wanted to check

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1V l (x_2)=(x_1)(x_3)} that is hard to parse thru, can you clear it up any?

TedG
 one year ago
Best ResponseYou've already chosen the best response.1\[ W:= \{(x_{1}, x_{2}, x_{3}) \in V \text{such that} x_{2}=x_{1}x_{3}\} ) \]

TedG
 one year ago
Best ResponseYou've already chosen the best response.1sorry wasnt sure on how to put a space after text

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1that is alot easier to read :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if you can prove that: \[c_1\vec X+c_2\vec Y\in W\]that should satisfy the addition and multiplcation requirements in one fell swoop

TedG
 one year ago
Best ResponseYou've already chosen the best response.1oh ok so i have done it seperately, but under multiplication can it be said that for cu=c(u1,u2,u3) then cu=(cu1,cu2,cu3) and if we let say v= cu that conditions are met? as cu2=cu1cu3 implies v2=v1v3 if that makes sense.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1cu1 cu2 = c^2 u1 u2 right?

TedG
 one year ago
Best ResponseYou've already chosen the best response.1yh see thats where i was not sure as then surely that wouldnt work

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im getting the indexes a little mixed up i see

TedG
 one year ago
Best ResponseYou've already chosen the best response.1sorry its just my latex typing isnt that quick

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\vec u=c_1\vec x_1+c_2\vec x_2+c_3\vec x_3\] \[\vec u=c_1\vec x_1+c_2\vec x_1\vec x_3+c_3\vec x_3\] \[k\vec u=k(c_1\vec x_1)+k(c_2\vec x_1\vec x_3)+k(c_3\vec x_3)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now im getting lost lol ... oy vey

TedG
 one year ago
Best ResponseYou've already chosen the best response.1ok i think i got it it let me type it in latex so it is clear.

TedG
 one year ago
Best ResponseYou've already chosen the best response.1Suppose \(c \in R \) and \(\mathbf{v} \in V \) \[c\mathbf{v} = c(v_{1}, v_{2}, v_{3}) =(cv_{1}, cv_{2}, cv_{3}) \]Now if we let \(\mathbf{u} \in V \) where \(\mathbf{u} = c\mathbf{v}\) which is valid as it is given that \( (x_{1}, x_{2}, x_{3}) \in V \) and \(V\) is the vector space \(R^3\) so multiplying by a scalar will still be in the reals. Thus, we have that \[ cv_{2} =cv_{1}cv_{3} \] and as \(\mathbf{u} = c\mathbf{v}\) it can be shown that \[ u_{2} =u_{1}u_{3} \]

TedG
 one year ago
Best ResponseYou've already chosen the best response.1i think that makes sense.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1with my limited recollection, that does make sense to me as well. but wouldnt we want: u in W? or is the line definition of v2 = v1v3 sufficient for that ....

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im not even sure what that definition would actually entail ... but for the sake of definitions :)

TedG
 one year ago
Best ResponseYou've already chosen the best response.1yh you are right. i want u in W not V

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1by defining u in W, u = cv, and W is of the defintion x2= x1x3; cv2 = c^2 v1v2 for all c seems to defy closure to me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1..but knowing me, id want a 2nd opinion :) @.Sam. you any decent at these things?

TedG
 one year ago
Best ResponseYou've already chosen the best response.1yh thats the reason i came on here to ask in the first place the c^2 might mean it is not a subspace.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if we could visualize a x1x3 product ... we would have something to compare with

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if its a "dot product" 3<1,1,1> 3<4,1,0>  3(4+1+0) doesnt seem to make sense to me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1cross product the only other thing i can thing of for a test

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1<1,1,1>x<4,1,0> = <1,4,3> <3,3,3>x<12,3,0> = <9,36,27> = 3<3,12,9>

TedG
 one year ago
Best ResponseYou've already chosen the best response.1in that case it definately doesnt hold right

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1thats what im wondering about we can still get <1,4,3> back again, by factoring out another 3, and since 3^2 is a real number ... im just not sure if thats acceptable or not :/

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1<1,4,3> , if we are defining the product of x1x3 as a cross, IS in W, and any real scalar of it is as well

TedG
 one year ago
Best ResponseYou've already chosen the best response.1hmm i guess i will just have to email my lecturer see what he says

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, good luck with it :) i think ive stared myself into the 50s bracket of an IQ test with this one ....

TedG
 one year ago
Best ResponseYou've already chosen the best response.1well thanks for the help anyway, much appreciated
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