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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
 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
 one year ago

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amistre64Best ResponseYou've already chosen the best response.1
there are certain criteria for "subspace"
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
do you recall what they are?
 one year ago

TedGBest ResponseYou've already chosen the best response.1
I 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
V l (x_2)=(x_1)(x_3)} that is hard to parse thru, can you clear it up any?
 one year ago

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

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

amistre64Best ResponseYou've already chosen the best response.1
that is alot easier to read :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
if 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
 one year ago

TedGBest ResponseYou've already chosen the best response.1
oh 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.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
cu1 cu2 = c^2 u1 u2 right?
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
im getting the indexes a little mixed up i see
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
\[\vec u=c_1\vec x_1+c_2\vec x_2+c_3\vec x_3\] \[\vec u=c_1\vec x_1+c_2\vec x_1\vec x_3+c_3\vec x_3\] \[k\vec u=k(c_1\vec x_1)+k(c_2\vec x_1\vec x_3)+k(c_3\vec x_3)\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
now im getting lost lol ... oy vey
 one year ago

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

TedGBest ResponseYou've already chosen the best response.1
Suppose \(c \in R \) 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 \[ u_{2} =u_{1}u_{3} \]
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
with 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 ....
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
im not even sure what that definition would actually entail ... but for the sake of definitions :)
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
by 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
..but knowing me, id want a 2nd opinion :) @.Sam. you any decent at these things?
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
if we could visualize a x1x3 product ... we would have something to compare with
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
if its a "dot product" 3<1,1,1> 3<4,1,0>  3(4+1+0) doesnt seem to make sense to me
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
cross product the only other thing i can thing of for a test
 one year ago

amistre64Best 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>
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
thats 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 :/
 one year ago

amistre64Best 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
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.1
well, good luck with it :) i think ive stared myself into the 50s bracket of an IQ test with this one ....
 one year ago

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