anonymous
  • anonymous
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)
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
there are certain criteria for "subspace"
amistre64
  • amistre64
do you recall what they are?
anonymous
  • anonymous
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

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amistre64
  • amistre64
V l (x_2)=(x_1)(x_3)} that is hard to parse thru, can you clear it up any?
anonymous
  • anonymous
\[ W:= \{(x_{1}, x_{2}, x_{3}) \in V \text{such that} x_{2}=x_{1}x_{3}\} ) \]
anonymous
  • anonymous
sorry wasnt sure on how to put a space after text
amistre64
  • amistre64
that is alot easier to read :)
amistre64
  • amistre64
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
anonymous
  • anonymous
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.
amistre64
  • amistre64
cu1 cu2 = c^2 u1 u2 right?
anonymous
  • anonymous
yh see thats where i was not sure as then surely that wouldnt work
amistre64
  • amistre64
im getting the indexes a little mixed up i see
anonymous
  • anonymous
sorry its just my latex typing isnt that quick
amistre64
  • amistre64
\[\vec u=c_1\vec x_1+c_2\vec x_2+c_3\vec x_3\] \[\vec u=c_1\vec x_1+c_2\vec x_1\vec x_3+c_3\vec x_3\] \[k\vec u=k(c_1\vec x_1)+k(c_2\vec x_1\vec x_3)+k(c_3\vec x_3)\]
amistre64
  • amistre64
now im getting lost lol ... oy vey
anonymous
  • anonymous
ok i think i got it it let me type it in latex so it is clear.
anonymous
  • anonymous
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 \[ u_{2} =u_{1}u_{3} \]
anonymous
  • anonymous
i think that makes sense.
amistre64
  • amistre64
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 ....
amistre64
  • amistre64
im not even sure what that definition would actually entail ... but for the sake of definitions :)
anonymous
  • anonymous
yh you are right. i want u in W not V
amistre64
  • amistre64
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
amistre64
  • amistre64
..but knowing me, id want a 2nd opinion :) @.Sam. you any decent at these things?
anonymous
  • anonymous
yh thats the reason i came on here to ask in the first place the c^2 might mean it is not a subspace.
amistre64
  • amistre64
if we could visualize a x1x3 product ... we would have something to compare with
amistre64
  • amistre64
if its a "dot product" 3<1,1,1> 3<4,1,0> ---------- 3(4+1+0) doesnt seem to make sense to me
amistre64
  • amistre64
cross product the only other thing i can thing of for a test
amistre64
  • amistre64
<1,1,1>x<4,1,0> = <-1,4,-3> <3,3,3>x<12,3,0> = <-9,36,-27> = 3<-3,12,-9>
anonymous
  • anonymous
in that case it definately doesnt hold right
amistre64
  • amistre64
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 :/
amistre64
  • amistre64
<-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
anonymous
  • anonymous
hmm i guess i will just have to email my lecturer see what he says
amistre64
  • amistre64
well, good luck with it :) i think ive stared myself into the 50s bracket of an IQ test with this one ....
anonymous
  • anonymous
well thanks for the help anyway, much appreciated

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