anonymous
  • anonymous
Help with proving W = (1,2,4x+3Y) not a vector
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
are there any more details to this question?
anonymous
  • anonymous
x and y are real numbers and V = R3
anonymous
  • anonymous
Help with proving W = (1x,2y,4x+3Y) not a vector , x and y are real numbers and V = R3

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anonymous
  • anonymous
not a vector or not a vector space?
anonymous
  • anonymous
not a vector sub space
anonymous
  • anonymous
okay, makes more sense now
anonymous
  • anonymous
it appears to pass the conditions to be a subspace of R^3
anonymous
  • anonymous
how to prove with addition
anonymous
  • anonymous
Let \[\vec{a}=(x_1,2y_2,4x_1+3y_1)\] and \[\vec{b}=(x_2,y_2,4x_2+3y_2)\] be in W
anonymous
  • anonymous
Now \[\vec{a}+\vec{b}=(x_1+x_2,y_1+y_2,4x_1+3y_1+4x_2+3y_2)\] group and factor the third component to get \[(x_1+x_2,y_1+y_2,4(x_1+x_2)+3(y_1+y_2))\]
anonymous
  • anonymous
since \[x_1,x_2,y_1,y_2\in \mathbb{R}\] clearly this vector sum of arbitrary vectors in W is in W
anonymous
  • anonymous
I should have made the statement about \[x_1,x_2,y_1,y_2\in \mathbb{R}\] first of course
anonymous
  • anonymous
Thank you I am reviewing to make sure I understand.
anonymous
  • anonymous
would it work the same if W = x,2y,4x - 3Y
anonymous
  • anonymous
it should, but I have not tried it, Proving a subspace is one of those "turn the crank" type proofs in linear algebra that may seem a little abstract at first but once yo get the process down is straight forward but sometimes tedious...
anonymous
  • anonymous
Thank You
anonymous
  • anonymous
IS W subspace of V, W is the set of all functions F(0)=1 , V=C(\[-\infty\],\[\infty\])

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