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Zero vector - Linear Subspace

  1. Jul 30, 2014 #1
    Lets say H = [3a+b, 4, a-5b] where a and b are any scalars.

    This not a vector space because Zero vector is not in H. I don't get what it means by zero vector is not in H? Can't you just multiply the vectors by zero and get a zero vector? I am confused.
     
  2. jcsd
  3. Jul 30, 2014 #2

    ehild

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    The components of the vectors of H are defined by a and b, but the second component is 4. You can make the first and third components zero, and have the vector [0, 4, 0], but it is not zero vector.



    ehild
     
  4. Jul 31, 2014 #3

    HallsofIvy

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    Yes, you can multiply any vector by 0 and get the 0 vector. That's why the 0 vector must be in any subspace.

    You are given that H is the set of all vectors of the form [3a+ b, 4, a- 5b]. Since a and b are any scalars, there is a solution to 3a+ b= x and a- 5b= y for all x, y. In other words, we can write any vector in H in the form [x, 4, y]. IF H were a subspace, then 0[x, 4, y]= [0, 0, 0] would be in H. But that is NOT of the form [x, 4, y] so is NOT in H. That is why H is not a subspace.

    (Although showing that the 0 vector is not in H is simplest, one can also show that the other requirements for a subspace are not true for H. Your point that "0 times any vector is the 0 vector" shows that H is NOT "closed under scalar multiplication". 0 times a vector in H is the 0 vector which is NOT in H.

    Also we can write any two vectors in H as [x, 4, y] and [x', 4, y'] for any scalars x, y, x', and y'. The sum of those two vectors is [x+ x', 8, y+ y']. That is NOT of the form [x, 4, y] because the middle component is not 4. Therefore H is not closed under addition so is not a subspace.)
     
  5. Aug 1, 2014 #4

    Fredrik

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    The result 0x=0 holds in every vector space, because for all x, we have 0x=(0+0)x=0x+0x, and now we can add -0x to both sides to get 0 on the left and 0x on the right.

    ##\mathbb R^3## is a vector space. H is a subset of ##\mathbb R^3##. So each element of H is an element of ##\mathbb R^3##. This means that if you multiply an element of H by the number 0, you get the 0 vector of ##\mathbb R^3##. But the 0 vector of ##\mathbb R^3## isn't an element of H.
     
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