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Is this subset a subspace

  1. Oct 27, 2011 #1
    Is the subset A a subspace of W

    [tex]W=\left \{ \begin{bmatrix}
    1 &1 \\
    a_{21}& a_{22} \\
    a_{31}& a_{32}
    \end{bmatrix} :a_{ij} \in \mathbb{C}\right \}[/tex]

    Let [tex]A=\begin{bmatrix}
    1 &1 \\
    a_{21}& a_{22}\\
    a_{31}& a_{32}
    \end{bmatrix}[/tex]

    [tex]A \in W[/tex]

    Then [tex]2A \in W[/tex] since

    [tex]2A=2\begin{bmatrix}
    1 &1 \\
    a_{21}& a_{22}\\
    a_{31}& a_{32}
    \end{bmatrix}[/tex]

    Is this correct? My notes is telling its NOT closed under scalar multiplication which I dont think is correct.
     
    Last edited by a moderator: Oct 27, 2011
  2. jcsd
  3. Oct 27, 2011 #2
    What exactly is A, a matrix? Or what do the two lines around the components mean? In any case, multiply 2A and write it out, so that you have it in the form with the two lines around it. You should then be able to see why this isn't a subspace.

    (Unless those two lines stand for something else than what I had in mind.)
     
  4. Oct 27, 2011 #3

    Mark44

    Staff: Mentor

    2A is NOT in W since the entries in the top row aren't both 1.

    Ryker, A is a matrix. I changed the OP's LaTeX so that it didn't look like a determinant.
     
  5. Oct 27, 2011 #4
    Thanks for the clarification on the lines.

    Ok, so this is true for any scalar [tex]\alpha \in \mathbb{R}[/tex]?
     
  6. Oct 27, 2011 #5
    No, it's not true for any scalar (it is true for all but one - can you tell, which one?), but to show this subset is not a subspace, you only need to find one scalar where this fails.
     
  7. Oct 27, 2011 #6
    Now I am confused...
    Based on post #3, it is NOT a subspace because the 1's are not present in the matrix due to multiplication of real scalar '2'. I thought this WOULD be true for [tex]\alpha \in (0,\infty)[/tex] where the round brackets mean excluding 0 and infinity.

    But you are saying its not true for all real scalars....?!
     
  8. Oct 27, 2011 #7

    Mark44

    Staff: Mentor

    I think you mean that it's true for only one scalar.
     
  9. Oct 27, 2011 #8

    Mark44

    Staff: Mentor

    What post #3 is saying is that, in general, if A [itex]\in[/itex] W, then cA [itex]\notin[/itex] W, so scalar multiplication is not closed in W.

    BTW, why are you limiting you scalars to positive real numbers? Set W has matrices with entries from C, the field of complex numbers. Any scalars would also come from this field.
     
  10. Oct 27, 2011 #9
    Ok,

    TO clarify...the subset is not a subspace for all scalars [tex]\in \mathbb{C}[/tex] except the scalar '1'...that it?
     
  11. Oct 27, 2011 #10
    I thought by "is this true for all scalars" he meant "is it true for all scalars a, that aA is not an element of W" :smile: So this would be true for all scalars, except for one, for which it would hold that aA is in W. So much confusion :smile:
     
  12. Oct 27, 2011 #11
    No, you can't say the subset is a subspace for all scalars except for one. Either it is a subspace or it isn't.
     
  13. Oct 27, 2011 #12
    ok, which scalar makes A a subspace....? It cant be 0, i, infinity.....I havent a clue....?
     
  14. Oct 27, 2011 #13

    Mark44

    Staff: Mentor

    None, and that's the point of this exercise. A is NOT a subspace of the space of 3 X 2 matrices with entries from C.

    Set A is not closed under scalar multiplication. This set is also not closed under addition.
     
  15. Oct 27, 2011 #14
    ok, cheers!
     
  16. Oct 27, 2011 #15
    As mentioned above, by "it is true for all but one", I meant that multiplying A by a particular scalar (call it "a"), will result in an element of W, i.e. aA is in W, but if you multiply A by any other scalar, say, "b", bA is not in W.

    Other than that, perhaps revisit the definition of a subspace. No scalar can make a subspace, since being a subspace is a property of the set itself. So if scalar multiplication with an element of your set results in another element of your set, then the set is closed under scalar multiplication. But if you can find one (or two or a thousand or infinitely many) scalar for which this statement is not true, then your set is not closed under scalar multiplication.
     
  17. Oct 27, 2011 #16
    but what is so special about 'a' that makes it a subspace when 'alpha' a scalar entry from C doesnt?
     
  18. Oct 27, 2011 #17
    Just to clear up confusion, I used "a" instead of your "alpha", because I didn't want to go into tex mode. But a = alpha, I just used a different variable name. So in my replies, a is also an element of C, i.e. a is a complex number, a scalar. And the whole point of my replies was that that special "a" or "alpha" does NOT make your set a subspace. That is, if your set is closed under multiplication by a particular scalar, then this is needed, but not sufficient for a set to be a subspace, because it needs to be closed under multiplication by all scalars. But if you do find one scalar, under multiplication by which your set is not closed, then that is sufficient in showing your set is not a subspace.
     
  19. Oct 27, 2011 #18

    Mark44

    Staff: Mentor

    Nothing, and the name of the scalar has no bearing on things. For scalar multiplication to be closed, aA has to be in the set for every scalar in whatever field is relevant to the problem.
     
  20. Oct 27, 2011 #19
    I am looking at my notes and found a counter example....!

    In R^3 where the vector x = (1,2,-4)

    apparently its closed under addition if we add it to another vector y etc...but 2x=(2,4,-8)...according to this post...this would imply x is NOT a subspace (because 1,2 and -4 is not in the new set).....but notes say it is a real vector space.....!???
     
  21. Oct 27, 2011 #20
    What do you mean? 2x is still in R^3, so everything is as it should be.
     
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