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Subspaces and Algebra

  1. Aug 29, 2011 #1
    Hi there. I started learning about subspaces in linear algebra and I came across a question which i'm unsure how to solve. I understand that there are 'rules' which need to be passed in order for something to be a subspace, but I have no idea how to start with this problem:

    Consider the set M23 of all 2 × 3 matrices with real entries under the usual operations of matrix addition and scalar multiplication.
    Let
    T=([a b c] : a + c= 0 and b + d + f =0)
    [d e f]
    Prove that T is a subspace of M23
    (T is a 2x3 matrix if i made it unclear)
    I know that T must contain a zero vector and I know that there must be closer of scalar multiplication and addition.

    Can anyone help?
     
  2. jcsd
  3. Aug 29, 2011 #2

    micromass

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    So there are three things that you need to do:

    • T contains the zero matrix. Is this true.
    • T is closed under composition. Thus if

      [tex]\left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right)~\text{and}~\left(\begin{array}{ccc} a^\prime & b^\prime & c^\prime \\ d^\prime & e^\prime & f^\prime\end{array}\right)[/tex]

      are in T, then

      [tex]\left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right) + \left(\begin{array}{ccc} a^\prime & b^\prime & c^\prime \\ d^\prime & e^\prime & f^\prime\end{array}\right)[/tex]

      is in T
    • T is closed under scalar multiplication, thus if

      [tex]\left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right)[/tex]

      is in T, then for each [itex]\alpha[/itex]

      [tex]\alpha\left(\begin{array}{ccc} a & b & c\\ d & e & f\end{array}\right)[/tex]

    Let's start with the first. Why is the zero matrix in T?
     
  4. Aug 29, 2011 #3
    Well i'd say that it has a zero matrix because a + c = 0 (which can be 0 + 0 = 0) and b + d + f = 0 (i.e 0 + 0 + 0 =0) i.e a and d = 0. Is this correct?
     
  5. Aug 29, 2011 #4

    micromass

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    Yes, that is correct.
    Now, why is it closed under addition?
     
  6. Aug 29, 2011 #5
    I'm not sure if this is the way to do it but I would say
    u=[a1]
    [d1]
    v=[a2]
    [d2]

    i.e if u and v are vectors in T, then u+v will be a vector in T.

    u+v=[a1+a2]
    [d1+d2]
    a1+a2+d1+d2=0
    (a1+a2)+(d1+d2)=0
    0 + 0 = 0

    I'm not sure if it can be written like that
     
  7. Aug 29, 2011 #6

    micromass

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    Why do you write u and v like that?? u and v are 3x2 matrices!! But you seem to write them as 2x1 matrices...
     
  8. Aug 29, 2011 #7
    Thats what i meant when i said i wasn't sure lol. So would it be correct if i wrote u and v as 3x2 matrices containing all the elements of T and writing them as a1+a b1 +b2 etc.. what would be the next step? can i still use this after:
    a1+a2+d1+d2=0
    (a1+a2)+(d1+d2)=0
    0 + 0 = 0
    or must i use all elements from u+v?
     
  9. Aug 29, 2011 #8

    micromass

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    You must write u and v as

    [tex]u= \left(\begin{array}{ccc} a_1 & b_1 & c_1\\ d_1 & e_1 & f_1\end{array}\right)~ \text{and}~ v=\left(\begin{array}{ccc} a_2 & b_2 & c_2\\ d_2 & e_2 & f_2 \end{array}\right)[/tex]

    Now, what is u+v??
     
  10. Aug 29, 2011 #9
    Yes thats what I meant to say xD

    u+v=[a1+a2 b1+b2 c1+c2]
    [d1+d2 e1+e2 f1+f2]
     
  11. Aug 29, 2011 #10

    micromass

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    Yes. Now, does

    [tex]\left(\begin{array}{ccc} a_1+a_2 & b_1+b_2 & c_1+c_2\\ d_1+d_2 & e_1+e_2 & f_1+f_2\\
    \end{array}\right)[/tex]

    belong to T? That is, is [itex](a_1+a_2)+(c_1+c_2)=0[/itex] and [itex](b_1+b_2)+(d_1+d_2)+(f_1+f_2)=0[/itex]??
     
  12. Aug 29, 2011 #11
    Yes because (a1+a2)=0, (c1+c2)=0, (b1+b2)=0, (d1+d2)=0, (f1+f2)=0
    0 + 0 = 0 and 0+0+0=0
     
  13. Aug 29, 2011 #12

    micromass

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    OK, now do the same thing with scalar multiplication.
     
  14. Aug 29, 2011 #13
    Hmm like this?

    ka + kc = 0
    k(a + c) = 0
    k(0) = 0

    kb + kd + kf = 0
    k(b + d + f) = 0
    k(0) = 0
     
  15. Aug 29, 2011 #14

    micromass

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    That's it!
     
  16. Aug 29, 2011 #15
    Awesome! Thanks so much! I understand this much better than i did several hours ago :D
     
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