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Proving matrices are subspaces

  1. Oct 3, 2012 #1
    Hi, I was wondering if someone could check my work for this linear algebra problem. I have attached the problem statement in the file "problem" and my work in the file "work." I would type out my work on here, but I couldn't figure out how to put matrices in a post so I just took a pic of my work.


    I only attached the work for proving S is a subspace.

    I basically checked the 3 conditions my professor gave me to determine if something is a subspace. They are (with respect to my problem):
    1. Is the 0 vector in S?
    2. If U and V are in S, is U+V in S?
    3. If V is in S, then is cV in S for some scalar c?


    I feel like I made this problem too complicated. It took me a whole page to just prove S is a subspace.

    Can someone verify if that's accurate?
    Is there a faster way to do it or a way that takes up less paper?
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2012 #2
    You messed up your V matrix, it's not a member of S at all. Likewise your proof that U+V is a member of S is not correct, as it clearly is not.

    But the proof is on the right tracks otherwise. I'd write a bit less text in the obvious parts, but otherwise you're covering it well.
     
  4. Oct 3, 2012 #3
    Crap! Stupid mistake.
    Should've been in the form
    u -u
    v w
    instead of
    u -v
    v w
    Changing this would make my answer correct.

    Also for part b) Describe all matrices in S+T. What is it asking?
    Do I just add S+T
    Giving
    x+a -x+b
    y-a z+c
    where a, b, c, x, y, z are in R.
    S+T includes all matrices of the above form.

    Is that what it's asking?
     
  5. Oct 4, 2012 #4
    The more interesting answer would be if you found out if there are some limitations to the matrices S+T. That is, if you can express any 2 by 2 matrix M in the form M=S+T, or only some subset of them.
     
  6. Oct 4, 2012 #5
    Hmm I'm not quite sure how to do that, but not all 2x2 matrices can be written in the form of S+T. This is because the 1st (x+a) and 2nd element (-x+b) are both dependent on x. The 1st and 3rd element are both dependent on the value of a.
     
  7. Oct 4, 2012 #6

    vela

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    Yeah, but you have six degrees of freedom and you only need to come up with four numbers. That seems to give you enough flexibility to represent any matrix. You need to check that though.
     
  8. Oct 4, 2012 #7
    Do you think I would be sufficiently answer the question by saying every matrix in S+T is of the form:

    x+a -x+b
    y-a z+c
     
  9. Oct 4, 2012 #8

    vela

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    No, I don't think so.
     
  10. Oct 4, 2012 #9
    I don't know what else I could add.
    I guess I could say that -a=b when x=0
    and -x=y when a=0.
     
  11. Oct 5, 2012 #10
    Is there any kind of matrix that you are not able to represent in the form S+T? And yeah, it might be useful to reduce your variables slightly. For example, you can see that z and c are completely redundant, so you can freely set one of them to zero. After that you have 5 variables and 4 matrix elements, so maybe set one more to zero to equalize the numbers. Does it matter which one you choose?
     
  12. Oct 6, 2012 #11
    Why are z and c redundant?
     
  13. Oct 6, 2012 #12

    vela

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    Look at it this way. Let ##P = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix}##. Can you find x, y, z, a, b, and c such that
    $$\begin{bmatrix} x & -x \\ y & z\end{bmatrix} + \begin{bmatrix} a & -a \\ b & c \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix}$$ Note that this implies that ##z+c = p_{22}##. What clamtrox means is you can satisfy this condition by setting, say, z=0 and c=p22.
     
  14. Oct 7, 2012 #13
    I think the 2nd matrix you have is not the form of T.


    So if i have the below:
    x+a -x+b
    y-a z+c

    I'll set x=0, z=0, y=0
    This gives:
    a b
    -a z+c

    the 1st and 3rd element are opposites of each other. So I don't think S+T represents all 2x2 matrices.
     
  15. Oct 7, 2012 #14

    vela

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    Sorry, you're right; it was supposed to be transposed. Anyway, can you show that there's always a solution given any matrix P?
     
  16. Oct 7, 2012 #15
    Hmm, I don't think I can represent all 2x2 matrices though because if i set x=z=y=0 then this gives:
    a b
    -a z+c

    the 1st and 3rd element are opposites of each other. So I don't think S+T represents all 2x2 matrices.
     
  17. Oct 8, 2012 #16
    Then don't set x=y=z=0! Why on earth would you make such a choice anyway? That's like choosing S to only contain the zero matrix.

    What we said is that you have 6 degrees of freedom, and need only 4 to represent an arbitrary 2x2 matrix. Now if you get rid of 3 of them, that leaves you with 6-3=3<4, which is not enough. If you get rid of 2 of them, then you have 6-2 = 4, which is just right. You still don't have a free choice though! For example, if you choose z=c=0, then you obviously can't represent all 2x2 matrices, since one of the components is always zero.
     
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