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Linear algebra : Doing a proof with a square matrix

  1. Feb 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that all square matrix (A whatever) can be written as the sum of a symmetric matrix and a anti symmetric matrix.

    2. Relevant equations

    I think this relation might be relevant : $$
    A=\frac{1}{2}*(A+A^{T})+\frac{1}{2}*(A-A^{T})
    $$
    3. The attempt at a solution

    I know that we have the following theorems : A square matrix "A" is symmetric if and only if $$A^T=A$$

    and

    A square matrix "A" is anti symmetric if and only if $$A^T=-A$$

    But besides that I'm not sure where to go.... Any hints ?
     
  2. jcsd
  3. Feb 15, 2016 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why don't you check to see if your proposed equation does it?
     
  4. Feb 15, 2016 #3
    Sorry I'm not following you. You mean I must reduce my equation ? We get for example :

    $$2A=(A+A^T)+(A-A^T)$$
     
  5. Feb 15, 2016 #4

    fresh_42

    Staff: Mentor

    That's not what LC said. Read it again, carefully. It contains already all that can be said.
     
  6. Feb 15, 2016 #5
    To be honest I'm not sure of understanding the problem. I think that I must get this as a result : $$A = A + A^T$$

    Am I right or not ?
     
  7. Feb 15, 2016 #6

    fresh_42

    Staff: Mentor

    Write it ##A=B+C## and see if you already have ##B## and ##C## with the desired properties.
     
  8. Feb 15, 2016 #7
    Ok, so if I use my equation I set the following equations which : $$B = \frac{1}{2}*(A+A^{T})$$

    and $$C=\frac{1}{2}*(A-A^{T})$$

    Is this where you go ?
     
  9. Feb 15, 2016 #8

    fresh_42

    Staff: Mentor

    Yep, just collect it.
     
  10. Feb 15, 2016 #9
    Ok, sorry but I'm not understanding the problem at all right now. I'm not even sure what I'm supposed to get as a final result... It says that I must have a sum of a symmetric and anti symmetric matrix. So... how does what I just did helpt exactly ?
     
  11. Feb 15, 2016 #10

    fresh_42

    Staff: Mentor

    Is ##B## symmetric and ##C## anti-symmetric as defined above?
     
  12. Feb 15, 2016 #11
    Ok I understand what you'Re saying. You'Re saying that B and C are respectively symmetric and anti symmetric. But how do I know that they are ? Must I prove that the sum in B is symmetric while the difference in C is anti symmetric ?
     
  13. Feb 15, 2016 #12

    fresh_42

    Staff: Mentor

    Yes. Put a big bracket around them, right a big T on the right one above and resolve the brackets.
     
  14. Feb 15, 2016 #13
    You want this, for example :

    $$B=\frac{1}{2}*(A+A^{T})$$

    then

    $$B^T=(\frac{1}{2}*(A+A^{T}))^T$$

    and I do the same for C.
     
  15. Feb 15, 2016 #14

    fresh_42

    Staff: Mentor

    Yes, but what is ##B^T=(\frac{1}{2}*(A+A^{T}))^T##? It must equal ##B##. And the version with ##C^T## has to be ##-C##. Do you know why this is the case?
     
  16. Feb 15, 2016 #15
    I didn't get B though.

    I got this : $$B^T=1/2*(A^T + A)$$


    That was my end result.
     
  17. Feb 15, 2016 #16

    fresh_42

    Staff: Mentor

    And if you put all together you have for ##A=B+C= 1/2(A+A^T)+1/2(A-A^T)## (in very detailed form):
    ## B^T=(1/2*(A + A^T))^T= 1/2*(A^T + (A^T)^T)=1/2*(A^T+A)=1/2*(A+A^T) = B##
    and similar with ##C##.
     
  18. Feb 15, 2016 #17
    Ok yeah I ended up with $$C^T=-C$$
     
  19. Feb 15, 2016 #18
    Ok,

    so i did the following $$A=1/2(A + A^T) + 1/2(A-A^T)$$

    I applied the T on the whole of this...

    I got $$A^T=1/2*(A+A^T) - 1/2 (A-A^T)$$
     
  20. Feb 15, 2016 #19

    fresh_42

    Staff: Mentor

    Ok then, you're probably blocked. That happens from time to time. (I have asked one of the silliest questions at all myself a couple of hours ago.)

    You have ##A= 1/2*(A+A^T) + 1/2*(A-A^T)##.
    You need to show that ##A## is a sum of a symmetric and a anti-symmetric matrix.
    You already have a sum (the plus sign in the middle).
    All you need to show is that the first term is symmetric, i.e. ##[1/2*(A+A^T)]^T=1/2*(A+A^T)##.
    We did this for ##B## which was only another name for ##1/2*(A+A^T)##.
    And you need to show that the second term is anti-symmetric, i.e. ##[1/2*(A-A^T)]^T=1/2*(A^T-A)= - 1/2*(A-A^T)##
    We did this for ##C## which was only another name for ##1/2*(A-A^T)##.

    It's nothing left to do.

    And if you still don't see what it is about, then let it rest for a day and have another look at it tomorrow.
     
  21. Feb 15, 2016 #20
    I think I understood. Just another question for C. We know that C is anti symmetric because when we tranpose it we obtain $$C^T=-C$$ and we know there's a theorem which says if we obtian this then our square matrix is anti symmetric, right ?

    After Ive tranposed the B and C I know that my sum is a sum of symmetric and anti symmetric.
     
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