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  • In my book, I've:

    Problem: Express the matrix [] as a sum of symmetric and skew symmetric matrices.

    Okay, as far as I know, symmetric matrix remains the same even if we transpose it.
    And the skew symmetric matrix is the same as it's -1 x the transpose of itself.

    How do I suppose to solve this?
    here [] =
    | 2 -4 9 |
    | 14 7 13 |
    | 3 5 11 |


    I think, first I should write the transpose of matrix which is the symmetric matrix and then write the transpose again multiplied by -1.

    After that I can add both matrix and prove that it is equal to the [] matrix.

    but in book, they used these formula,

    B = 1/2 (A+A')
    C = 1/2 (A-A')

    I'm not sure what is this... and where it came from 😨

    Anyone can explain me? My method is right or the method used in the book? ☕
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  • ErAnushka

    Member • Sep 3, 2012

    No problem I got 😀

    A direct proof is as follows:

    Let X be the given square matrix

    Let X=A+B where A is symmetric and B is skew symmetric

    That is A'=A and B' = - B .......(1)
    Therefore,
    X'=A'+B' ( using the property of transpose of a matrix)
    or, X'=A - B from (1) above

    Now, as X is square , therefore, X+X' and X-X' are defined.

    Therefore X+X'=A+B +A-B= 2A

    and X-X'=A+B- (A-B) = 2B

    Therefrore A=(X+X')/2

    and B= (X-X')/2

    It is easy to verify that A'=A and B'= - B

    That is A is symmetric and B is skew symmetric

    Therefore , X can be uniquely expressed as sum of a symmetric matrix and a skew symmetric matrix, which is

    X =(X+X')/2 + (X-X')/2




    I understood the green part 😀 , but the red part is little confusing😕 .. A'=A and B' = -B but how? we got only A=(X+X')/2 and B= (X-X')/2.... ??? 😨
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  • silverscorpion

    Member • Sep 4, 2012

    I'm not sure what you are not getting here.. A = A' and B = -B' are assumptions that you made in the beginning, right? (ie., you have assumed that A is symmetric and B is skew symmetric)

    You can probably take the transpose of (X+X')/2 and (X-X')/2, and check that they are indeed the same as A and -B.. but, here it's actually assumed that it is true.
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  • ErAnushka

    Member • Sep 4, 2012

    silverscorpion
    I'm not sure what you are not getting here.. A = A' and B = -B' are assumptions that you made in the beginning, right? (ie., you have assumed that A is symmetric and B is skew symmetric)

    You can probably take the transpose of (X+X')/2 and (X-X')/2, and check that they are indeed the same as A and -B.. but, here it's actually assumed that it is true.
    Omg wait... I got it 😀

    We are actually proving that any square matrix is a sum of symmetric matrix and skew symmetric matrix, so here, A is symmetric matrix and B is skew symmetric matrix.

    We got both value, so,

    A+B = (X+X')/2 + (X-X')/2
    A+B = (X+X' + X-X') /2
    A+B = 2X/2
    A+B = X

    Here X is a square matrix! OMG! I'm so stupid 😁
    By the by thanks for the reply, I got the idea at the moment of reading this again 😀
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