I have this square matrix:


and I want to find its symmetric and skew-symmetric parts but I am confuse because it is already a skew symmetric matrix, and when finding the symmetric part I get a zero matrix. Is that possible?


1 Answer 1


In general, you can decompose $A$ uniquely as


with $A_{\rm sym}=\frac{A+A^{T}}{2}$ the symmetric part and $A_{\rm anti-sym}=\frac{A-A^{T}}{2}$ the anti-symmetric part. If $A$ is anti-symmetric then $A^{T}=-A$ and you get

$$A_{\rm sym}=0$$

$$A_{\rm anti-sym}=A$$

Thus, the symmetric part of an anti-symmetric matrix is indeed zero. There is no problem with your conclusion.

  • 1
    $\begingroup$ ok good ! then what about the anti-symmetric part of A ? because it is already anti-symmertic I am not sure if I have to calculate it or just state it is already a anti-symmertic matrix. $\endgroup$ Commented Dec 10, 2017 at 12:39
  • $\begingroup$ @WonderWomen See my updated answer. It is up to you to decide if calculation is needed in this case. If it is your first time dealing with this decomposition, it might be good for you to calculate and verify the result. $\endgroup$
    – eranreches
    Commented Dec 10, 2017 at 12:42
  • $\begingroup$ very helpful. Thank you so much $\endgroup$ Commented Dec 10, 2017 at 12:47
  • $\begingroup$ @WonderWomen You're welcome. A good exercise to do is to prove the uniqueness of this factorization. Then if $A$ is anti-symmetric, you can argue that $A=0+A$ is a decomposition into symmetric part $0$ and anti-symmetric part $A$, and because of uniqueness it must be that $A_{\rm sym}=0$ and $A_{\rm anti-sym}=A$. $\endgroup$
    – eranreches
    Commented Dec 10, 2017 at 12:51
  • $\begingroup$ one more question, in my lecture notes it was written that sym part is A + transpose(A) and the anti-sym part is A - transpose(A) then A is equal to the summation of the half of each part but you divided them by 2 from the beginning. Is it the same thing ? or my lecture notes is wrong. Sorry for the wording but I am not sure how to write it using codes. $\endgroup$ Commented Dec 10, 2017 at 13:00

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