Skew-symmetric matrix problem

[Kolahal Bhattacharya]

Homework Statement

I am to show that the trace of the product of a symmetric and a skew-symmetric matrix is zero.Please check what I did is corect:

Homework Equations

The Attempt at a Solution

Let me assume:A~=A and B~=-B

(I will use # sign to denote the sum process)

trace(AB)=[#(i)](AB)_ii=[#(i)] [#(j)] a_ij*b_ji

trace(AB)=-[#(j)] [#(i)] b_ji*a_ij using conditions on A and B
=-[#(j)](AB)_jj
Since i and j are equivalent,
what we have is 2trace(AB)=0
hence,conclusion

 

[Hurkyl]

A~ means the transpose of A?


I think your proof is right, although you skipped some steps and did not provide justification for what was skipped, so I can't be sure.

I will suggest, though, that you don't need to bother with summations at all: you can just use the algebraic properties of the trace instead.

 

[Kolahal Bhattacharya]

Yes, ~ means transpose.
What are the algebraic properties of trace you are referring to?
also I do not uderstand which steps have I jumped?
Thank you.

 

[Hurkyl]

What are the algebraic properties of trace you are referring to?

The first few equations here

also I do not uderstand which steps have I jumped?

These are the two equalities I take issue with

trace(AB)=-[#(j)] [#(i)] b_ji*a_ij

-[#(j)] [#(i)] b_ji*a_ij=-[#(j)](AB)_jj

They are certainly true, but I don't think they're adequately explained.