Let A in M(nxn) (F). Under what conditions is det(-A)=det(A)

• Aug 25th 2009, 04:04 PM
sofwhere
Let A in M(nxn) (F). Under what conditions is det(-A)=det(A)
Hi All(Hi),
Have a quick question

Let A in M(nxn) (F). Under what conditions is det(-A)=det(A)

Thanks
sofwhere
• Aug 25th 2009, 05:31 PM
NonCommAlg
Quote:

Originally Posted by sofwhere
Hi All(Hi),
Have a quick question

Let A in M(nxn) (F). Under what conditions is det(-A)=det(A)

Thanks
sofwhere

since $\displaystyle \det(-A)=(-1)^n \det A,$ we have $\displaystyle \det(-A)=\det A$ if and only if at least one of theses conditions holds:

1) $\displaystyle \det A = 0,$

2) $\displaystyle 2 \mid n,$

3) $\displaystyle \text{char}(F)=2.$
• Aug 25th 2009, 06:05 PM
sofwhere
Quote:

Originally Posted by NonCommAlg
since $\displaystyle \det(-A)=(-1)^n \det A,$ we have $\displaystyle \det(-A)=\det A$ if and only if at least one of theses conditions holds:

1) $\displaystyle \det A = 0,$

2) $\displaystyle 2 \mid n,$

3) $\displaystyle \text{char}(F)=2.$

I thought of conditions 1 & 2. But the 3rd one is which i didnt think of. Thanks for timely reply.(Bow)