Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it true that $\mathrm{det}(A-B)\neq0$ if and only if $\mathrm{det}(A+B)\neq0$?

(For $A,B$ real $n\times n$ matrices, say.)

share|improve this question
17  
No, let $A = B = I$ –  Vectk Jul 26 '12 at 3:45
3  
@Brian: That's more of an answer than a comment. –  user26872 Jul 26 '12 at 3:46
10  
I am curious to know where this bizarre conjecture comes from. The implication doesn't even hold for $1\times 1$ matrices, i.e. real numbers. –  Erick Wong Jul 26 '12 at 4:10
4  
This result would be true in characteristic $2$. Otherwise the guess is without a chance: you can easily make $A+B=P$ and $A-B=Q$ for any square maretices $P,Q$ of your choosing. –  Marc van Leeuwen Jul 26 '12 at 4:44
2  
-1 This is the kind of questions I can't understand as it takes about 3 minutes of serious work, try-and-fail, examples with 1x1, 2x2 matrices, etc. to realize what's going on. Besides all this, the question has no background, what's it related to, no own effort... –  DonAntonio Jul 26 '12 at 6:09

1 Answer 1

It is not true, even if you require $A\ne B$. For instance, take $$ A=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $\det(A-B)=0$, $\det(A+B)=2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.