Tell me more ×
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.
up vote 2 down vote favorite
3

If $A$ and $B$ are $3\times 3$ matrices and $A$ is invertible, then can we say that there exist an integer $n$ such that $A+nB$ invertible? I was trying by choosing n such that eigne values of $A+nB$ is non zero. In case of $B = I$ we can find out eigen value of $A+nB$ that would be $\lambda +n B$ (though i am not cleared about its proof). This choosing $n$ such that $\lambda$ not equal to -$n$times eigen value of $B$ will serve the purpose. But i am not sure about general $B$. What if i take any matrix $A$ and $B$.

share|improve this question
2  
Hint: $p(n) := \det(A + nB)$ is a polynomial of degree $\le 3$ in $n$, which isn't zero as $p(0) \ne 0$. – martini May 10 '12 at 6:15
1  
If $-n$ is not an eigenvalue of $A^{-1}B$ then $I+nA^{-1}B$ is invertible so $A+nB$ is also invertible. – Joel Cohen May 10 '12 at 6:15
@martini is it a polynomial of degree 3 or $\leq 3$? – srijan May 10 '12 at 6:42
still i haven't got a complete answer. I need little more help. – srijan May 10 '12 at 6:46

1 Answer

up vote 4 down vote accepted

Since $\det(A+xB)$ is a polynomial in $x$, it either has finitely many zeroes or is $0$ for all $x$. Since $A$ is invertible it is not zero when $x=0$, thus all but finitely many integers $n$ are such that $\det(A+nB)\neq 0$ so $A+nB$ is invertible.

share|improve this answer
I need explanation for $\det(A+xB)$ has finitely many zeros or is 0 for all $x$. – srijan May 10 '12 at 6:23
It's a polynomial (when $A$,$B$ are $3\times 3$ it has degree $3$). All nonzero polynomials have only finitely many zeroes (in fact, no more than their degree). – Alex Becker♦ May 10 '12 at 6:30

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.