Let A and B be n*n matrices such that trace(A)<0<trace(B).

Question

Let A and B $n\times n$ such that trace(A)$\lt0\lt$trace(B).

Then, $f(t)=1-det(e^{tA+(1-t)B})$ has

1) infinitely many zeros in $0\lt t\lt1$

2) at least one zero in $\Bbb R$

3) no zeros

4) either no zeros or infinitely many zeros in $\Bbb R$.

According to my knowledge determinant is defined for square matrices not for exponential functions. So, I think something is wrong in the question. If this question is correct then how can I solve it? If it is not correct then try to correct the question and give a hint to solve it.

Thank you

Answer 1

The exponent of a matrix is defined by a convergent infinite sum.

Use $\det e^A=e^{\text{Tr }A}$.

http://en.wikipedia.org/wiki/Matrix_exponential