Hi, I need to prove this statement, but my teacher wants me to so without using and I kind of stuck because using Cramer's rule to prove that is almost like using the .
" therefore
prove that: if is nonsingular in , is also nonsingular in "
Hi, I need to prove this statement, but my teacher wants me to so without using and I kind of stuck because using Cramer's rule to prove that is almost like using the .
" therefore
prove that: if is nonsingular in , is also nonsingular in "
I assume that $M_n(\mathbb{R})$ is the set of all $n\times n$ matrices having real entries. If that is correct then $M_n(\mathbb{Q}) \subset M_n(\mathbb{R})$. Any nonsingular square matrix has non-zero determinate. Can you explain what you are and are not allowed to do for a proof.
One way to do this is to transform the problem to vector spaces. Here's the relevant theorem:
Let V be a vector space of finite dimension n over a field F and T a linear transformation from V to V. Then $T^{-1}$ exists if and only if Ker(T)=0. If $T^{-1}$ exists, $T^{-1}$ is a linear transformation.
You can use this theorem to prove your statement (without determinants at all). If you don't see how to do this or need help on the theorem, post again.