Relationship between determinants of these matrices

Question

I have two basis for $\mathbb{R}^n$, namely $F_1=\{\mathbf{e}_1,...,\mathbf{e}_n\}$ and $F_2=\{\mathbf{b}_1,...,\mathbf{b}_n\}$ and I know that:

$$\mathbf{e}_i=M\cdot \mathbf{b}_i,\quad i=1,...,n$$

where $M$ is an invertible given $n\times n$ matrix. I would find the transition matrix from one frame to the other, that is the matrix:

$$\left\{\begin{matrix} \mathbf{b}_1=a_{11}\mathbf{e}_1+...+a_{1n}\mathbf{e}_n \\ ...\\ \mathbf{b}_n=a_{n1}\mathbf{e}_1+...+a_{nn}\mathbf{e}_n \end{matrix}\right.\iff A=\left(\begin{matrix}a_{11} & ...&a_{1n}\\ ...&...&...\\ a_{n1} & ...&a_{nn}\end{matrix}\right)$$

and understand if there is a relationship between the sign of the determinant of $A$ and the sign of the determinant of $M$.

All I managed to write is:

$\left\{\begin{matrix} \mathbf{b}_1=\left(\frac{\mathbf{b}_1\cdot \mathbf{e}_1}{|\mathbf{b}_1||\mathbf{e}_1|}\right)\mathbf{e}_1+...+\left(\frac{\mathbf{b}_1\cdot \mathbf{e}_n}{|\mathbf{b}_1||\mathbf{e}_n|}\right)\mathbf{e}_n \\ ...\\ \mathbf{b}_n=\left(\frac{\mathbf{b}_n\cdot \mathbf{e}_1}{|\mathbf{b}_n||\mathbf{e}_1|}\right)\mathbf{e}_1+...+\left(\frac{\mathbf{b}_n\cdot \mathbf{e}_n}{|\mathbf{b}_n||\mathbf{e}_n|}\right)\mathbf{e}_n \end{matrix}\right.$

$\left\{\begin{matrix} \mathbf{b}_1=\left(\frac{\mathbf{b}_1\cdot M\mathbf{b}_1}{|\mathbf{b}_1||M\mathbf{b}_1|}\right)\mathbf{e}_1+...+\left(\frac{\mathbf{b}_1\cdot M\mathbf{b}_n}{|\mathbf{b}_1||M\mathbf{b}_n|}\right)\mathbf{e}_n \\ ...\\ \mathbf{b}_n=\left(\frac{\mathbf{b}_n\cdot M\mathbf{b}_1}{|\mathbf{b}_n||M\mathbf{b}_1|}\right)\mathbf{e}_1+...+\left(\frac{\mathbf{b}_n\cdot M\mathbf{b}_n}{|\mathbf{b}_n||M\mathbf{b}_n|}\right)\mathbf{e}_n \end{matrix}\right.$

and so:

$A=\left(\begin{matrix}\left(\frac{\mathbf{b}_1\cdot M\mathbf{b}_1}{|\mathbf{b}_1||M\mathbf{b}_1|}\right) & ...&\left(\frac{\mathbf{b}_1\cdot M\mathbf{b}_n}{|\mathbf{b}_1||M\mathbf{b}_n|}\right)\\ ...&...&...\\ \left(\frac{\mathbf{b}_n\cdot M\mathbf{b}_1}{|\mathbf{b}_n||M\mathbf{b}_1|}\right) & ...&\left(\frac{\mathbf{b}_n\cdot M\mathbf{b}_n}{|\mathbf{b}_n||M\mathbf{b}_n|}\right)\end{matrix}\right)$

At this point I see no efficient way to calculate the determinant of A and relate it to that of M. Did I take a wrong way?

Answer 1

Working in the basis $F_1$:

The columns of $A$ are the vectors $\mathbf b_1, \dots, \mathbf b_n$.

The columns of the product $MA$ are therefore the vectors $M\mathbf b_1, \dots, M\mathbf b_n$, and we're given then that these are $\mathbf e_1, \dots, \mathbf e_n$. In other words, $MA = I$.

Therefore $\det(M) \det(A) = \det(I) = 1$, which in particular means that $\det(M)$ and $\det(A)$ must have the same sign.