Determinant/Transposed matrix/Universal property/Fact/Proof

Proof

If ${}M$ is not invertible, then, due to fact, the determinant is ${}0$ and the rank is smaller than ${}n$. This does also hold for the transposed matrix, so that its determinant is again ${}0$. So suppose that ${}M$ is invertible. We reduce the statement in this case to the corresponding statement for the elementary matrices, which can be verified directly, see exercise. Because of fact, there exist elementary matrices ${}E_{1},\ldots ,E_{s}$ such that

${}D=E_{s}\cdots E_{1}M\,$

is a diagonal matrix. Due to exercise, we have

${}{D^{\text{tr}}}={M^{\text{tr}}}{E_{1}^{\text{tr}}}\cdots {E_{s}^{\text{tr}}}\,$

and

${}{M^{\text{tr}}}={D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\,.$

The diagonal matrix ${}D$ is not changed under transposing it. Since the determinants of the elementary matrices are also not changed under transposition, we get, using fact,

${}{\begin{aligned}\det {M^{\text{tr}}}&=\det {\left({D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\right)}\\&=\det {D^{\text{tr}}}\cdot \det({E_{s}^{\text{tr}}})^{-1}\cdots \det({E_{1}^{\text{tr}}})^{-1}\\&=\det D\cdot \det {\left(E_{s}^{-1}\right)}\cdots \det {\left(E_{1}^{-1}\right)}\\&=\det {\left(E_{1}^{-1}\right)}\cdots \det {\left(E_{s}^{-1}\right)}\cdot \det D\cdot \\&=\det {\left(E_{1}^{-1}\cdots E_{s}^{-1}\cdot D\right)}\\&=\det M.\end{aligned}}$

To fact