Determinant of Transpose

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Theorem

Let $\mathbf A = \left[{a}\right]_{n}$ be a square matrix of order $n$.

Let $\det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.


Then:

$\det \left({\mathbf A}\right) = \det \left({\mathbf A^\intercal}\right)$


Proof

Let $\mathbf A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$.

Then $\mathbf A^\intercal = \begin{bmatrix} a_{11} & a_{21} & \ldots & a_{n1} \\ a_{12} & a_{22} & \cdots & a_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{nn} \\ \end{bmatrix}$.


Let $b_{r s} = a_{s r}$ for $1 \le r, s \le n$.

We need to show that $\det \left({\left[{a}\right]_n}\right) = \det \left({\left[{b}\right]_n}\right)$.

By the definition of determinant and Permutation of Determinant Indices, we have:

\(\displaystyle \det \left({\left[{b}\right]_n}\right)\) \(=\) \(\displaystyle \sum_\lambda \operatorname{sgn} \left({\lambda}\right) b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{\lambda \left({1}\right) 1} a_{\lambda \left({2}\right) 2} \cdots a_{\lambda \left({n}\right) n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \det \left({\left[{a}\right]_n}\right)\) $\quad$ $\quad$

$\blacksquare$


Comment

Thus there is symmetry between rows and columns of a determinant.

So, if we can prove something about a determinant's rows, it follows that this also applies to the determinant's columns, and vice versa.