Determinant of Transpose

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Theorem

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

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


Then:

$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$


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 $\map \det {\sqbrk a_n} = \map \det {\sqbrk b_n}$.

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

\(\displaystyle \map \det {\sqbrk b_n}\) \(=\) \(\displaystyle \sum_\lambda \map {\sgn} \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_\lambda \map {\sgn} \lambda a_{\map \lambda 1 1} a_{\map \lambda 2 2} \cdots a_{\map \lambda n n}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \det {\sqbrk a_n}\) $\quad$ $\quad$

$\blacksquare$


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.