18
Feb 19

## Determinant of a transpose

Exercise 1. $\det (A^T)=\det A.$

Proof. The proof is similar to the derivation of the Leibniz formula. Using the notation from that derivation, we decompose rows of $A$ into linear combinations of unit row-vectors $A_i=\sum_{j=1}^na_{ij}e_j$. Hence $A_i^T=\sum_{j=1}^na_{ij}e_j^T.$ Therefore by multilinearity in columns

(1) $\det (A^{T})=\sum_{j_{1},...,j_{n}=1}^{n}a_{1j_{1}}...a_{nj_{n}}\det \left( e_{j_{1}}^{T},...,e_{j_{n}}^{T}\right)$

$=\sum_{j_{1},...,j_{n}:\det P_{j_{1},...,j_{n}}^{T}\neq 0}a_{1j_{1}}...a_{nj_{n}}\det P_{j_{1},...,j_{n}}^{T}.$

Now we want to relate $\det P_{j_1,...,j_n}^T$ to $\det P_{j_1,...,j_n}$.

(2) $\det P_{j_1,...,j_n}^T=\det (P_{j_1,...,j_n}^{-1})$ (the transpose of an orthogonal matrix equals its inverse)

$=\frac{1}{\det (P_{j_1,...,j_n})}$          (the determinant of the inverse is the inverse of the determinant)

$=\det (P_{j_1,...,j_n})$      (because $P_{j_1,...,j_n}$ is orthogonal and its determinant is either 1 or -1 ).

(1) and (2) prove the statement.

Apart from being interesting in its own right, Exercise 1 allows one to translate properties in terms of rows to properties in terms of columns, as in the next corollary.

Corollary 1. $\det A=0$ if two columns of $A$ are linearly dependent.

Indeed, columns of $A$ are rows of its transpose, so Exercise 1 and Property III yield the result.