Raising the bar

Determinant of a transpose

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

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

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

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

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

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

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

$latex =\det (P_{j_1,...,j_n})$      (because $latex P_{j_1,...,j_n}$ 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. $latex \det A=0$ if two columns of $latex A$ are linearly dependent.

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

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