Raising the bar

Multilinearity in columns

The axioms and properties of determinants established so far are asymmetric in that they are stated in terms of rows, while similar statements hold for columns. Among the most important properties is the fact that the determinant of $latex A$ is a multilinear function of its rows.  Here we intend to establish multilinearity in columns.

Exercise 1. The determinant of $latex A$ is a multilinear function of its columns.

Proof. To prove linearity in each column, it suffices to prove additivity and homogeneity. Let $latex A,B,C$ be three matrices that have the same columns, except for the $latex l$th column. The relationship between the $latex l$th columns is specified by $latex A^{(l)}=B^{(l)}+C^{(l)}$ or in terms of elements $latex a_{il}=b_{il}+c_{il}$ for all $latex i=1,...,n.$ Alternatively, in Leibniz' formula

(1) $latex \det A=\sum_{j_{1},...,j_{n}:\det P_{j_{1},...,j_{n}}\neq
0}a_{1j_{1}}...a_{nj_{n}}\det P_{j_{1},...,j_{n}}$

for all $latex i$ such that $latex j_i=l$ we have

(2) $latex a_{ij_i}=b_{ij_i}+c_{ij_i}.$

If $latex \det P_{j_1,...,j_n}\neq 0,$ for a given set $latex j_1,...,j_n$ there exists only one $latex i$ such that $latex j_i=l.$ Therefore by (2) we can continue (1) as

$latex \det A=\sum_{j_{1},...,j_{n}:\det P_{j_{1},...,j_{n}}\neq
0}a_{1j_{1}}...(b_{ij_{i}}+c_{ij_{i}})a_{nj_{n}}\det
P_{j_{1},...,j_{n}}=\det B+\det C.$

Homogeneity is proved similarly, using equations $latex A^{(l)}=kB^{(l)}$ or in terms of elements $latex a_{il}=kb_{il}$ for all $latex i.$

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