17
Feb 19

Determinant of a product

Determinant of a product

The derivation is very similar to the proof of the Leibniz rule.

Consider matrices A,B and let B_j denote the jth row of B. Row C_i of the product C=AB, obviously, can be written as C_i=\sum_{j=1}^na_{ij}B_j. Using Property V n times, we have

(1) \det C=\det \left(\begin{array}{c}\sum_{j_1=1}^na_{1j_1}B_{j_1}\\C_2\\...\\C_n\end{array}\right) =\sum_{j_1=1}^na_{1j_1}\det \left(\begin{array}{c}B_{j_1}\\C_2\\...\\C_n\end{array}\right)

=\sum_{j_1=1}^na_{1j_1}\det \left(  \begin{array}{c}B_{j_1}\\\sum_{j_2=1}^na_{2j_2}B_{j_2}\\...\\C_n\end{array}\right)=\sum_{j_1,j_2=1}^na_{1j_1}a_{2j_2}\det\left(\begin{array}{c}B_{j_1}\\B_{j_2}\\...\\  C_{n}\end{array}\right)

=...=\sum_{j_1,...,j_n=1}^na_{1j_1}...a_{nj_n}\det\left(\begin{array}{c}  B_{j_1}\\B_{j_2}\\...\\B_{j_n}\end{array}\right).

By analogy with permutation matrices denote

B_{j_1,...,j_n}=\left(\begin{array}{c}B_{j_1}\\B_{j_2}\\...\\B_{j_n}\end{array}\right).

Recall the link between B_{j_1,...,j_n} and B:

(2) B_{j_1,...,j_n}=P_{j_1,...,j_n}B.

If we had proved the multiplication rule for (2), we would have

(3) \det B_{j_1,...,j_n}=(\det P_{j_1,...,j_n})(\det B).

In the theory of permutations (3) is proved without relying on the multiplication rule. I am going to use (3) as a shortcut that explains the idea. Combining (1) and (3) we obtain by the Leibniz formula

\det C=\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})(\det B)=(\det A)(\det B).

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