Page 242 - Linear Algebra Done Right

Determinant of a Matrix
...
...
a n ],
det A = det[ a 1
a k
th
column a k . This
and we can think of det A as a function of the k
function, which takes a k to the determinant above, is a linear map 233
from the vector space of n-by-1 matrices with entries in F to F. The
linearity follows easily from 10.25, where each term in the sum contains
precisely one entry from the k th column of A.
Now we are ready to prove one of the key properties about determi-
nants of square matrices. This property will enable us to connect the
determinant of an operator with the determinant of its matrix. Note
that this proof is considerably more complicated than the proof of the
corresponding result about the trace (see 10.9).
10.31 Theorem: If A and B are square matrices of the same size, This theorem was first
then proved in 1812 by the
det(AB) = det(BA) = (det A)(det B). French mathematicians
Jacques Binet and
Proof: Let A = [ a 1 ... a n ], where each a k is an n-by-1 column Augustin-Louis Cauchy.
of A. Let
 
b 1,1 ... b 1,n
 . . 
B =  . . . .  = [ b 1 ... b n ],
 
b n,1 ... b n,n
where each b k is an n-by-1 column of B. Let e k denote the n-by-1 matrix
that equals 1 in the k th row and 0 elsewhere. Note that Ae k = a k and
n
Be k = b k . Furthermore, b k = m=1 b m,k e m .
First we will prove that det(AB) = (det A)(det B). A moment’s
thought about the definition of matrix multiplication shows that AB =
[ Ab 1 ... Ab n ]. Thus

det(AB) = det[ Ab 1 ... Ab n ]
n n
= det[ A( m 1 =1 b m 1 ,1 e m 1 ) . . . A( m n =1 b m n ,n e m n ) ]
n n
= det[ m 1 =1 b m 1 ,1 Ae m 1 ... m n =1 b m n ,n Ae m n ]
n n

= ··· b m 1 ,1 ...b m n ,n det[ Ae m 1 ... Ae m n ],
m 1 =1 m n =1

where the last equality comes from repeated applications of the linear-
ity of det as a function of one column at a time. In the last sum above,