Page 241 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
232
We need to introduce notation that will allow us to represent a ma-
trix in terms of its columns. If A is an n-by-n matrix
 
a 1,1 ... a 1,n
 . . 
A =  . . . .  ,
 
a n,1 ... a n,n
then we can think of the k th column of A as an n-by-1 matrix
 
a 1,k
 . 
a k =  . .  .
 
a n,k
We will write A in the form
[ a 1 ... a n ],
with the understanding that a k denotes the k th column of A. With this
notation, note that a j,k , with two subscripts, denotes an entry of A,
whereas a k , with one subscript, denotes a column of A.
The next lemma shows that a permutation of the columns of a matrix
changes the determinant by a factor of the sign of the permutation.

Some texts define the 10.30 Lemma: Suppose A = [ a 1 ... a n ] is an n-by-n matrix.
determinant to be the If (m 1 ,...,m n ) is a permutation, then
function defined on the
... ] = sign(m 1 ,...,m n ) det A.
square matrices that is det[ a m 1 a m n
linear as a function of
each column separately Proof: Suppose (m 1 ,...,m n ) ∈ perm n. We can transform the
... ] into A through a series of steps. In each
and that satisfies 10.30 matrix [ a m 1 a m n
and det I = 1. To prove step, we interchange two columns and hence multiply the determinant
that such a function by −1 (see 10.28). The number of steps needed equals the number
exists and that it is of steps needed to transform the permutation (m 1 ,...,m n ) into the
unique takes a permutation (1,...,n) by interchanging two entries in each step. The
nontrivial amount of proof is completed by noting that the number of such steps is even if
work. (m 1 ,...,m n ) has sign 1, odd if (m 1 ,...,m n ) has sign −1 (this follows
from 10.23, along with the observation that the permutation (1,...,n)
has sign 1).

Let A = [ a 1 ... a n ]. For 1 ≤ k ≤ n, think of all columns of A
except the k th column as fixed. We have