Determinants
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
1 / 25
Notation
The determinant of a square matrixn×n
A is denoted det(A) or |A|.
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
2 / 25
Definition of Determinant
The determinant ofn×n
A = [aij ] is defined as
n!
n
X
Y
m(pk )
|n×n
A|=
(−1)
aipk (i) ,
k=1
i=1
where p1 , . . . , pn! are the n! distinct bijections (one-to-one and onto
functions) from {1, . . . , n} to {1, . . . , n} and
m(pk ) =
n−1 X
n
X
11[pk (i) > pk (i∗ )].
i=1 i∗ =i+1
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
3 / 25
Example: n = 2
Suppose
"
A=
a11 a12
#
.
a21 a22
Then n! = 2,
p1 (1) = 1 p1 (2) = 2
m(p1 ) = 0
p2 (1) = 2 p2 (2) = 1 m(p2 ) = 1,
and
|A| =
2
2
X
Y
(−1)m(pk )
aipk (i)
k=1
i=1
= a11 a22 − a12 a21 .
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
4 / 25
Example: n = 3
Find an expression for the determinant of


a11 a12 a13


.
A=
a
a
a
21
22
23


a31 a32 a33
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
5 / 25
Example: n = 3
n! = 6,
p1 (1) = 1 p1 (2) = 2 p1 (3) = 3 m(p1 ) = 0
p2 (1) = 1 p2 (2) = 3 p2 (3) = 2 m(p2 ) = 1
p3 (1) = 2 p3 (2) = 1 p3 (3) = 3 m(p3 ) = 1
p4 (1) = 2 p4 (2) = 3 p4 (3) = 1 m(p4 ) = 2
, and
p5 (1) = 3 p5 (2) = 1 p5 (3) = 2 m(p5 ) = 2
p6 (1) = 3 p6 (2) = 2 p6 (3) = 1 m(p6 ) = 3
|A| = a11 a22 a33 − a11 a23 a32 − a12 a21 a33
+a12 a23 a31 + a13 a21 a32 − a13 a22 a31 .
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
6 / 25
Example: n = 1
Suppose A = [a11 ]. Then n! = 1,
p1 (1) = 1 m(p1 ) = 0,
and
|A| =
1
X
(−1)m(pk )
k=1
Copyright c 2012 Dan Nettleton (Iowa State University)
1
Y
aipk (i) = a11 .
i=1
Statistics 611
7 / 25
Verbal Description of Determinant Computation
1
Form all possible products of n matrix elements that contain
exactly one element from each row and exactly one element from
each column.
2
For each product, count the number of pairs of elements for which
the “lower” element is to the “left” of the “higher” element. (For
example, a31 is “lower” than a12 because a31 is in the third row
while a12 is in the first row. Also, a31 is to the “left” of a12 because
a31 is in the first column while a12 is in the second column.) If the
number of such pairs is odd, multiply the product by −1. If the
number of pairs is even, multiply the product by 1.
3
Add the signed products determined in steps 1 and 2.
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
8 / 25
Prove that if A is an n × n matrix and c ∈ R, then
Copyright c 2012 Dan Nettleton (Iowa State University)
|cA| = cn |A|.
Statistics 611
9 / 25
Proof:
|cA| =
=
n!
X
k=1
n!
X
(−1)m(pk )
n
Y
(caipk (i) )
i=1
m(pk ) n
(−1)
c
k=1
= c
n
n!
X
k=1
(−1)m(pk )
n
Y
aipk (i)
i=1
n
Y
aipk (i)
i=1
= cn |A|.
2
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
10 / 25
Prove that ifn×n
A is lower or upper triangular, then
|A| =
n
Y
aii .
i=1
(Note that this result implies |A| =
Copyright c 2012 Dan Nettleton (Iowa State University)
Qn
i=1 aii
for a diagonal matrix.)
Statistics 611
11 / 25
Proof for A Lower Triangular:
Suppose A is lower triangular. Then aij = 0 whenever i < j.
Thus,
Qn
i=1 aipk (i)
= 0 if ∃ i 3 pk (i) > i.
Because pk is a bijection,
pk (i) ≤ i ∀ i = 1, . . . , n ⇒ pk (i) = i ∀ i = 1, . . . , n.
Therefore,
|A| =
n!
n
n
X
Y
Y
(−1)m(pk )
aipk (i) =
aii .
k=1
Copyright c 2012 Dan Nettleton (Iowa State University)
i=1
i=1
2
Statistics 611
12 / 25
Suppose A is an n × n matrix. Prove that |A| = |A0 |.
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
13 / 25
Proof:
Let qk = p−1
k ; i.e., qk (j) = i ⇐⇒ pk (i) = j.
Note that
i < i∗ , pk (i) = j > pk (i∗ ) = j∗ ⇐⇒ j∗ < j, qk (j∗ ) = i∗ > qk (j) = i.
Thus, m(pk ) = m(qk ).
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
14 / 25
Proof:
It follows that
n!
n
X
Y
m(pk )
|A| =
(−1)
aipk (i)
=
=
k=1
n!
X
i=1
n
Y
k=1
j=1
(−1)m(pk )
n!
X
(−1)
k=1
0
m(qk )
n
Y
aqk (j)j
aqk (j)j
j=1
= |A |.
2
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
15 / 25
Find an expression for


I
n×n
0
n×m
B C
m×n m×m
Copyright c 2012 Dan Nettleton (Iowa State University)

 =
I
n×n
0
n×m
B C
m×n m×m
.
Statistics 611
16 / 25
Copyright c 2012 Dan Nettleton (Iowa State University)
I
n×n
0
n×m
B C
m×n m×m
= |C|
Statistics 611
17 / 25
Likewise
B
0
n×n
n×m
C
m×n
I
m×m
B
C
n×n
n×m
0
m×n
I
m×m
= |B|,
= |B|,
and
I
n×n
B
n×m
0 C
m×n m×m
Copyright c 2012 Dan Nettleton (Iowa State University)
= |C|.
Statistics 611
18 / 25
More generally, it can be shown that
B
n×n
C
m×n
0
n×m
D
= |B||D|
m×m
and
B
n×n
C
n×m
0 D
m×n m×m
Copyright c 2012 Dan Nettleton (Iowa State University)
= |B||D|
Statistics 611
19 / 25
Result A.18 (b):
If A and B are each n × n matrices, then
|AB| = |A||B|.
(We will not go through a proof here. Most graduate-level linear
algebra textbooks contain a proof of this result.)
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
20 / 25
Use Result A.18 (b) to prove
A nonsingular ⇐⇒ |A| =
6 0.
n×n
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
21 / 25
Proof (=⇒):
|A||A−1 | = |AA−1 | = |I| = 1 ⇒ |A| =
6 0.
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
22 / 25
Proof (⇐=):
Suppose A is singular. We will show |A| = 0.
A singular ⇒ ∃ b 6= 0 3 Ab = 0.
Letn×n
B be a nonsingular matrix whose first column is b. (We know such
a matrix exists by Fact V5.)
Then |A||B| = |AB| = 0 because the first column of AB is 0.
2
Dividing both sides by |B| =
6 0 yields |A| = 0.
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
23 / 25
Determinant of a Partitioned Matrix
If A is nonsingular,
A B
C D
= |A||D − CA−1 B|.
If D is nonsingular,
A B
C D
Copyright c 2012 Dan Nettleton (Iowa State University)
= |D||A − BD−1 C|.
Statistics 611
24 / 25
Proof:
A B
C D
= |I|
= |A|
= |A|
A B
C D
A B
C D
I
CA
−1
= |AA−1 |
A B
C D
|A−1 | = |A|
0
D − CA
−1
B
A B
C D
= |A||A−1 |
A B
C D
A−1 −A−1 B
0
I
= |A||D − CA−1 B|.
2
A similar argument proves the second result.
Copyright c 2012 Dan Nettleton (Iowa State University)
Statistics 611
25 / 25