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[V, §8]
SYLVESTER'S THEOREM
139
2. Let V be a finite dimensional space over R, and let ( , ) be a scalar product
on V. Show that V admits a direct sum decomposition
where Vo is defined as in Theorem 6.1, and where the product is positive defi
nite on V + and negative definite on V -. (This means that
(v v) > 0
(v, v) < 0
for all v E V+, v#-o
for all v E V-, v #- 0.)
Show that the dimensions of the spaces V +, V - are the same in all such de
com positions.
3. Let V be the vector space over R of 2 x 2 real symmetric matrices.
(a) Given a symmetric matrix
show that (x, y, z) are the coordinates of A with respect to some basis of
the vector space of all 2 x 2 symmetric matrices. Which basis?
(b) Let
f(A) = xz - yy = xz - y2.
If we view (x, y, z) as the coordinates of A then we see that f is a quad
ratic form on V. Note that f(A) is the determinant of A, which could be
defined here ad hoc in a simple way.
Let W be the subspace of V consisting of all A such that tr(A) = O.
Show that for A E Wand A#-O we have f(A) < o. This means that the
quadratic form is negative definite on W. CHAPTER VI
Determ i nants
We have worked with vectors for some time, and we have often felt the
need of a method to determine when vectors are linearly independent.
Up to now, the only method available to us was to solve a system of
linear equations by the elimination method. In this chapter, we shall
exhibit a very efficient computational method to solve linear equations,
and determine when vectors are linearly independent.
The cases of 2 x 2 and 3 x 3 determinants will be carried out sepa
rately in full, because the general case of n x n determinants involves no
tation which adds to the difficulties of understanding determinants. In a
first reading, we suggest omitting the proofs in the general case.
VI, §1. DETERMINANTS OF ORDER 2
Before stating the general properties of an arbitrary determinant, we shall
consider a special case.
Let
be a 2 x 2 matrix in a field K. We define its determinant to be
ad - be. Thus the determinant is an element of K. We denote it by
a b = ad - be.
e d [VI, §1]
DETERMINANTS OF ORDER 2
141
For example, the determinant of the matrix
lS equal to 2·4 - 1 . 1 = 7. The determinant of
is equal to (- 2)·5 - ( - 3)·4 = -10 + 12 = 2.
The determinant can be viewed as a function of the matrix A. It can
also be viewed as a function of its two columns. Let these be A 1 and A 2
as usual. Then we write the determinant as
D(A),
Det(A),
or
The following properties are easily verified by direct computation,
which you should carry out completely.
As a function of the column vectors, the determinant is linear.
This means: let b', d' be two numbers. Then
Furthermore, if t is a number, then
Det(a tb)
= t Det(a
b)
c
td cd·
The analogous properties also hold with respect to the first column.
We give the proof for the additivity with respect to the second column
to show how easy it is. Namely, we have
a(d + d') - c(b + b') = ad + ad' - cb - cb'
= ad - bc + ad' - b' c,
which is precisely the desired additivity. Thus in the terminology of
Chapter V, §4 we may say that the determinant is bilinear.
If the two columns are equal, then the determinant is equal to O. 142
DETERMINANTS
[VI, §1]
If A is the unit matrix,
then Det(A) = 1.
The determinant also satisfies the following additional properties.
If one adds a multiple of one column to the other, then the value of the
determinant does not change.
In other words, let t be a number. The determinant of the matrix
(
a + tb b)
c + td d
is the same as D(A), and similarly when we add a multiple of the first
column to the' second.
If the two columns are interchanged, then the determinant changes by a
sign.
In other words, we have
The determinant of A is equal to the determinant of its transpose, i.e.
D(A) = D(~).
Explicitly, we have
The vectors (:) and (~) are linearly dependent if and only if the deter
minant ad - bc is equal to O.
We give a direct proof for this property. Assume that there exists
numbers x, y not both 0 such that
xa + yb = 0,
xc + yd = O. [VI, §2]
EXISTENCE OF DETERMINANTS
143
Say x =1= O. Multiply the first equation by d, multiply the second by b,
and subtract. We obtain
xad - xbe = 0,
whence x(ad - be) = O. It follows that ad - be = O. Conversely, assume
that ad - be = 0, and assume that not both vectors (a, e) and (b, d) are
the zero vectors (otherwise, they are obviously linearly dependent). Say
a =1= O. Let y = - a and x = b. Then we see at once that
xa + yb = 0,
xe + yd = 0,
so that (a, e) and (b, d) are linearly dependent, thus provIng our asser
tion.
VI, §2. EXISTENCE OF DETERMINANTS
We shall define determinants by induction, and give a formula for com
puting them at the same time. We first deal with the 3 x 3 case.
We have already defined 2 x 2 determinants. Let
al2 al3 )
a22 a23
a32 a33
be a 3 x 3 matrix. We define its determinant according to the formula
known as the expansion by a row, say the first row. That is, we define
(*)
Det(A) = all
a22 a23
a
2l
a23
+ al 3 a
2l
a22
- al2
a32
a33
a
3l a
33
a
3l a
32
all
al2 al 3
a2l a22 a23
a3l a32 a33
We may describe this sum as follows. Let Aij be the matrix obtained
from A by deleting the i-th row and the j-th column. Then the sum ex
pressing Det(A) can be written 144
DETERMINANTS
[VI, §2]
In other words, each term consists of the product of an element of the
first row and the determinant of the 2 x 2 matrix obtained by deleting
the first row and the j-th column, and putting the appropriate sign to
this term as shown.
Example 1. Let
A = (
~ ! ~).
-3 2 5
Then
and our formula for the determinant of A yields
1 4 1 4 1 1
Det( A) = 2 2 5 - 1 _ 3 5 + 0 - 3 2
= 2(5 - 8) - 1(5 + 12) + 0
= -23.
The determinant of a 3 x 3 matrix can be written as
We use this last expression if we wish to consider the determinant as a
function of the columns of A.
Later we shall define the determinant of an n x n matrix, and we use
the same notation
IAI = D(A) = Det(A) = D(Al, ... ,An).
Already in the 3 x 3 case we can prove the properties expressed in the
next theorem, which we state, however, in the general case.
Theorem 2.1. The determinant satisfies the following properties:
1. As a function of each column vector, the determinant is linear, i.e. if
the j-th column Ai is equal to a sum of two column vectors, say
Ai = C + C', then
D(A 1, ... ,C + C', ... ,An)
= D( A 1, ... ,C, ... ,A n) + D( A 1 , ... ,C', ... ,A n). [VI, §2]
EXISTENCE OF DETERMINANTS
145
Furthermore, if t is a number, then
D(A 1, ... ,tAi, ... ,An) = tD(A 1, ... ,Ai, ... ,An).
2. If two adjacent columns are equal, i.e. if Ai = Ai+ 1
for some
j = 1, ... ,n - 1, then the determinant D( A) is equal to O.
3. If I is the unit matrix, then D(I) = 1.
Proof (in the 3 x 3 case). The proof is by direct computations. Sup
pose say that the first column is a sum of two columns:
Al = B + C,
Substituting in each term of (*), we see that each term splits into a sum
of two terms corresponding to Band C. For instance,
a22
a23
= bl
a
22
a
23
+ C1
a
22
a
23
all a32
a33
a32
a33
a32
a33
b
2
+
c
2
a
23
b2
a
23
+ a12
c
2
a
23
a 12
h3
+
c 3
= a12
h3
a33
a33
a33
c 3
and similarly for the third term. The proof with respect to the other
column is analogous. Furthermore, if t is a number, then
because each 2 x 2 determinant is linear in the first column, and we can
take t outside each one of the second and third terms. Again the proof
is similar with respect to the other columns. A direct substitution shows
that if two adjacent columns are equal, then formula (*) yields 0 for the
determinant. Finally, one sees at once that if A is the unit matrix, then
Det(A) = 1. Thus the three properties are verified.
In the above proof, we see that the properties of 2 x 2 determinants
are used to prove the properties of 3 x 3 determinants. 146
DETERMINANTS
[VI, §2]
Furthermore, there is no particular reason why we selected the expan
sion according to the first row. We can also use the second row, and
write a similar sum, namely:
Again, each term is the product of a2j times the determinant of the 2 x 2
matrix obtained by deleting the second row and j-th column, and putting
the appropriate sign in front of each term. This sign is determined ac
cording to the pattern:
One can see directly that the determinant can be expanded according to
any row by multiplying out all the terms, and expanding the 2 x 2 deter
minants, thus obtaining the determinant as an alternating sum of six
terms:
Furthermore, we can also expand according to columns following the
same principle. For instance, expanding out according to the first
column:
yields precisely the same six terms as in (**).
The reader should now look at least at the general expression given
for the expansion according to a row or column in Theorem 2.4, inter
preting i, j to be 1, 2, or 3 for the 3 x 3 case.
Since the determinant of a 3 x 3 matrix is linear as a function of its
columns, we may say that it is trilinear; just as a 2 x 2 determinant IS
bilinear. In the n x n case, we would say n-linear, or multilinear.
In the case of 3 x 3 determinants, we have the following result.
Theorem 2.2. The determinant satisfies the rule for expansion according
to rows and columns, and Det(A) = Det(~). In other words, the deter
minant of a matrix is equal to the determinant of its transpose. [VI, §2]
EXISTENCE OF DETERMINANTS
147
This last assertion follows because taking the transpose of a matrix
changes rows into columns and vice versa.
Example 2. Compute the determinant
301
125
-1 4 2
by expanding according to the second column.
The determinant is equal to
2
3
-1
1 3
-4
2 1
1
5
=2(6-(-1»)-4(15-1)= -42.
Note that the presence of a 0 in the second column eliminates one term
in the expansion, since this term would be O.
We can also compute the above determinant by expanding according
to the third column, namely the determinant is equal to
The n x n case
Let
1
+ 1 -1
23030
4 - 5 -1 4 + 2 1 2 = -42.
F: K n x ... x K n ~ K
be a function of n variables, where each variable ranges over Kn. We say
that F is multilinear if F satisfies the first property listed in Theorem 2.1,
that is
F(Al, ... ,C + C', ... ,An) = F(Al, ... ,C, ... ,An) + F(Al, ... ,C', ... ,An),
F(A 1, ... ,tC, ... ,An) = tF(A 1, ... ,C, ... ,An).
This means that if we consider some index j, and fix Ak for k i= j, then
the function Xi H
F(A 1, ... ,X i, ... ,An) is linear in the j-th variable.
We say that F is alternating if whenever Ai = Ai+ 1 for some j we
have
F(A 1, ... ,Ai,Ai, ... ,An) = O.
This is the second property of determinants.
One fundamental theorem of this chapter can be formulated as fol
lows. 148
DETERMINANTS
[VI, §2]
Theorem 2.3. There exists a multilinear alternating function
F:Knx···xKn~K
such that F(J) = 1. Such a function is uniquely determined by these
three properties.
The uniqueness proof will be postponed to Theorem 7.2. We have al
ready proved existence in case n = 2 and n = 3. We shall now prove the
existence in general.
The general case of n x n determinants is done by induction. Suppose
that we have been able to define determinants for (n - 1) x (n - 1)
matrices. Let i, j be a pair of integers between 1 and n. If we cross out
the i-th row and j-th column in the n x n matrix A, we obtain an
(n - 1) x (n - 1) matrix, which we denote by Aij• It looks like this:
j
all
aij
anl
We give an expression for the determinant of an n x n matrix in terms
of determinants of (n - 1) x (n - 1) matrices. Let i be an integer,
1 < i < n. We define
Each Aij is an (n - 1) x (n - 1) matrix.
This sum can be described in words. For each element of the i-th
row, we have a contribution of one term in the sum. This term is equal
to + or - the product of this element, times the determinant of the
matrix obtained from A by deleting the i-th row and the corresponding
column. The sign + or - is determined according to the chess-board
pattern:
+
+
...
)
...
+
+
+
+ [VI, §2]
EXISTENCE OF DETERMINANTS
149
This sum is called the expansion of the determinant according to the i-th
row. We shall prove that this function D satisfies properties 1, 2, and 3.
Note that D(A) is a sum of the terms
L (-IY+jaij Det(Aij)
as j ranges from 1 to n.
1. Consider D as a function of the k-th column, and consider any
term
If j i= k, then aij does not depend on the k-th column, and Det(Aij)
depends linearly on the k-th column. If j = k, then aij depends linearly
on the k-th column, and Det(Aij) does not depend on the k-th column.
In any case, our term depends linearly on the k-th column. Since D(A)
is a sum of such terms, it depends linearly on the k-th column, and
property 1 follows.
2. Suppose two adjacent columns of A are equal, namely Ak = Ak+ 1.
Let j be an index i= k or k + 1. Then the matrix Aij has two adjacent
equal columns, and hence its determinant is equal to O. Thus the term
corresponding to an index j i= k or k + 1 gives a zero contribution to
D(A). The other two terms can be written
(-1) i+k aik Det(Aik) + (-1) i+k+1 ai,k+1 Det(Ai,k+ 1).
The two matrices Aik and Ai,k+ 1 are equal because of our assumption
that the k-th column of A is equal to the (k + 1)-th column. Similarly,
aik = ai, k + 1· Hence these two terms cancel since they occur with opposite
signs. This proves property 2.
3. Let A be the unit matrix. Then aij = 0 unless i = j, in which case
aii = 1. Each Aij is the unit (n - 1) x (n - 1) matrix. The only term in
the sum which gives a non-zero contribution is
which is equal to 1. This proves property 3.
Example 3. We wish to compute the determinant
121
-1 3 1.
015