Page 338 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes One important class of symmetric bilinear forms consists of the inner products on real vector
spaces discussed earlier. If V is a real vector space, an inner product on V is a symmetric bilinear
form f on V which satisfies
f( , ) > 0 if 0. ...(6)
A bilinear form satisfying (6) is called positive definite. Thus, an inner product on a real vector
space is a positive definite, symmetric bilinear form on that space. Note that an inner product is
non-degenerate. Two vectors , are called orthogonal with respect to the inner product f if
f( , ) = 0 . The quadratic form q( ) = f( , ) takes only non-negative values, and q( ) is usually
thought of as the square of the length of . Of course, these concepts of length and orthogonality
stem from the most important example of an inner product – the dot product.
If f is any symmetric bilinear form on a vector space V, it is convenient to apply some of the
terminology of inner products to f. It is especially convenient to say that and are orthogonal
with respect to f if f( , ) = 0. It is not advisable to think of f( , ) as the square of the length of ;
for example if V is a complex vector space, we may have f( , ) = 1 or on a real vector space,
f( , ) = –2.
Theorem 3: Let V be n finite-dimensional vector space over a field of characteristic zero, i.e. if n
is a positive integer the sum 1 + 1 + ... + 1 (n times) in F is not zero, and let f be a symmetric
bilinear form on V. Then there is an ordered basis for V in which f is represented by a diagonal
matrix.
Proof: What we must find is an ordered basis
= { , ..., }
1 n
such that f( , ) = 0 for i j. If f = 0 or n = 1, the theorem is obviously true. Thus we may suppose
i j
f 0 and n > 1. If f( , ) = 0 for every in V, the associated quadratic form q is identically 0, and
the polarization identity (5) shows that f = 0. Thus there is a vector in V such that f( , ) =
q( ) 0. Let W be the one-dimensional subspace of V which is spanned by , and let W be the set
of all vectors in V such that f( , ) = 0. Now we claim that V = W W . Certainly the subspaces
W and W are independent. A typical vector in W is c , where c is a scalar. If c is also in W , then
2
f(c , c ) = c f( , ) = 0. But f( , ) 0, thus c = 0. Also, each vector in V is the sum of a vector in W
and a vector in W . For, Let be any vector in V, and put
f ( , )
= – .
f ( , )
Then

f ( , )
( , ) = f( , ) – f( , )
f ( , )
and since f is symmetric, f( , ) = 0. Thus is in the subspace W . The expression
f ( , )
= +
f ( , )
shows us that V = W + W .
The restriction of f to W is a symmetric bilinear form on W . Since W has dimension (n – 1), we
may assume by induction that W has a basis { , ..., } such that
2 n
f( , ) = 0, i j (i, 2, j 2)
i j
Putting = , we obtain a basis { , ..., } for V such that f( , ) = 0 for i j.
1 1 n i j

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