Page 339 - DMTH502_LINEAR_ALGEBRA

Unit 30: Bilinear Forms and Symmetric Bilinear Forms

Corollary: Let F be a subfield of the complex numbers, and let A be a symmetric n n matrix Notes
t
over F. Then there is an invertible n n matrix P over F such that P AP is diagonal.
In case F is the field of real numbers, the invertible matrix P in this corollary can be chosen to be
t
–1
an orthogonal matrix, i.e., P = P . In other words, if A is a real symmetric n n matrix, there is
a real orthogonal matrix P such that P AP is diagonal.
t
Theorem 4: Let V be a finite-dimensional vector space over the field of complex numbers. Let f be
a symmetric bilinear form on V which has rank r. Then there is an ordered basis = { , ..., } for
1 n
V such that
(i) the matrix of f in the ordered basis is diagonal
1, j 1, ..., r
(ii) f( , ) =
j j 0, j . r
Proof: By Theorem 3, there is an ordered basis ( , ..., ) for V such that
1 n
f( , ) = 0 for i j.
i j
Since f has rank r, so does its matrix is the ordered basis { , ... , }. Thus we must have
1 n
f( , ) 0 for precisely r values of j. By reordering the vectors , we may assume that
j j j
f( , ) 0, j = 1, ..., r.
j j
Now we use the fact that the scalar field is the field of complex numbers. If ( f , j j ) denotes
any complex square root of f( , ), and if we put
j j
1
j , j 1, ..., r
= ( f , j j )
j
j , j r
the basis { , ..., } satisfies conditions (i) and (ii).
1 n
Of course, Theorem 4 is valid if the scalar field is any subfield of the complex numbers in which
each element has a square root. It is not valid, for example, when the scalar field is the field of
real numbers. Over the field of real numbers, we have the following substitute for Theorem 4.
Theorem 5: Let V an n-dimensional vector space over the field of real numbers, and let f be a
symmetric bilinear form on V which has rank r. Then there is an ordered basis { , , ..., } for
1 2 n
V in which the matrix of f is diagonal and such that
f( , ) = 1, j = 1, ... r.
j j
Furthermore, the number of basis vectors for f( , ) = 1 is independent of the choice of basis.
j j j
Proof: There is a basis { , ..., } for V such that
1 n
f( , ) = 0, i j
i j
f( , ) 0, 1 j r
j j
f( , ) = 0, j > r.
j j
Let
= |f( , )| –1/2 , 1 j r
j j j j
= , j > r.
j j
Then ( , ..., ) is a basis with the stated properties.
1 n

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