Page 340 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Let p be the number of basis vectors, for which f( , ) = 1; we must show that the number p is
j j j
+
independent of the particular basis we have, satisfying the stated conditions. Let V be the
subspace of V spanned by the basis vectors for which f( , ) = 1, and let V’ be the subspace
j j j
+
spanned by the basis vectors for which f( , ) = –1. Now p = dim V , so it is the uniqueness of
j j j
the dimension of V which we must demonstrate. It is easy to see that if is a non-zero vector in
+
+
V then f( , ) > 0; in other words, f is positive definite on the subspace V . Similarly, if is a non-
+
–
zero vector in V , then f( , ) < 0, i.e., f is negative definite on the subspace V . Now let V be the
–
subspace spanned by the basis vectors for which f( , ) = 0. If is in V , then f( , ) = 0 for all
j j j
in V.
Since { , ..., } is a basis for V, we have
1 n
–
+
V = V V V .
Furthermore, we claim that if W is any subspace of V on which f is positive definite, then the
subspaces W, V , and V are independent. For, suppose is in W, is in V , is in V , and +
–
–
+ = 0. Then
0 = f( , + + ) = f( , ) + f( , ) + f( , )
0 = f( , + + ) = f( , ) + f( , ) + f( , )
Since is in V , f( , ) = f( , ) = 0; and since f is symmetric, we obtain
0 = f( , ) + f( , )
0 = f( , ) + f( , )
hence f( , ) = f( , ). Since f( , ) 0 and f( , ) 0, it follows that
f( , ) = f( , ) = 0
–
But f is positive definite on W and negative definite on V . We conclude that = = 0, and hence
that = 0 as well.
Since
V = V V V
–
+
and W, V , V are independent, we see that dim W dim V . That is, if W is any subspace of V on
+
–
+
which f is positive definite, the dimension of W cannot exceed the dimension of V . If is
1
another ordered basis for V which satisfies the conditions of the theorem, we shall have
+
–
+
+
corresponding subspaces V , V , and V and, the argument above shows that dim V dim V .
1
1
1
1
+
+
Reversing the argument, we obtain dim V dim V , and consequently
1
+
+
dim V = dim V .
1
There are several comments we should make about the basis ( , ..., } of Theorem 5 and the
1 n
associated subspaces V , V , and V First, note that V is exactly the subspace of vectors which are
+
–
'orthogonal' to all of V. We noted above that V is contained in this subspace; but,
–
+
dim V = dim V – (dim V + dim V ) = dim V – rank f
so every vector a such that f( , ) = 0 for all must be in V . Thus, the subspace V is unique. The
subspaces V and V are not unique; however, their dimensions are unique. The proof of Theorem
–
+
+
5 shows us that dim V is the largest possible dimension of any subspace on which f is positive
–
definite. Similarly, dim V is the largest dimension of any subspace on which f is negative
definite.
Of course
–
dim V + dim V = rank f.
+
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