Diagonalization of a quadratic form
can be transformed to
under appropriate change of coΓΆrdinates where
Proof
Without loss of generality it can be assumed that
. For if some π‘ 0 0 β 0 then we can permute coΓΆrdinates. If π‘ π π β 0 for all π‘ π π = 0 , then we may assume π by the same token. Let π‘ 0 1 β 0 , π 0 = π 0 , and otherwise π 1 = π π 0 + π 1 for π π = π π . Then π > 1 where πΉ ( π ) = β π π , π = 1 π π π π π π π . Hence we can choose π 0 0 = π‘ 0 1 π + π‘ 1 0 π whence π = π‘ β 1 0 1 . π 0 0 = 2 β 0 Under this assumption, it follows
πΉ ( π ) = π‘ 0 0 π 2 0 + π 0 π β π = 1 π‘ 0 π π π + π 0 π β π = 1 π‘ π 0 π π + π β π , π = 1 π‘ π π π π π π = π‘ β 1 0 0 ( π β π = 0 π‘ π 0 π π ) ( π β π = 0 π‘ 0 π π π ) + π β π , π = 1 π‘ β² π π π π π π for some
. Let π‘ β² π π and otherwise π 0 = π 0 β π‘ β 1 0 0 β π π = 1 π‘ 1 π π π for π π = π π . Then π > 0 πΉ ( π ) = π‘ 0 0 π 2 0 + π β π , π = 1 π‘ β² π π π π π π = π‘ 0 0 π 2 0 + πΉ β² ( π ) One can then repeat the same steps for
&c. until one has a quadratic form πΉ β² πΉ ( π ) = π β π = 0 π π π 2 π where
iff the corresponding quadric is singular. π < π
It follows that A quadric is singular iff its matrix is singular away from 2.
Footnotes
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2020. Finite geometries, ΒΆ4.25, p. 91 β©