Mathematical Physics Vol 1

1.6 Gram-Schmidt orthogonalization procedure

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- step three: prove that it is true for k also. Let us assume the opposite, i.e. that p k is a zero vector ( p k = 0 ), then the vector u k can be represented in terms of vectors p 1 ,..., p k − 1 u k = proj ( u k , p 1 )+ ··· + proj ( u k , p k − 1 ) , and it follows that the vector u k belongs to a space whose basis is { p 1 ,..., p k − 1 } , and which is a sub-space of a space whose basis is { u 1 ,..., u k − 1 } , which is contradictory to the assumption that the set { u 1 ,..., u k } is linearly independent, which proves that the vector p k is not a zero-vector. Given that vectors p 1 ,..., p k − 1 belong to space wˇ hose basis is { u 1 ,..., u k − 1 } it follows that p k = u k − c 1 p 1 −···− c k − 1 p k − 1 . (1.76) From here, it follows that p k belong to a space whose basis is { u 1 ,..., u k − 1 } . Part two. It needs further to be proved that these vectors are mutually orthogonal, i.e. that p k ⊥ p n , for each n = 1 ,..., k − 1. Given that p k is not a zero-vector it can be represented in the form (see (1.76)) p k = u k − c 1 p 1 −···− c k − 1 p k − 1 and by scalar multiplication of the equation by vector p n the following expression is obtained p k · p n = u k · p n − c n ( p n · p n ) . Vector u k can be represented in the form u k = c 1 p 1 + ··· + c k p k , and, on basis of Theorem 2, it follows that u k · p n = c n p n · p n ⇒ c n = .

u k · p n p n · p n

Substituting c n from this equation into the previous one, we obtain

( p n · p n ) ( p n · p n )

p k · p n = u k · p n − u n · p k

=

= u k · p n − c n ( p n · p n )= = u k · p n − u k · p n p n · p n

( p n · p n )= 0 . Thus, the scalar product of these vectors is ( p k · p n )= 0, which means that these two vectors are mutually orthogonal ( p k ⊥ p n ) .

Definition The Basis of space R k is orthonormalized , when this basis is composed of a orthonor malized set of vectors.