Mathematical Tools for Physics

12—Tensors 358

Let β=

−αf(~ω 1 )

f(~ω 2 )

.

With this choice then,f(α~ω 1 +β~ω 2 ) = 0, meaning that the combination is inM. But if a vector is inM, and
at the same time it perpendicular to every vector inM, then it must be the zero vector.

α~ω 1 +β~ω 2 = 0,

or,~ω 2 is a constant times~ω 1. This is just the statement thatM⊥is one dimensional.
Any vector inM⊥is then of the formα~ω, and

A~.(α~ω)=(α~ω).~ωf(~ω)

|~ω|^2

=

α~ω.~ωf(~ω)

|~ω|^2

=αf(~ω) =f(α~ω),

so the theorem is true for these vectors.
Let~v be any vector in M. By the definition ofM, f(~v) = 0. Also ~ω is perpendicular to M, so

~ω.~v= 0 =A~.~v, and you have agreement here too.
Finally,anyvector can be written as the unique sum of a vector inMand one inM⊥. This is just a matter
finding the projection of~v 1 alongM⊥and subtracting it from~vto get the vector inM.

~v=~v 1 +~v 2

f(~v) =f(~v 1 +~v 2 ) =f(~v 1 ) +f(~v 2 )

=A~.v 1 + 0

=A~.v. 2

v

v

v

M^

1

For an easier proof, see problem 3.

Multilinear Functionals
Functionals can be generalized to more than one variable. A bilinear functional is a scalar valued function of two
vector variables, linear in each

T(~v 1 , ~v 2 ) =a scalar

T(α~v 1 +β~v 2 , ~v 3 ) =αT(~v 1 , ~v 3 ) +βT(~v 2 , ~v 3 )

T(~v 1 , α~v 2 +β~v 3 ) =αT(~v 1 , ~v 2 ) +βT(~v 1 , ~v 3 ).

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