Mathematical Tools for Physics

12—Tensors 357

Now obviously the function defined byA~.~v, whereA~is a fixed vector, is a linear functional. The burden
of this theorem is that all linear functionals are of precisely this form.
Proof: Case I: It’s possible thatf(~v)equals zero for all~v. This is the trivial case whereA~= 0.
Case II:f is not identically zero. Even here however there will generally be some vectors around where
f(~v) = 0. Denote byMthe set of all vectors wheref(~v) = 0. Mis itself a vector space. This follows because
the functionfis linear.I.e., if~v 1 and~v 2 are inM [f(~v 1 ) =f(~v 2 ) = 0], then

f(α~v 1 +β~v 2 ) =αf(~v 1 ) +βf(~v 2 ) = 0.

Denote byM⊥the set of vectors that are perpendicular to every vector inM. A picture of what’s happening is
this:

M

M⊥

or maybe

M

M⊥

In three dimensionsM could form a plane (two dimensional) and thenM⊥is the one dimensional line
perpendicular to this plane. Alternatively,M could be one dimensional and thenM⊥is the plane perpendicular
to this line. (In fact, the first case will turn out to be correct.) Not only isM a vector space, butM⊥is too,
because if both~v 1 and~v 2 are perpendicular to some given vector (inM) then so isα~v 1 +β~v 2.
Since by assumptionf(~v)is not zero for all vectors, there is at least one vector inM⊥. Call it~ω 6 = 0.
Consequently all scalar multiples of~ωare inMtoo;f(c~ω) =cf(~ω)

Let A~=

~ωf(~ω)

|~ω|^2

. (6)

Claim: This does the job as suggested in the statement of the theorem: f(~v) =A~.~vfor all~v. To show
this, first demonstrate thatM⊥ is one dimensional. Let~ω 1 and~ω 2 be any two vectors in M⊥ (that is,f is
non-zero for each). Consider
f(α~ω 1 +β~ω 2 ) =αf(~ω 1 ) +βf(~ω 2 ).