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B) P T = P , thus not perpendicular projection. (c) Perpendicular projection. (d) Perpendicular projection. (e) P T = P but P P = P , thus not perpendicular projection. 11 For contradiction assume that there are two tensors A and B such that Su · v = u · Av = u · Bv, ∀u, v. Then u · (A − B)v = 0 for all u, v and hence (A − B)v = 0 for all v, which implies A = B. Thus there can be only one tensor S T such that Su · v = u · S T v, ∀u, v, which proves uniqueness. For the second result notice that [S]Tij = [S]j i .

13) yields Aik Alk = δil . 4. Let [v] = (v1 , v2 , v3 )T be its representation in {ei }, let [v] = (v1 , v2 , v3 )T be its representation in {ei }, and let A be the change of basis tensor from {ei } to {ei }. 3 Second-Order Tensors 21 vector they are related through Aij . To see this, notice that by definition of components we have v = vi ei = vj ej , and by definition of Aij we have v = vi ei = vj ej = vj Aij ei , which implies vi = Aij vj . Similarly, vk = Aik vi . In matrix notation we write this as [v] = [A][v] and [v] = [A]T [v].

Analogous to the zero vector, we define a zero tensor O with the property Ov = 0 for all v ∈ V, and we define an identity tensor I with the property Iv = v for all v ∈ V. Two second-order tensors S and T are said to be equal if and only if Sv = T v for all v ∈ V. 12 Tensor Algebra Examples: (1) Projections. The action of projecting a vector v onto a given unit vector e can be expressed in terms of a second-order tensor P , namely P (v) = (v · e)e. v P(v) e2 e o e3 e1 (2) Rotations. The action of rotating a vector v through a given angle θ about a fixed axis, say e3 , can be expressed in terms of a second-order tensor S.

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