Part 2 · Tensors

2.2 Transpose and composition

Source: Favata’s notes, appendix A.2, p. 12–13

Objectives

Understand the transpose as the tensor that moves $A$ across a scalar product.
Understand the product of tensors as doing one map after the other.
Recipe: compute $(AB)_{ij}$ and transposes of products and dyads.
Pictures: $Au \cdot v = u \cdot A^{⊤} v$ for every $u, v$ ; a composition in three panels.
Vocabulary: transpose, composition (product).

Transpose

Definition 2.2.1Transpose

$A^{⊤}$ is the unique tensor with $Au \cdot v = u \cdot A^{⊤} v \forall u, v \in V .$

It moves a tensor from one side of a scalar product to the other. The definition never mentions a basis; the matrix comes after.

Uniquenessbeyond the notes

If $B$ and $B^{'}$ both qualify, then $u \cdot (B - B^{'}) v = 0$ for all $u$ . Take $u = (B - B^{'}) v$ : then $∣ (B - B^{'}) v ∣^{2} = 0$ for all $v$ , so $B = B^{'}$ .

Theorem 2.2.2Transpose in components

$(A^{⊤})_{ij} = A_{ji}$ : the matrix of $A^{⊤}$ is the transposed matrix. And $(a \otimes b)^{⊤} = b \otimes a$ .

Proofbeyond the notes

$(A^{⊤})_{ij} = A^{⊤} e_{j} \cdot e_{i} = e_{j} \cdot A e_{i} = A_{ji}$ . For the dyad, $(a \otimes b) u \cdot v = (b \cdot u) (a \cdot v) = u \cdot (a \cdot v) b = u \cdot (b \otimes a) v$ .

Also $I^{⊤} = I$ .

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Figure 2.2.1 Drag

u

on the left and

v

on the right. The left panel also shows

A^{⊤} v

, the right one

Au

. Wherever you put them,

Au \cdot v

and

u \cdot A^{⊤} v

agree: the transpose is defined by exactly that.

Warning

$A^{⊤} v$ is not “ $Av$ turned around”. In the demo the two dashed and solid arrows have nothing to do with each other individually; only the two scalar products match.

Composition

Definition 2.2.3Composition (product)

$(AB) v := A (Bv) =: ABv \forall A, B \in Lin, v \in V .$ First apply $B$ , then $A$ : a product is read right to left.

Theorem 2.2.4Product in components

$(AB)_{ij} = A_{ik} B_{kj} = A_{i 1} B_{1 j} + A_{i 2} B_{2 j} + A_{i 3} B_{3 j} .$ The free indices $i, j$ match on both sides; the dummy $k$ is summed. This is the row-times-column product of matrices.

Proofbeyond the notes

$(AB)_{ij} = A (B e_{j}) \cdot e_{i} = A (B_{kj} e_{k}) \cdot e_{i} = B_{kj} A e_{k} \cdot e_{i} = A_{ik} B_{kj}$ .

The column picture explains it: column $j$ of $[AB]$ is $A$ applied to column $j$ of $[B]$ .

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Figure 2.2.2 A simple shear

A

, then a rotation by

9 0^{\circ}

B

. The middle panel is

Ax

, the right one

B (Ax) = (BA) x

. The readout multiplies the matrices; compare its columns with the arrows on the right.

Non-example 2.2.5Products do not commutebeyond the notes

$(e_{1} \otimes e_{2}) (e_{2} \otimes e_{1}) = e_{1} \otimes e_{1}$ , but $(e_{2} \otimes e_{1}) (e_{1} \otimes e_{2}) = e_{2} \otimes e_{2}$ . Shear then rotate is not rotate then shear: swap the two in your head in the demo above and the letter ends up leaning the other way.

Identities

Theorem 2.2.6Rules for transposes and products

For all $A, B \in Lin$ and $a, b, c, d \in V$ :

$(a \otimes b) A = a \otimes A^{⊤} b$ ;
$(A^{⊤})^{⊤} = A$ ;
$(AB)^{⊤} = B^{⊤} A^{⊤}$ (the order reverses).

Proof (the notes call these easy to verify)beyond the notes

$(a \otimes b) Av = (b \cdot Av) a = (A^{⊤} b \cdot v) a = (a \otimes A^{⊤} b) v$ .
$A^{⊤} u \cdot v = v \cdot A^{⊤} u = Av \cdot u = u \cdot Av$ , so $A$ does the job that defines $(A^{⊤})^{⊤}$ .
$ABu \cdot v = Bu \cdot A^{⊤} v = u \cdot B^{⊤} A^{⊤} v$ .

Proposition 2.2.7More dyad rulesbeyond the notes

$A (a \otimes b) = Aa \otimes b$ , since $A (b \cdot v) a = (b \cdot v) Aa$ .
$(a \otimes b) (c \otimes d) = (b \cdot c) a \otimes d$ , since $(a \otimes b) (d \cdot v) c = (d \cdot v) (b \cdot c) a$ .
So $(e_{i} \otimes e_{j}) (e_{h} \otimes e_{k}) = δ_{jh} e_{i} \otimes e_{k}$ .

The first rule, with $I = e_{i} \otimes e_{i}$ , gives a useful trick: $A = AI = A e_{i} \otimes e_{i}$ , a tensor as the sum of its columns paired with the basis. It is how the deformation gradient gets its Cartesian form (§3.2 The deformation gradient).

Exercises

Exercise 2.2.1beyond the notes

Show that $A^{⊤} A$ and $A A^{⊤}$ are symmetric for every $A$ .

Solution

$(A^{⊤} A)^{⊤} = A^{⊤} (A^{⊤})^{⊤} = A^{⊤} A$ by rules 3 and 2; the same for $A A^{⊤}$ . This is why the right Cauchy–Green tensor $F^{⊤} F$ is symmetric (§3.3 Change in length and angle).

Exercise 2.2.2beyond the notes

Compute $[AB]$ and $[BA]$ for $[A] = [1011]$ , $[B] = [01 - 1 0]$ .

Solution

$[AB] = [11 - 1 0]$ , $[BA] = [01 - 1 1]$ . Different: the order matters.

Exercise 2.2.3beyond the notes

Show $u \cdot Av = A \cdot u \otimes v$ , with the tensor–dyad scalar product of Definition 2.1.5.

Solution

By definition $A \cdot u \otimes v = Av \cdot u$ . In components both are $u_{i} A_{ij} v_{j}$ .