Part 2 · Tensors

2.4 Trace and the scalar product of tensors

Source: Favata’s notes, appendix A.2, p. 13–14 (trace, scalar product, norm, three properties, Exercise 1.1)

Objectives

Define the trace without a basis, and compute it as the sum of the diagonal.
Use the scalar product of tensors, and see it as a sum over any orthonormal frame.
Understand why $Sym$ and $Skw$ are orthogonal, and what that buys in mechanics.
Recipe: compute $tr A$ , $A \cdot B$ , $∣ A ∣$ in components.
Picture: $A \cdot B = \sum_{i} A q_{i} \cdot B q_{i}$ , the same for every frame ${q_{i}}$ .
Theorem: $Lin = Sym \oplus Skw$ , an orthogonal sum.
Vocabulary: trace, scalar product of tensors, norm of a tensor.

Trace

Definition 2.4.1Trace

The trace is the linear map $tr : Lin \to R$ fixed on dyads by $tr (u \otimes v) := u \cdot v \forall u, v \in V .$

Every tensor is a combination of dyads, so linearity fixes the trace of all of them: $tr A = tr (A_{ij} e_{i} \otimes e_{j}) = A_{ij} tr (e_{i} \otimes e_{j}) = A_{ij} δ_{ij} = A_{ii} .$

Key observation 2.4.2The trace is the same in every basis

$tr A = A_{11} + A_{22} + A_{33}$ , the sum of the diagonal of the matrix. Since the definition does not mention a basis, that sum is the same in every orthonormal basis, although the diagonal entries themselves change.

Properties, each a line of indices:

$tr I = δ_{ii} = 3$ ;
$tr A^{⊤} = A_{ii} = tr A$ ;
$tr (AB) = A_{ik} B_{ki} = B_{ki} A_{ik} = tr (BA)$ , and so the trace of a product is unchanged by cyclic shifts: $tr (ABC) = tr (BCA) = tr (CAB)$ ;
$tr W = 0$ for skew $W$ , since $tr W = tr W^{⊤} = - tr W$ .

Warning

Cyclic, not arbitrary: $tr (ABC) \neq = tr (ACB)$ in general.

Scalar product and norm

Definition 2.4.3Scalar product and norm of tensors

$A \cdot B := tr (A^{⊤} B), ∣ A ∣ := A \cdot A .$

The notes use the same dot for vectors and tensors (some books write $A : B$ ). In components, $A \cdot B = (A^{⊤} B)_{jj} = (A^{⊤})_{ji} B_{ij} = A_{ij} B_{ij}, ∣ A ∣ = A_{ij} A_{ij} = \sum_{i, j} A_{ij}^{2} :$ multiply corresponding entries and add, exactly as for vectors with nine components. It extends the dyad rule $a \otimes b \cdot c \otimes d = (a \cdot c) (b \cdot d)$ (Exercise 2.4.1), and $A \cdot B = A^{⊤} \cdot B^{⊤}$ .

It is a genuine scalar product: symmetric ( $A_{ij} B_{ij} = B_{ij} A_{ij}$ ), linear in each argument, and positive definite ( $A \cdot A = \sum A_{ij}^{2} > 0$ unless $A = 0$ ). Handy identities: $A \cdot I = tr A$ , $∣ I ∣ = 3$ , $A \cdot B = tr (A B^{⊤})$ .

What it measures

The component formula hides the geometry. Take any orthonormal basis ${q_{i}}$ ; then $I = q_{i} \otimes q_{i}$ , and

Theorem 2.4.4The scalar product as a sum over a framebeyond the notes

For every orthonormal basis ${q_{i}}$ , $A \cdot B = tr (A^{⊤} B) = i \sum q_{i} \cdot A^{⊤} B q_{i} = i \sum A q_{i} \cdot B q_{i} .$

Map an orthonormal frame by both tensors and add up the scalar products of matching images. Each term depends on the frame; the sum does not.

$∣ A ∣^{2} = \sum_{i} ∣ A q_{i} ∣^{2}$ : the squared lengths of the images of any orthonormal frame.
$tr A = A \cdot I = \sum_{i} A q_{i} \cdot q_{i}$ : how far each image points back along its own frame vector.

Loading the demo…

Figure 2.4.1 Drag the tip of

q_{1}

q_{2}

to turn the frame (the third follows); drag the background to turn the view. The three terms change as the frame turns, their sum does not. Hover over or tap a term to see it as a projection. In the controls, the other cases: the norm, the trace, and a symmetric against a skew tensor.

Three properties

Theorem 2.4.5Three properties of the scalar product

$A \cdot B = 0$ for all $B \in Lin$ $⟹$ $A = 0$ .
$A \in Sym$ , $W \in Skw$ $⟹$ $A \cdot W = 0$ .
$A \cdot B = 0$ for all $B \in Sym$ $⟹$ $A \in Skw$ .

Proof

Choose $B = A$ : $∣ A ∣^{2} = 0$ .
Transpose both: $A \cdot W = A^{⊤} \cdot W^{⊤} = A \cdot (- W) = - A \cdot W$ , so $A \cdot W = 0$ .
For any $B \in Lin$ , $B + B^{⊤} \in Sym$ . So for all $B$ , $0 = A \cdot (B + B^{⊤}) = A \cdot B + A^{⊤} \cdot B = (A + A^{⊤}) \cdot B,$ and by 1, $A + A^{⊤} = 0$ . (The middle step is $A \cdot B^{⊤} = A^{⊤} \cdot B$ , the rule $A \cdot B = A^{⊤} \cdot B^{⊤}$ applied to $B^{⊤}$ .)

By 2, the split of Theorem 2.3.2 is orthogonal: $Lin = Sym \oplus Skw .$

A symmetric tensor only sees the symmetric part of what it is paired with: $S \cdot B = S \cdot sym B$ for $S \in Sym$ . In mechanics, the stress is symmetric, so the power it spends on a velocity gradient sees only the stretching, never the spin.

More consequences:

$A \cdot B = sym A \cdot sym B + skw A \cdot skw B$ : the cross terms vanish by 2.
$∣ A ∣^{2} = ∣ sym A ∣^{2} + ∣ skw A ∣^{2}$ (Pythagoras).
$u \cdot Au = u \cdot (sym A) u$ , since $u \cdot Au = A \cdot u \otimes u$ and $u \otimes u \in Sym$ .

Exercises

Exercise 2.4.1notes p. 14 (Exercise 1.1)

Prove

$A \cdot BC = B^{⊤} A \cdot C = A C^{⊤} \cdot B$ ;
$u \cdot Av = A \cdot u \otimes v$ ;
$a \otimes b \cdot c \otimes d = (a \cdot c) (b \cdot d)$ .

Solution

$A \cdot BC = A_{ij} B_{ik} C_{kj} = (B^{⊤})_{ki} A_{ij} C_{kj} = B^{⊤} A \cdot C$ , and $= A_{ij} (C^{⊤})_{jk} B_{ik} = A C^{⊤} \cdot B$ .
$u \cdot Av = u_{i} A_{ij} v_{j} = A_{ij} (u_{i} v_{j}) = A \cdot u \otimes v$ .
$a \otimes b \cdot c \otimes d = a_{i} b_{j} c_{i} d_{j} = (a_{i} c_{i}) (b_{j} d_{j}) = (a \cdot c) (b \cdot d)$ .

Exercise 2.4.2beyond the notespractice (A.1)

For the $A$ of Exercise 2.3.1, find $tr A$ and check $∣ A ∣^{2} = ∣ sym A ∣^{2} + ∣ skw A ∣^{2}$ .

Solution

$tr A = 1 + 3 + 5 = 9 = tr sym A$ (the skew part has no trace). $∣ A ∣^{2} = 1 + 4 + 9 + 16 + 4 + 25 = 59$ ; $∣ sym A ∣^{2} = 47$ ; $∣ skw A ∣^{2} = 12$ ; $47 + 12 = 59$ .

Exercise 2.4.3beyond the notespractice (A.2)

Show that $u \cdot Au = u \cdot (sym A) u$ for all $u$ , and deduce $u \cdot Wu = 0$ for skew $W$ .

Solution

By Exercise 1.1.2, $u \cdot Au = A \cdot u \otimes u$ . Since $(u \otimes u)^{⊤} = u \otimes u$ , it is symmetric, so by property 2 the skew part of $A$ contributes nothing. For skew $W$ , $sym W = 0$ .

Exercise 2.4.4beyond the notespractice (A.4)

Show that $tr (SW) = 0$ for $S \in Sym$ , $W \in Skw$ .

Solution

$tr (SW) = tr (S^{⊤} W) = S \cdot W = 0$ by property 2.