Part 2 · Tensors

2.1 Tensors as linear maps

Source: Favata’s notes, appendix A.2, p. 11–12 (second-order tensors, tensor product, components); projections are added here

Primary goal

A second-order tensor is a machine that takes a vector and returns a vector, linearly. This part asks three questions of it, the same three that kinematics will ask of the deformation gradient: where does it send a vector (this section and the next), what does it do to lengths, areas and volumes (§2.4 Trace and the scalar product of tensors, §2.5 Inverse and determinant, §2.6 Orthogonal tensors), and what is it made of (§2.3 Symmetric and skew parts, §2.7 Eigenvalues and the spectral theorem).

Objectives

Think of a tensor as a map first, and of its matrix as one way of writing it down.
Build tensors from dyads, and read a dyad as a projection followed by a stretch.
Recipe: write the matrix of a tensor by asking where it sends each basis vector.
Pictures: a tensor applied to a whole figure; a dyad flattening the plane onto a line.
Vocabulary: $Lin$ , zero tensor, identity, components, projection.
Essential vocabulary: second-order tensor, dyad (tensor product).

Second-order tensors

Definition 2.1.1Second-order tensorEssential

A second-order tensor is a linear map $A : V \to V$ : $A (α u + β v) = α Au + β Av \forall u, v \in V, \forall α, β \in R .$ The set of all of them is $Lin$ .

Tensors are bold uppercase. Linearity is a strong constraint: a linear map sends lines to lines, the origin to itself, parallel lines to parallel lines, and the midpoint of a segment to the midpoint of its image. So it is fixed by what it does to a few vectors.

Tensors are added and scaled pointwise: for all $A, B \in Lin$ , $v \in V$ , $c \in R$ , $(A + B) v = Av + Bv, (c A) v = c (Av) =: c Av .$ With these, $Lin$ is itself a vector space, of dimension 9 (below). Two tensors have names: the zero tensor, $0v = 0$ , and the identity, $Iv = v$ .

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Figure 2.1.1 A tensor of the plane at work. Drag

x

on the left;

Ax

moves on the right. The whole grid, the unit square and the letter are carried along: straight lines stay straight, parallel ones stay parallel. The dashed path shows

Ax

built from the columns. Try the other tensors in the controls.

Columns: where the basis goes

Write $v = v_{j} e_{j}$ . By linearity, $Av = v_{1} A e_{1} + v_{2} A e_{2} + v_{3} A e_{3} .$ So $A$ is known as soon as $A e_{1}, A e_{2}, A e_{3}$ are known, and $Av$ is the combination of them with the components of $v$ as coefficients.

Recipe 2.1.2The matrix of a tensor

For each $j$ , find $A e_{j}$ , the image of the $j$ -th basis vector.
Write its components as column $j$ of a $3 \times 3$ array: $A_{ij} = A e_{j} \cdot e_{i}$ .
Then $(Av)_{i} = A_{ij} v_{j}$ : the usual matrix–vector product.

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Figure 2.1.2 Drag the tips of

A e_{1}

and

A e_{2}

on the right: you are writing the matrix one column at a time, and the readout updates. The columns snap to the half-grid. Line them up (they snap onto each other’s line) and see the square collapse.

Example 2.1.3beyond the notes

The rotation by $9 0^{\circ}$ of the plane sends $e_{1} \mapsto e_{2}$ and $e_{2} \mapsto - e_{1}$ , so its columns are $(0, 1)$ and $(- 1, 0)$ : $[R] = [01 - 1 0] .$ No computation of angles was needed: only the images of the basis.

Warning

A tensor is the map itself, not a matrix. A matrix only represents it once a basis is fixed, and it changes when the basis does (§2.8 Change of basis). Transpose, trace and determinant will be defined without a basis; that is what makes statements about them hold in every frame.

The tensor product

The simplest tensors are built from two vectors.

Definition 2.1.4Dyad (tensor product)Essential

The dyadic (tensor) product of $a, b$ is the tensor $(a \otimes b) v := (b \cdot v) a \forall v \in V .$

Read it right to left: $b$ reads the input (how much of $v$ lies along $b$ ), and $a$ is the output direction. Every vector lands on the line of $a$ .

It is linear, since $(a \otimes b) (α u + β v) = (b \cdot (α u + β v)) a = α (a \otimes b) u + β (a \otimes b) v$ . So any combination of dyads is a tensor too.
$\otimes$ is linear in each factor, but not symmetric: $a \otimes b \neq = b \otimes a$ in general.
Every vector orthogonal to $b$ is sent to $0$ . A dyad flattens space onto a line: it has rank one.

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Figure 2.1.3 The dyad

a \otimes b

with

a = (1, 2)

b = (1, 0.5)

. Whatever you do with

x

Ax

stays on one line, the range (both columns are multiples of

a

). Find the direction that is sent to

0

: the dashed red kernel, orthogonal to

b

A basis for Lin

For the dyad $e_{1} \otimes e_{2}$ :

$(e_{1} \otimes e_{2}) v = (e_{2} \cdot v) e_{1} = v_{2} e_{1}, that is [e_{1} \otimes e_{2}] v_{1} v_{2} v_{3} = v_{2} 00 \forall v,$

so, since $v$ is arbitrary,

$[e_{1} \otimes e_{2}] = 000100000 .$

In general $[e_{i} \otimes e_{j}]$ has a single $1$ , in row $i$ , column $j$ . So the nine dyads build every $3 \times 3$ matrix: $[M] = M_{ij} [e_{i} \otimes e_{j}]$ .

To show they are an orthonormal basis, give dyads a scalar product:

Definition 2.1.5Scalar product of dyads

$a \otimes b \cdot c \otimes d := (a \cdot c) (b \cdot d),$ extended to combinations of dyads by bilinearity. For $V \in Lin$ and a dyad, $V \cdot a \otimes b := Vb \cdot a$ .

The two rules agree: for $V = c \otimes d$ , $(c \otimes d) b \cdot a = (d \cdot b) (c \cdot a)$ . (Notation: $\otimes$ binds tighter than $\cdot$ .) Then

$e_{i} \otimes e_{j} \cdot e_{h} \otimes e_{k} = (e_{i} \cdot e_{h}) (e_{j} \cdot e_{k}) = δ_{ih} δ_{jk} :$

each dyad has unit norm and is orthogonal to the other eight. Orthonormal vectors are independent, so the nine dyads are a basis and $dim Lin = 9$ .

Components of a tensor

Theorem 2.1.6Components of a tensor

Every $V \in Lin$ is $V = V_{ij} e_{i} \otimes e_{j}, V_{ij} := V \cdot e_{i} \otimes e_{j} = V e_{j} \cdot e_{i} .$ The $V_{ij}$ are its components in the dyad basis, and the entries of its matrix in ${e_{i}}$ .

Proofbeyond the notes

Write $V e_{j}$ in components: $V e_{j} = (V e_{j} \cdot e_{i}) e_{i} = V_{ij} e_{i}$ . Then for any $v = v_{j} e_{j}$ : $Vv = v_{j} V e_{j} = V_{ij} v_{j} e_{i} = V_{ij} (e_{j} \cdot v) e_{i} = (V_{ij} e_{i} \otimes e_{j}) v .$

Read off from this:

Column $j$ of $[V]$ holds the components of $V e_{j}$ (Recipe 2.1.2).
Action in components: $(Vv)_{i} = V_{ij} v_{j}$ .
Components of a dyad: $(a \otimes b)_{ij} = (a \otimes b) e_{j} \cdot e_{i} = a_{i} b_{j}$ . Its matrix is the column $a$ times the row $b$ ; every column is a multiple of $a$ .
Identity: $I = e_{i} \otimes e_{i}$ , $I_{ij} = δ_{ij}$ . Check: $(e_{i} \otimes e_{i}) v = (e_{i} \cdot v) e_{i} = v_{i} e_{i} = v$ .

Projections

The projections of Recipe 1.2.6 are dyads.

Proposition 2.1.7Projection tensorsbeyond the notes

For a versor $e$ :

$P := e \otimes e$ gives $Pv = (e \cdot v) e$ , the projection on the line of $e$ ;
$I - e \otimes e$ gives $v - (e \cdot v) e$ , the projection on the plane normal to $e$ .

Both are symmetric and idempotent, $PP = P$ : projecting twice changes nothing. Every $v$ splits as $v = (e \otimes e) v + (I - e \otimes e) v$ .

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Figure 2.1.4

P = e \otimes e

with

e

3 0^{\circ}

. Hunt for its eigen-directions by dragging

x

(it snaps when you find one, and the line appears): along

e

nothing changes (

λ = 1

), across it everything vanishes (

λ = 0

A first equation to solve

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Figure 2.1.5 Solve

Ax = b

by dragging

x

until

Ax

lands on the green ring. Think in columns: how many

A e_{1}

and how many

A e_{2}

make

b

? Questions like this, whether a solution exists and whether it is unique, are settled in §2.5 Inverse and determinant.

Exercises

Exercise 2.1.1beyond the notes

Write the matrix of the tensor of the plane that reflects across the line $x_{1} = x_{2}$ , from where it sends $e_{1}$ and $e_{2}$ . Then write it with dyads.

Solution

$e_{1} \mapsto e_{2}$ and $e_{2} \mapsto e_{1}$ , so $[A] = [0110]$ and $A = e_{2} \otimes e_{1} + e_{1} \otimes e_{2}$ .

Exercise 2.1.2beyond the notes

For $a = (1, 0, 2)$ , $b = (0, 1, 1)$ , write $[a \otimes b]$ , apply it to $v = (3, 1, 1)$ both ways, and find every $v$ with $(a \otimes b) v = 0$ .

Solution

$[a \otimes b] = 000102102$ . Matrix times $(3, 1, 1)$ : $(2, 0, 4)$ . From the definition, $(b \cdot v) a = 2 (1, 0, 2)$ , the same. The kernel is ${v : b \cdot v = 0} = {v_{2} + v_{3} = 0}$ , a plane.

Exercise 2.1.3beyond the notes

Show that $P = e \otimes e$ is idempotent using the matrix $[e \otimes e]_{ij} = e_{i} e_{j}$ .

Solution

$(PP)_{ij} = P_{ik} P_{kj} = e_{i} e_{k} e_{k} e_{j} = e_{i} e_{j} (e_{k} e_{k}) = e_{i} e_{j}$ , since $e_{k} e_{k} = ∣ e ∣^{2} = 1$ .