Part 2 · Tensors

2.3 Symmetric and skew parts

Source: Favata’s notes, appendix A.2, p. 13 (Sym, Skw, (A3) axial vector)

Objectives

Split any tensor into a symmetric and a skew part, uniquely.
Understand a skew tensor in 3D as a cross product with a fixed vector: an infinitesimal rotation.
Recipe: find the axial vector of a skew tensor, and the skew tensor of a vector.
Picture: a skew tensor sends every vector to one orthogonal to it.
Vocabulary: symmetric, skew-symmetric, $Sym$ , $Skw$ , symmetric part, skew part, axial vector.

The split

Definition 2.3.1Symmetric and skew tensors

$A$ is symmetric if $A = A^{⊤}$ , skew-symmetric (antisymmetric) if $A = - A^{⊤}$ : $Sym := {A \in Lin : A = A^{⊤}}, Skw := {A \in Lin : A = - A^{⊤}} .$ The symmetric part and skew part of any $A$ are $sym A := \frac{1}{2} (A + A^{⊤}), skw A := \frac{1}{2} (A - A^{⊤}) .$

Theorem 2.3.2Every tensor splits uniquelybeyond the notes

$A = sym A + skw A$ , with $sym A \in Sym$ and $skw A \in Skw$ , and this is the only way to write $A$ as a symmetric plus a skew tensor.

Proofbeyond the notes

The sum is clear. Transpose to check the parts: $(sym A)^{⊤} = \frac{1}{2} (A^{⊤} + A) = sym A$ , and likewise for the skew part. Uniqueness: if $A = S + W = S^{'} + W^{'}$ , then $S - S^{'} = W^{'} - W$ is both symmetric and skew, hence equal to minus itself, hence $0$ .

In components: a symmetric matrix has 6 independent entries. A skew one has a zero diagonal ( $W_{ii} = - W_{ii}$ ) and 3 independent entries. $6 + 3 = 9 = dim Lin$ .

beyond the notes

In mechanics this split carries meaning. The symmetric part of the velocity gradient is the stretching, the skew part the spin; the same split of the displacement gradient gives the infinitesimal strain and the infinitesimal rotation.

Loading the demo…

Figure 2.3.1 A skew tensor

A

of the plane. Drag

x

anywhere:

Ax

is always at a right angle to it, so

Ax \cdot x = 0

. Switch to the general tensor in the controls: the readout shows its two parts.

Axial vector

In 3D a skew tensor is a cross product in disguise.

Theorem 2.3.3Axial vector (A3)

Every skew $W$ has an axial vector $w$ with $Wa = w \times a \forall a \in V .$

Recipe 2.3.4From W to w and backbeyond the notes

$[W] = 0 w_{3} - w_{2} - w_{3} 0 w_{1} w_{2} - w_{1} 0, w = (W_{32}, W_{13}, W_{21}) .$ In indices, $W_{ij} = - ϵ_{ijk} w_{k}$ and $w_{k} = - \frac{1}{2} ϵ_{ijk} W_{ij}$ .

Derivationbeyond the notes

A skew matrix and the components of $w \times a$ are $[W] = 0 - W_{12} - W_{13} W_{12} 0 - W_{23} W_{13} W_{23} 0, w \times a = w_{2} a_{3} - w_{3} a_{2} w_{3} a_{1} - w_{1} a_{3} w_{1} a_{2} - w_{2} a_{1} .$ Match the coefficients of $a_{1}, a_{2}, a_{3}$ row by row: $W_{12} = - w_{3}$ , $W_{13} = w_{2}$ , $W_{23} = - w_{1}$ . The index form of $w_{k}$ follows from $W_{ij} = - ϵ_{ijk} w_{k}$ and the contraction $ϵ_{ijk} ϵ_{ij l} = 2 δ_{k l}$ (Theorem 1.3.4).

Consequences:

$Ww = w \times w = 0$ . The axis is not moved, so a nonzero skew tensor is not invertible in 3D.
$a \cdot Wa = a \cdot (w \times a) = 0$ for every $a$ : $Wa$ is always orthogonal to $a$ . This is the velocity field of a rigid body spinning about $w$ with angular velocity $∣ w ∣$ .
$tr W = 0$ (all diagonal entries vanish).

Warning

In the plane a skew tensor is $[0 w - w 0]$ : one number, and the “axis” is the normal to the plane. The demo above is this case; it has no real eigenvector, and the hunt for one fails on purpose.

Exercises

Exercise 2.3.1beyond the notespractice (A.1)

For $[A] = 102230045$ , find $sym A$ , $skw A$ and the axial vector of $skw A$ .

Solution

$[sym A] = 111132125, [skw A] = 0 - 1 1 10 - 2 - 1 20 .$ Axial vector $w = (W_{32}, W_{13}, W_{21}) = (- 2, - 1, - 1)$ . Check one entry with $W_{ij} = - ϵ_{ijk} w_{k}$ : $W_{12} = - w_{3} = 1$ .

Exercise 2.3.2beyond the notes

Write the skew tensor whose axial vector is $w = ω e_{3}$ , and apply it to $e_{1}$ and $e_{2}$ . What motion does $v (x) = Wx$ describe?

Solution

$W = ω (e_{2} \otimes e_{1} - e_{1} \otimes e_{2})$ ; $W e_{1} = ω e_{2}$ , $W e_{2} = - ω e_{1}$ . Every point moves at right angles to its position in the $e_{1} e_{2}$ plane, at a speed proportional to its distance from the $e_{3}$ axis: a rigid spin about $e_{3}$ .