Part 2 · Tensors

2.8 Change of basis

Source: beyond the notes (on the syllabus; standard, as in Gurtin, in the notes’ notation)

Objectives

Understand that changing basis changes components, never the vector or the tensor.
Recipe: $[v]^{'} = [Q]^{⊤} [v]$ and $[A]^{'} = [Q]^{⊤} [A] [Q]$ .
Picture: turning the basis until the matrix of a symmetric tensor is diagonal.
Vocabulary: change of basis, invariant.

Everything so far was defined without a basis, then written in ${e_{i}}$ . Nothing forces that choice. Take a second orthonormal basis ${e_{i}^{'}}$ : how do the components change?

The tensor that turns one basis into the other

Definition 2.8.1Change of basisbeyond the notes

The change of basis from ${e_{i}}$ to ${e_{i}^{'}}$ is the tensor $Q := e_{i}^{'} \otimes e_{i}, Q e_{j} = e_{j}^{'}, Q_{ij} = e_{i} \cdot e_{j}^{'} .$ Column $j$ of $[Q]$ is $e_{j}^{'}$ written in the old basis.

It is orthogonal: $Q^{⊤} Q = (e_{i} \otimes e_{i}^{'}) (e_{j}^{'} \otimes e_{j}) = δ_{ij} e_{i} \otimes e_{j} = I$ (Proposition 2.2.7). Its columns are orthonormal, as Recipe 2.6.3 requires.

Theorem 2.8.2New componentsbeyond the notes

$[v]^{'} = [Q]^{⊤} [v], v_{i}^{'} = Q_{ji} v_{j};$ $[A]^{'} = [Q]^{⊤} [A] [Q], A_{ij}^{'} = Q_{hi} Q_{kj} A_{hk} .$

Proofbeyond the notes

$v_{i}^{'} = v \cdot e_{i}^{'} = v \cdot Q e_{i} = Q^{⊤} v \cdot e_{i}$ . And $A_{ij}^{'} = A e_{j}^{'} \cdot e_{i}^{'} = AQ e_{j} \cdot Q e_{i} = Q^{⊤} AQ e_{j} \cdot e_{i}$ (Theorem 2.1.6).

Loading the demo…

Figure 2.8.1 Drag

q_{1}

to turn the basis. The tensor stays put: the same ellipse, the same images of every vector. Only its matrix changes. Find the angle where the matrix is diagonal; for a symmetric tensor there always is one, and its axes are the axes of the ellipse. The general tensor and the shear only let you zero one entry; the rotation not even that.

Key observation 2.8.3Invariantsbeyond the notes

The vector and the tensor do not change; only their components do. Anything defined without a basis gives the same number from either matrix: $tr$ , $det$ , $I_{2}$ , the eigenvalues, $∣ A ∣$ . For instance $tr ([Q]^{⊤} [A] [Q]) = tr ([A] [Q] [Q]^{⊤}) = tr [A]$ .

Choosing ${e_{i}^{'}}$ as the eigenbasis of a symmetric $A$ makes $[A]^{'}$ diagonal: the spectral theorem in matrix language.

Warning

Do not confuse $Q^{⊤} AQ$ in components with the tensor $QA Q^{⊤}$ . The first is the same tensor, described from a turned basis. The second is a different tensor: $A$ physically rotated, as when a stretched body is turned. That one appears with frame indifference.

Non-example 2.8.4Turning the basis is not turning the bodybeyond the notes

In the demo, turning $q_{1}$ by $3 0^{\circ}$ leaves the ellipse where it is. Rotating the body by $3 0^{\circ}$ would turn the ellipse, $QA Q^{⊤}$ , and leave ${e_{i}}$ alone.

Exercises

Exercise 2.8.1beyond the notes

In the plane, take $e_{1}^{'} = (e_{1} + e_{2}) / 2$ , $e_{2}^{'} = (- e_{1} + e_{2}) / 2$ . Find the components of $[A] = [2112]$ in the new basis.

Solution

$[Q] = \frac{1}{2} [11 - 1 1]$ . $[A] [Q] = \frac{1}{2} [33 - 1 1]$ , and $[Q]^{⊤}$ times that is $\frac{1}{2} [6002] = [3001]$ . Diagonal: the new basis is the eigenbasis of Exercise 2.7.1. Trace $4$ and determinant $3$ in both.

Exercise 2.8.2beyond the notes

Show $∣ A ∣$ is the same in every orthonormal basis, using the change-of-basis formula.

Solution

$∣ [A]^{'} ∣^{2} = tr ([A]^{' ⊤} [A]^{'}) = tr ([Q]^{⊤} [A]^{⊤} [Q] [Q]^{⊤} [A] [Q]) = tr ([Q]^{⊤} [A]^{⊤} [A] [Q]) = tr ([A]^{⊤} [A] [Q] [Q]^{⊤}) = ∣ [A] ∣^{2}$ . The same as Exercise 2.6.1.