Part 2 · Tensors

2.7 Eigenvalues and the spectral theorem

Source: Favata’s notes, appendix A.2, p. 16 (eigenvalues, characteristic polynomial, spectral theorem); invariants, positive definite tensors and square roots are beyond the notes

Objectives

Understand an eigenvector as a direction a tensor does not turn.
Understand a symmetric tensor as a pure stretch along three orthogonal directions.
Recipe: find eigenvalues from $det (A - λ I) = 0$ ; build the spectral decomposition.
Pictures: hunting for the directions that are only stretched; the circle mapped to an ellipse whose axes are the eigenvectors.
Theorem: the spectral theorem; positive definite tensors have a unique positive definite square root.
Vocabulary: characteristic polynomial, principal invariants, spectral decomposition, positive definite, square root.
Essential vocabulary: eigenvalue, eigenvector.

Directions that are not turned

Definition 2.7.1Eigenvalue and eigenvectorEssential

$λ \in R$ is an eigenvalue of $A \in Lin$ if there is a unit vector $v$ , an eigenvector, with $Av = λ v .$

$A$ only stretches $v$ , by the factor $λ$ , without turning it. A negative $λ$ reverses it; $λ = 0$ crushes it. The eigenvectors with eigenvalue $λ$ , together with $0$ , are the eigenspace of $λ$ ; along it, $A$ acts as $λ I$ .

Loading the demo…

Figure 2.7.1 Hunt for the eigenvectors. Turn

x

around the origin and watch

Ax

: when the two point along the same line,

x

snaps and the orange line stays. For the symmetric tensors you will find two, at right angles, and they are the axes of the ellipse. The simple shear has only one; the rotation has none.

Finding them

$Av = λ v$ is $(A - λ I) v = 0$ with $v \neq = 0$ . By Theorem 2.5.2, that happens iff $A - λ I$ is not invertible:

Recipe 2.7.2Eigenvalues and eigenvectors

Solve the characteristic equation $det (A - λ I) = 0$ , a cubic in $λ$ (a quadratic in the plane). Its roots are the eigenvalues.
For each root, solve $(A - λ I) v = 0$ and normalise.

Definition 2.7.3Principal invariantsbeyond the notes

Expanding the $3 \times 3$ determinant, $det (A - λ I) = - λ^{3} + I_{1} λ^{2} - I_{2} λ + I_{3},$ $I_{1} = tr A, I_{2} = \frac{1}{2} [(tr A)^{2} - tr (A^{2})], I_{3} = det A .$

They are invariants: trace and determinant are defined without a basis, so the $I_{k}$ depend only on $A$ , never on the frame. If the eigenvalues are $λ_{1}, λ_{2}, λ_{3}$ , the polynomial is $- (λ - λ_{1}) (λ - λ_{2}) (λ - λ_{3})$ , and comparing coefficients $I_{1} = λ_{1} + λ_{2} + λ_{3}, I_{2} = λ_{1} λ_{2} + λ_{2} λ_{3} + λ_{3} λ_{1}, I_{3} = λ_{1} λ_{2} λ_{3} .$

Warning

A real cubic always has a real root, but the other two may be complex. A rotation by $θ \neq = 0, π$ about an axis has only one real eigenvalue, $1$ , with the axis as eigenvector: every other direction is turned. In the plane a rotation has none at all; the hunt above fails, and is meant to.

The spectral theorem

For a symmetric tensor nothing can go wrong.

Theorem 2.7.4Spectral theorem

Let $A \in Sym$ . Then all eigenvalues of $A$ are real, there is an orthonormal basis ${a_{i}}$ of eigenvectors of $A$ , and $A = i = 1 \sum 3 λ_{i} a_{i} \otimes a_{i},$ the spectral decomposition, where $λ_{i}$ is the eigenvalue of $a_{i}$ . Eigenvectors of distinct eigenvalues are orthogonal.

Key observation 2.7.5What a symmetric tensor does

In its eigenbasis, $[A] = diag (λ_{1}, λ_{2}, λ_{3})$ . A symmetric tensor is a pure stretch by $λ_{i}$ along three orthogonal directions $a_{i}$ : it takes the unit sphere to an ellipsoid with semi-axes $∣ λ_{i} ∣ a_{i}$ .

Proof (the notes say it can be shown)beyond the notes

Distinct eigenvalues give orthogonal eigenvectors. If $Aa = λ a$ , $Ab = μ b$ with $λ \neq = μ$ , then $λ a \cdot b = Aa \cdot b = a \cdot A^{⊤} b = a \cdot Ab = μ a \cdot b,$ so $(λ - μ) a \cdot b = 0$ .
The eigenvalues are real. Allow a complex pair $Av = λ v$ and let $s := \overset{v}{ˉ}_{i} A_{ij} v_{j}$ . Its conjugate is $v_{i} A_{ij} \overset{v}{ˉ}_{j} = \overset{v}{ˉ}_{j} A_{ji} v_{i} = s$ , by symmetry, so $s$ is real. But $s = λ \sum_{i} ∣ v_{i} ∣^{2}$ with $\sum_{i} ∣ v_{i} ∣^{2} > 0$ , so $λ$ is real.
The eigenbasis exists. Take a real eigenvalue $λ_{1}$ and a unit eigenvector $a_{1}$ . The plane normal to $a_{1}$ is mapped into itself: if $v \cdot a_{1} = 0$ then $Av \cdot a_{1} = v \cdot A a_{1} = λ_{1} v \cdot a_{1} = 0$ . On that plane $A$ is still symmetric; repeat there to find $a_{2}$ , and $a_{3}$ completes the basis. Repeated eigenvalues are no obstacle.
The decomposition. $I = a_{i} \otimes a_{i}$ for any orthonormal basis, so with Proposition 2.2.7 $A = AI = i \sum A a_{i} \otimes a_{i} = i \sum λ_{i} a_{i} \otimes a_{i} .$

Cases by multiplicity (beyond the notes):

$λ_{1}, λ_{2}, λ_{3}$ distinct: the eigen-directions are unique, each eigenvector up to sign.
$λ_{1} \neq = λ_{2} = λ_{3}$ : $A = λ_{1} a_{1} \otimes a_{1} + λ_{2} (I - a_{1} \otimes a_{1})$ , and every unit vector normal to $a_{1}$ is an eigenvector.
all equal: $A = λ I$ , and every vector is an eigenvector.

Non-example 2.7.6A shear is not a stretchbeyond the notes

The simple shear $I + γ e_{1} \otimes e_{2}$ is not symmetric. Its only eigenvalue is $1$ , with the single eigen-direction $e_{1}$ : no basis of eigenvectors exists. The ellipse it makes of the circle still has two orthogonal axes, but they are not eigenvectors of $F$ ; they belong to $F^{⊤} F$ (§3.6 Strain measures).

Worked example

Example 2.7.7A spectral decompositionbeyond the notes

$[A] = 210120003 .$

Invariants: $I_{1} = 7$ , $I_{3} = 3 (4 - 1) = 9$ ; $[A^{2}] = 540450009$ , so $I_{2} = \frac{1}{2} (49 - 19) = 15$ .
$- λ^{3} + 7 λ^{2} - 15 λ + 9 = - (λ - 1) (λ - 3)^{2}$ : $λ = 1, 3, 3$ . Check: $1 + 3 + 3 = 7$ , $3 + 9 + 3 = 15$ , $1 \cdot 3 \cdot 3 = 9$ .
$λ = 1$ : $a_{1} = (e_{1} - e_{2}) / 2$ . $λ = 3$ : every vector normal to $a_{1}$ , for instance $(e_{1} + e_{2}) / 2$ and $e_{3}$ .
$A = a_{1} \otimes a_{1} + 3 (I - a_{1} \otimes a_{1})$ : $\frac{1}{2} 1 - 1 0 - 1 10 000 + 3 \frac{1}{2} \frac{1}{2} 0 \frac{1}{2} \frac{1}{2} 0 001 = 210120003 .$

Positive definite tensors and square roots

The kinematics needs one more fact: how to take the square root of $F^{⊤} F$ .

Definition 2.7.8Positive definitebeyond the notes

$S \in Sym$ is positive definite if $u \cdot Su > 0$ for every $u \neq = 0$ .

Theorem 2.7.9Positive definite means positive eigenvaluesbeyond the notes

$S \in Sym$ is positive definite iff all its eigenvalues are positive.

Proofbeyond the notes

⇒: $λ_{i} = λ_{i} a_{i} \cdot a_{i} = a_{i} \cdot S a_{i} > 0$ . ⇐: write $u = u_{i} a_{i}$ in the eigenbasis; $u \cdot Su = \sum_{i} λ_{i} u_{i}^{2} > 0$ unless $u = 0$ .

So a positive definite tensor stretches along $a_{1}, a_{2}, a_{3}$ without ever reversing a direction, and $det S = λ_{1} λ_{2} λ_{3} > 0$ .

Theorem 2.7.10Square rootbeyond the notes

A positive definite $S$ has exactly one positive definite $U$ with $U^{2} = S$ : $S := i \sum λ_{i} a_{i} \otimes a_{i} .$

Proof that it works (uniqueness: Gurtin)beyond the notes

With Proposition 2.2.7, $S S = \sum_{i, j} λ_{i} λ_{j} (a_{i} \cdot a_{j}) a_{i} \otimes a_{j} = \sum_{i} λ_{i} a_{i} \otimes a_{i} = S$ , and its eigenvalues $λ_{i}$ are positive. For uniqueness: a positive definite $U$ with $U^{2} = S$ has, on each of its eigenvectors, $Su = μ^{2} u$ , so its eigenspaces are those of $S$ and its eigenvalues are forced to be the positive roots.

Recipe 2.7.11Taking a square rootbeyond the notes

Diagonalise, take the root of each eigenvalue, put it back together. Same eigenvectors, rooted eigenvalues. The same trick gives $S^{- 1}$ , $ln S$ , any function of a symmetric tensor.

This is how the right stretch tensor $U = F^{⊤} F$ is built.

Exercises

Exercise 2.7.1beyond the notes

Find the eigenvalues and an orthonormal eigenbasis of $[S] = [2112]$ (the “symmetric” tensor of the demo), and write $S$ .

Solution

$det (S - λ I) = (2 - λ)^{2} - 1 = 0$ : $λ = 1, 3$ . $λ = 3$ : $a_{1} = (1, 1) / 2$ ; $λ = 1$ : $a_{2} = (1, - 1) / 2$ . Orthogonal, as promised. $[S] = 3 \cdot \frac{1}{2} [1111] + 1 \cdot \frac{1}{2} [1 - 1 - 1 1] = \frac{1}{2} [3 + 1 3 - 1 3 - 1 3 + 1] .$ Check: the $(1, 1)$ entry of its square is $\frac{1}{4} [(3 + 1)^{2} + (3 - 1)^{2}] = \frac{1}{4} \cdot 8 = 2$ .

Exercise 2.7.2beyond the notes

Show that $F^{⊤} F$ is positive definite for every invertible $F$ .

Solution

It is symmetric (Exercise 2.2.1), and $u \cdot F^{⊤} Fu = Fu \cdot Fu = ∣ Fu ∣^{2}$ . That is $\geq 0$ , and $= 0$ only if $Fu = 0$ , which for invertible $F$ means $u = 0$ (Theorem 2.5.2, condition 2).

Exercise 2.7.3beyond the notes

Show that the principal invariants of $A$ and of $QA Q^{⊤}$ coincide for $Q \in Orth$ , and that if $v$ is an eigenvector of $A$ then $Qv$ is one of $QA Q^{⊤}$ .

Solution

$det (QA Q^{⊤} - λ I) = det (Q (A - λ I) Q^{⊤}) = det Q det (A - λ I) det Q = det (A - λ I)$ : same characteristic polynomial, same coefficients. And $QA Q^{⊤} Qv = QAv = λ Qv$ .