Eigenvalues and Eigenvectors of a Euclidean Second-Order TensorSecond-Order Tensor

A real numberis aneigenvalueof the tensorTif there exists a nonvanishing vector u called an eigenvectorbelonging to the eigenvalue , satisfying theeigenvalue equation

TuDu: (1.57)

Equation (1.57) reflects the following problem, which has many applications: the tensorTassociates to any vectoruthe vectorv, generally different fromu, both in norm and direction. The vectors which are not rotated by the linear applicationT are just the solutions of (1.57).

We easily recognize that the eigenvectors belonging to the same eigenvalue form a vector subspace of E3, called thecharacteristic space V ./ associated with the eigenvalue. Thegeometric multiplicityofis the dimensionmof this subspace. ThespectrumofTis the list12: : :of its eigenvalues.

Theorem 1.6. The following properties hold:

1. the eigenvalues of a positive definite tensor

vTv> 0 8v2E3; are positive;

2. the characteristic spaces of symmetric tensors are orthogonal to each other.

Proof. IfTis a symmetric positive definite tensor,its eigenvalue, anduone of the eigenvectors corresponding to, we have

uTuDjuj²:

Sincejuj² > 0, the previous equation implies > 0. Moreover, if₁ and₂ are distinct eigenvalues ofT, andu1,u2are the two corresponding eigenvectors, we can write

Tu1D₁u1; Tu2D₂u2;

1.8 Eigenvalues and Eigenvectors of a Euclidean Second-Order Tensor 21 so that

u2Tu1D1u1u2; u1Tu2D2u1u2:

From the symmetry.TDT^T/ofT, we deriveu2Tu1 Du1Tu2. Consequently, we find

.21/u1u2D0;

and the theorem is proved, since21¤0. ut

Let.ei/be a fixed basis; when the representation of Tin (1.46)1 is used, the eigenvalue equation becomes

T^ijei˝ej.u/DT^ijei ej uDuⁱei; or equivalently,

T^ijujei Dg^ijujei:

The unknowns of this homogeneous system are thecovariant components of the eigenvectoru:

.T^ij g^ij/u_j D0: (1.58)

This system has nonvanishing solutions if and only if the eigenvalueis a solution of thecharacteristic equation

det.T^ij g^ij/D0; (1.59)

which is a third-degree equation in the unknown. The multiplicity of, as a root of (1.59), is called thealgebraic multiplicityof the eigenvalue.

If the representation ofTin (1.46)2is used, we have T_jⁱu^jei Duⁱei Dı_jⁱu^jei;

so that the eigenvalue equation, written in mixed components, becomes

.T_jⁱı_jⁱ/u^j D0; (1.60)

and the characteristic equation is

det.T_jⁱıⁱ_j/D0: (1.61)

Finally, using the representation (1.46)3, we have

.Tij gij/u^j D0; (1.62)

det.Tij gij/D0: (1.63)

Remark 1.3. Developing (1.61), we obtain a third-degree equation having the following form:

³CI1²I2CI3D0; (1.64)

where the coefficientsI1; I2, andI3are called thefirst, second, andthird invariant, respectively, and are defined as follows:

I1DT_iⁱ; I2D 1

2.T_iⁱT_j^jT_jⁱT_i^j/; I3Ddet.T_jⁱ/: (1.65) It is clear that the roots of equation (1.64) are independent of the basis; that is, the eigenvalues are characteristic of the tensorTif and only if the coefficients of the characteristic equation (1.64) are invariant with respect to base changes. We leave to reader to prove this property (see hint in Exercise 1). If another kind of components is used, the expressions for the invariant coefficients change. For example, for covariant components, the expressions become

I₁Dg^ijT_ij; I₂D 1

2.g^ijT_ijg^hkT_hkg^ihT_hjg^{j k}T_{i k}/; (1.66) I3Ddet.g^ihThj/Ddet.g^ih/det.Thj/: (1.67) Theorem 1.7. IfTis a symmetric tensor, then

1. the geometric and algebraic multiplicities of an eigenvaluecoincide;

2. it is always possible to determine at least one orthonormal basis .ui/whose elements are eigenvectors ofT;

3. in these bases, one of the following representation ofTis possible:

i) if all the eigenvalues are distinct, then

TD X3 iD1

iui˝uiI (1.68)

ii) if1D2¤3, then

TD1.Iu3˝u3/C3u3˝u3I (1.69)

1.8 Eigenvalues and Eigenvectors of a Euclidean Second-Order Tensor 23 iii) if₁D₂D₃, then

TDI; (1.70)

whereIis the identity tensor ofE3.

Proof. We start by assuming that the property (1) has already been proved.

Consequently, three cases are possible: the three eigenvalues are distinct, two are equal but different from the third one, or all of them are equal. In the first case, the eigenvector subspaceV ._i/belonging to the eigenvalue_i is one-dimensional and orthogonal to the others (see Theorem1.5). Therefore, if a unit eigenvectoru_i is fixed in each of these subspaces, an orthonormal basis is obtained.

In the second case, the eigenvalue₁ D₂is associated with a two-dimensional eigenvector subspace V .₁/, so that it is always possible to find an orthonormal eigenvector pair .u₁;u2/ in V .₁/. If a third unit vector u3 is chosen in the one-dimensional eigenvector subspace V .₃/, orthogonal to the previous one, an orthonormal eigenvector basis.u_i/is obtained.

Finally, if all the eigenvalues are equal to , the subspace V ./ is three-dimensional, that is, it coincides with the whole spaceE3and it is certainly possible to determine an orthonormal eigenvector basis.

In conclusion, an orthonormal eigenvector basis.u₁;u2;u3/can always be found.

Since this basis is orthonormal, the contravariant, mixed, and covariant components of any tensorTcoincide; moreover, the reciprocal basis is identical to.u₁;u₂;u₃/ and the tensorg_ij has componentsı_ij.

Finally, all types of components of the tensorTin the basis.u₁;u₂;u₃/are given by the diagonal matrix

.T^ij/D 0

@1 0 0 0 ₂ 0 0 0 3

1

A: (1.71)

In fact, using any one of the representation formulae (1.46), the eigenvalue equation TuhDhuhcan be written in the form

T^ijıhjui Dhı_hⁱui; from which we derive

T^ihDhı_hⁱ; and (1.68) is proved. In the case (ii) (1.68) becomes

TD₁.u₁˝u1Cu2˝u2/C₃u3˝u3:

But the unit matrix can be written as

IDu1˝u1Cu2˝u2Cu3˝u3 (1.72) and (1.69) is also verified. Finally, (1.70) is a trivial consequence of (1.68)

and (1.72). ut

The relation (1.68) is called aspectral decompositionofT.

The following square-root theorem is important in finite deformation theory (see Chap.3):

Theorem 1.8. IfTis a positive definite symmetric tensor, then one and only one positive definite symmetric tensorUexists such that

U²DT: (1.73)

Proof. Since T is symmetric and positive definite from the previous theorem it follows that

TD X3 iD1

iui˝ui;

wherei > 0and.ui/is an orthonormal basis. Consequently, we can define the symmetric positive definite tensor

UD X3 iD1

piui ˝ui: (1.74)

In order to verify thatU²DT, it will be sufficient to prove that they coincide when applied to eigenvectors. In fact,

U²uhD X3 iD1

piui ˝ui

! ₃ X

iD1

piui˝ui

!

uh (1.75)

D X3 iD1

p_iui ˝ui

!p

_huhD_huhDTuh: (1.76)

To verify that Uis uniquely defined, suppose that there is another symmetric positive definite tensorVsuch thatV²DT. Then we have

.U²V²/.u_i/D0)U²DV²: