Section 1.11 Solid Mechanics Part III Kelly 961.11 The Eigenvalue Problem and Polar Decomposition 1.11.1 Eigenvalues, Eigenvectors and Invariants of a Tensor Consider a second-order tensor A. Suppose that one can find a scalar λand a (non-zero) normalised, i.e. unit, vector nˆsuch that nnAˆˆλ=(1.11.1) In other words, Atransforms the vector nˆinto a vector parallel to itself, Fig. 1.11.1. If this transformation is possible, the scalars are called the eigenvalues(or principal values) of the tensor, and the vectors are called the eigenvectors(or principal directionsor principal axes) of the tensor. It will be seen that there are threevectors nˆ(to each of which corresponds some scalar λ) for which the above holds. Figure 1.11.1: the action of a tensor A on a unit vector Equation 1.11.1 can be solved for the eigenvalues and eigenvectors by rewriting it as ()0ˆ=−nIAλ(1.11.2) or, in terms of a Cartesian coordinate system, ()()()0ˆˆ0ˆˆ0ˆˆ=−→=−→=⊗−⊗iijijrrijijrrqppqkkjiijnnAnnAnnAeeeeeeeeeλλλδIn full, 0ˆ)(ˆˆ0ˆˆ)(ˆ0ˆˆˆ)(333323213123232221211313212111=−++=+−+=++−eeenAnAnAnAnAnAnAnAnAλλλ(1.11.3) Dividing out the base vectors, this is a set of three homogeneous equations in three unknowns (if one treats λas known). From basic linear algebra, this system has a solution (apart from 0ˆ=in) if and only if the determinant of the coefficient matrix is zero, i.e. if Anˆnˆλ
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Section 1.11 Solid Mechanics Part III Kelly 970det)det(333231232221131211=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−=−λλλλAAAAAAAAAIA(1.11.4) Evaluating the determinant, one has the following cubic characteristic equationof A, 0IIIIII23=−+−AAAλλλTensor Characteristic Equation(1.11.5) where ()AAAAAAAdetIII)tr()(trIItrI321222121==−=−===kjiijkijjijjiiiiAAAAAAAAε(1.11.6) It can be seen that there are three roots 321