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A.1.14 Eigenvalues and Eigenvectors of Real Symmetric Tensors
A second-order tensor has three pairs of eigenvalues and eigenvectors that each satisfy The eigenvalues are the roots of the characteristic equation of , which is the cubic polynomial produced by setting , where are called invariants of .
According to the Cayley-Hamilton theorem, a tensor satisfies its own characteristic equation, Therefore, the cubic power of can be expressed in terms of its lower powers according to . Taking the trace of this equation allows us to solve for as Multiplying eq.(A.1.14-4) by also produces Using all these relations, we may differentiate the three invariants of with respect to to get
The eigenvalues of real symmetric tensors are real (proof not provided here).
- If the eigenvalues of a real symmetric tensor are all distinct, the eigenvectors are orthogonal to each other.
Proof: Given , , , then and ,
When two of the eigenvalues are repeated (a double root of the characteristic equation), the resulting eigenvectors are not necessarily orthogonal to each other; however, they remain orthogonal to the third eigenvector. This means that any vector lying in the plane normal to the third eigenvector is an eigenvector corresponding to the double root. Similarly, when all three eigenvalues are repeated (a triple root), any vector becomes an eigenvector of .
In hydrostatics the stress tensor is , where is the hydrostatic pressure. In this case, is a triple root of the characteristic equation of . Any vector satisfies , and is thus an eigenvector of .
In continuum mechanics the eigenvectors of a tensor are generally normalized, Thus, we can always find a set of three orthonormal eigenvectors for any real symmetric tensor , even when the eigenvalues are repeated. Given a tensor with eigenvalues and eigenvectors , the components of in the orthonormal basis can be obtained from Thus, Since , we find that This is known as the spectral representation of the tensor . In particular, since the eigenvalues of the identity tensor are , and since any vector is an eigenvector of , we can select the basis vectors so that the spectral representation of may be given by