Relative eigenvalues and the Rayleigh quotient in tensor notation

eigenvalues-eigenvectorsindex-notationlinear algebramatricestensors

I'm working through Pavel Grinfeld's Introduction to Tensor Analysis and the Calculus of Moving Surfaces and I'm very stuck on exercise 118, which reads:

Show that the eigenvalues of the generalized equation (7.71) are given by the Rayleigh quotient
$$\lambda = A_{ij} x^i x^j.$$

This references an equation from a preceding section of text:

In this section, we show that, much like $A x = b$, the eigenvalue problem
$$A x = \lambda M x\tag{7.66}$$
can be formulated as a variational problem. The matrix $M$ is assumed to be symmetric and positive define [sic]. The variational formulation is to find the extrema of
$$f(x) = A_{ij} x^i x^j\tag{7.67}$$
subject to the constraint that
$$M_{ij} x^i x^j = 1.\tag{7.68}$$
Geometrically, equation (7.68) states that the vector $x^i$ unit length [sic]. Use a Lagrange multiplier $\lambda$ to incorporate the constraint on the augmented function $E(x, \lambda)$
$$E(x, \lambda) = A_{ij} x^i x^j – \lambda (M_{ij} x^i x^j – 1).\tag{7.69}$$
Following earlier analysis,
$$\frac{1} {2} \frac{\partial E(x)} {\partial x^i} = A_{ij} x^j – \lambda M_{ij} x^j.\tag{7.70}$$
Equating the partial derivatives to zero yields
$$A_{ij} x^j = \lambda M_{ij} x^j\tag{7.71}$$
which is equivalent to the eigenvalue problem (7.66).

First I tried plugging the expression for $\lambda$ (with renamed indices) into the RHS of equation (7.71) to try and obtain the LHS, but I didn't really get any further than plugging it in. Then I considered solving the variational problem for $\lambda$ supposing we'd found an $x$ that works, but had no idea where to begin.

Best Answer

You can just multiply both sides of $(7.71)$ by $x^i$ and use the constraint:

$$ A_{ij} x^j x^i = \lambda M_{ij} x^i x^j = \lambda \,.$$

Related Solutions

[Math] Eigenvalues and Spectrum

Spectral theory in infinite-dimensional spaces is quite a bit more complicated than in the finite-dimensional case. In particular, we have to distinguish between the spectrum $\sigma(A)$ of an operator and its eigenvalues. Let $A$ be a linear operator on a Banach space $X$ over the scalar field $C$. We have $$ \sigma(A) = \{ \lambda \in C: (\lambda I - A) \text{ does not have a bounded inverse} \}. $$ An eigenvalue $\lambda$ of $A$ is a value such that there exists a nonzero eigenvector $x \in X$ such that $$ A x = \lambda x, $$ or equivalently, $\operatorname{ker}(\lambda I - A) \neq \emptyset$. We then call $\dim \operatorname{ker}(\lambda I - A)$ the geometric multiplicity of the eigenvalue $\lambda$.

An eigenvalue is always in the spectrum, as you can see from the definition, but not every element of the spectrum is an eigenvalue in general.

In increasing order of "complicatedness", we could say:

Matrices (linear bounded operators on finite-dimensional vector spaces): the spectrum is finite, and each of its elements is an eigenvalue.
Compact self-adjoint operators on a Hilbert space: almost as nice as matrices. The spectrum is a compact set and countable, and it is contained in the reals. Every nonzero element of the spectrum is an eigenvalue with finite multiplicity. There is a spectral decomposition of the operator much as one would have for a matrix.
Bounded operators: the spectrum is still compact, but may be uncountable. In fact, for any compact set in $C$, you can find an operator which has this set as its spectrum. Yet it's still quite possible that no element of the spectrum is an eigenvalue; see the example of T.A.E. given in another answer.
Unbounded operators: the spectrum is in general unbounded.

This list could of course be refined with more specific conditions. I'm still studying the theory myself and will go back and add details as I learn about them. (Suggestions are welcome.) If you want to learn more, I found some relatively digestible lecture notes by E. Kowalski, ETH Zürich.

[Math] Multidimensional Tensor Inverse – Index Notation

I was able to answer the question for myself well enough. The answer is that yes, for a given tensor $R_{ijkl}$ It is possible to find another tensor $L_{ijmn}$ with the property that

$$ L_{ijmn}R_{ijkl} = \delta_{mk}\delta_{nl} $$

The point is that $L_{ijmn}$ has $N^4$ components and the equation above stands for $N^4$ equations where $N$ is the length of the indices. We should then be able to solve the above $N^4$ equations for the $N^4$ components of $L_{ijmn}$ to find the desired tensor.

I implemented some code in Mathematica to convince myself which I'll post here for those curious.. I don't know the best way to input code here so apologies if it doesn't look great.

R = Table[RandomReal[], {i, 1, 6}, {j, 1, 6}, {k, 1, 6}, {l, 1, 6}];

L = Table[a[i, j, k, l], {i, 1, 6}, {j, 1, 6}, {k, 1, 6}, {l, 1, 6}];

K = Table[
   KroneckerDelta[m, k]*KroneckerDelta[n, l], {k, 1, 6}, {l, 1, 
    6}, {m, 1, 6}, {n, 1, 6}];

Flatten[
  Table[
   Sum[L[[i, j, m, n]]*R[[i, j, k, l]], {i, 1, 6}, {j, 1, 6}] == 
    K[[k, l, m, n]]
   , {k, 1, 6}, {l, 1, 6}, {m, 1, 6}, {n, 1, 6}]];

s = Solve[Flatten[
    Table[
     Sum[L[[i, j, m, n]]*R[[i, j, k, l]], {i, 1, 6}, {j, 1, 6}] == 
      K[[k, l, m, n]]
     , {k, 1, 6}, {l, 1, 6}, {m, 1, 6}, {n, 1, 6}]]];

Chop[Table[
   Sum[L[[i, j, m, n]]*R[[i, j, k, l]], {i, 1, 6}, {j, 1, 6}]
   , {k, 1, 6}, {l, 1, 6}, {m, 1, 6}, {n, 1, 6}] /. s[[1]]]

The final line shows the result of multiplying the tensors $L$ and $R$ when the components of $L$ are replaced with the components found in the above line. The result is the double Kronecker delta tensor $K$ as desired.

Fortunately for my application the code can run in a few seconds for me (on my laptop) with tensors with $6^4 = 1296$ components.

I'm not sure what to call $L_{ijmn}$. I could call it $R^{-1}_{ijmn}$, but I can also pose another problem:

Find a tensor $G_{ikmj}$ with the property that

$$G_{ikmj}R_{ijkl} = \delta_{ml}$$

This new tensor could similarly be called $R^{-1}_{ikmj}$ and so could a number of other tensors with different combinations and permutations of indices. Therefore, for the time being I'll just continue to use unique names for these inverse tensors and explicitly state the relevant property in terms of the Kronecker deltas.

Best Answer

Related Solutions

[Math] Eigenvalues and Spectrum

[Math] Multidimensional Tensor Inverse – Index Notation

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