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Suppose an $n\times n$ matrix has real entries and has $n$ real eigenvalues and its eigenvectors span $\mathbb R^n.$ Are there any interesting conditions under which $k$ of its eigenvectors are mutually orthogonal, for $k<n$? If so, might that conclusion be useful?

PS: Here's an example. This matrix is not symmetric but would become so if the $(1,2)$ entry were changed. Six of its seven eigenvectors are all orthogonal to each other. $$ \left[ \begin{array}{rrrrrrrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 \end{array} \right] $$

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    $\begingroup$ A smaller example would be $$\pmatrix{1&0&1\cr0&2&2\cr0&0&3\cr}$$ with orthogonal eigenvectors $(1,0,0)$ and $(0,1,0)$, and nonorthogonal eigenvector $(1,4,2)$. I don't know whether that's interesting. Perhaps interest is in the eye of the beholder. $\endgroup$ Commented Jun 13 at 11:44
  • $\begingroup$ See : archive.org/details/… "...of the coefficients of system (105) to be equal to $(n — u_{r})$, where py is the multiplicity of the root A, of the secular equation. When this condition is satisfied, system (105) defines ux linearly independent vectors ..." $\endgroup$
    – The Tiler
    Commented Jun 13 at 12:02
  • $\begingroup$ @GerryMyerson : You example just mentions one particular example, of which the conclusion is true. But the question is about what sorts of assumptions entail that conclusion. $\endgroup$ Commented Jun 13 at 18:52
  • $\begingroup$ @TheTiler : I haven't yet understood your comment. But here we see a MathJax hazard: The code (n—u_r) gets rendered as $(n—u_r),$ whereas the code (n-u_r) gets rendered as $(n-u_r).$ Which shows you why the latter is considered correct in MathJax (and in LaTeX). $\endgroup$ Commented Jun 13 at 18:55
  • $\begingroup$ I'm aware that I have only given one example, although I think it points the way to many more examples; a diagonal matrix $D$ in the upper left corner, zeroes below $D$, and some more-or-less arbitrary columns to the right (keeping the matrix upper triangular, for simplicity). It doesn't answer your question, but it adds more information than any other user has added so far (and may I point out that you, too, have just mentioned one particular example?). $\endgroup$ Commented Jun 13 at 22:23

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I encountered this situation when I was studying a weird kind of radiative corrections in my formal analyses of relativistic quantum field theories with weak violations of Lorentz symmetry (arXiv version). I had previously observed that certain kinds of radiative corrections that had provoked a lot of controversy at leading order in the small Lorentz violation parameter $b$ were actually much more problematic if the methods described by various authors were extended to higher orders, especially $\mathcal{O}(b^{2})$. In particular, there tended to be violations of the gauge symmetries of the theories—something that could be seen as an indication of deep problems in the structure of the theory, meaning that the whole notion that these $\mathcal{O}(\hbar)$ corrections could be related to interesting new physics was fairly unlikely.

However, I subsequently looked at what the $\mathcal{O}(b^{2})$ results might mean if they were really interpreted physically. A breaking of gauge symmetry is a major problem for a relativistic quantum field theory, but not necessarily an insuperable one. The standard model of particle physics has some of its gauge symmetries broken spontaneously. (That is, the field equations are invariant under the full symmetry group, but the ground state of the theory is not.) Also, for an Abelian gauge theory (quantum electrodynamics, essentially) gauge symmetry may be explicitly broken without totally destroying the theory's ability to make predictions. The typical way in which is is done is with a photon mass term.

What the $\mathcal{O}(b^{2})$ radiative corrections I had found in the course of my earlier analyses looked like were very much like a photon mass term, but a mass term that was not symmetric under $SO(3,1)$ symmetry. The free photon Lagrange density with a generalized mass term looks is $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+M^{\mu\nu}A_{\mu}A_{\nu},$$ where $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$ is the electromagnetic field strength (or "Faraday") tensor, whose elements are the components of the electric and magnetic fields, expressed in terms of a 4-vector potential $A^{\mu}$, $\mu=0,1,2,3$; and the Einstein summation convention is in operation. The first term is the Maxwell Lagrange density, which, in terms of the field strengths, is simply $\frac{1}{2}\left(\vec{E}^{2}-\vec{B}^{2}\right)$. The usual Proca Lagrange density for massive photons has a diagonal mass matrix $M^{\mu\nu}=\frac{1}{2}m^{2}\eta^{\mu\nu}$, where $\eta^{\mu\nu}=I_{1,3}$ is the metric tensor in Minkowski space. (Lowering one index gives a $4\times4$ matrix $M^{\mu}{}_{\nu}=\frac{1}{2}m^{2}I_{4}$ proportional to the identity, with a little abuse of notation.)

The theories I had looked at had radiatively-generated mass-like terms with a different structure, in which the mass matrix, written in an appropriate basis, was diagonal, but with one of the diagonal elements of $M^{\mu}{}_{\nu}$ different from the others. A similar general structure had previously been studied with different motivations. As an outgrowth of my interest in these theories, I looked at their propagating wave solutions, with a $\exp\left(\vec{k}\cdot\vec{x}-\omega t\right)$ spatiotemporal Ansatz for the the solutions; and here was where I found something relevant to the question.

Normally, in the Proca theory with massive but Lorentz-invariant electromagnetic equations of motion, the number of physical polarization states of propagating radiation changes from two (in the standard Maxwell theory) to three. The apparent discontinuity between the $m^{2}=0$ and $m^{2}\neq0$ theories is resolved by the fact that the third mode is continuously decoupled from the charge matter in the theory as $m^{2}\rightarrow0$. We would naturally expect the same to occur as any $M^{\mu}{}_{\nu}\rightarrow0$. This was correct, by the Lorentz symmetry breaking complicated things. Although the theory I was considering had two different "wild" aspects—breaking of gauge symmetry and breaking of Lorentz symmetry—the wild effects were not "doubly" small. The eigenvalue of the $4\times4$ mass matrix $M^{\mu}{}_{\nu}$ were small, because photon masses are known to be small; however, their ratios were not small. When looking at specifically-mass-related effects, the Lorentz violation was $o(1)$, small only because of the overall smallness of the mass terms, not any further suppressed by the Lorentz violation.

Normally, even in the massive Proca theory, we can find a natural orthonormal basis for the polarization vectors of the electromagnetic field modes. However, what I found was that this was not the case in this theory. If the (spacelike) eigenvector along which $M^{\mu}{}_{\nu}$ has its single different eigenvalue is taken to be along the $\hat{e}_{3}$-direction, then the polarization eigenmodes for propagation along oblique directions have a different structure. Quoting my paper:

The basis of polarization states we have found is not orthogonal. If the mass parameters are negligible compared to the components of $\vec{k}$, then the mode with the unconventional dispersion relation is essentially longitudinally polarized. However, the other modes are not necessarily transverse; their polarization vectors are normal to $\hat{e}_{3}$, not to $\vec{k}$.

In other words, in the weak coupling limit (small $M^{\mu}{}_{\nu}$), the eigenvectors fall into three types. One lies along the timelike direction $\hat{e}_{0}$ (and is less interesting from a physics point of view, since it is actually non-propagating). Another lies close to $\hat{k}$, the propagation direction; in the massless theory, this one would also be unphysical, with the two physically meaning polarizations spanning the remaining two-dimensional subspace. However, in this theory the other two are orthogonal to $\hat{e}_{3}$, not to $\hat{k}$. One of them may be taken to be normal to the other two, but the second may not, and the process of finding the modes that physically dominate as the masses go to zero involves taking some tricky linear combinations.

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