In some 'fiddling about' with 3x3 real symmetric matrices, related to wave propagation, I noticed that: Real symmetric matrix S with eigenvalue L (one of the three).

   **T**= **S** - L.

   **D** is matrix of co-factors of **T**.

  The matrix **D***= **D**/ trace(**D**) is the outer product of the eigenvector **n** associated with the value L. i.e. D*ij = ni nj

So, the eigenvector component values can be found from the square roots of the diagonal components of D*, and the signs ( + and - of each eigenvector are equally valid of course) from the other components. The trace of D must not be zero, of course. Computationally this is probably of no value compared with using 'standard' methods but I wondered how well known this was- and is there a simple reason? Please bear in mind that I'm an Engineer and a manifold to me is something that bolts onto a cylinder head...

  • $\begingroup$ Sorry, missed the identity matrix, I, out of the definition of T. T= S - I L i.e T is the matrix S minus the eigenvalue for each leading diagonal element. $\endgroup$ – Pete_Bate Jun 13 '15 at 16:41
  • $\begingroup$ the matrix of co-factors is the transpose of the adjoint, so its rows and coloumns are eigenvectors for T and for S $\endgroup$ – Exodd Jun 13 '15 at 16:43
  • $D$ is the matrix of co-factors of a symmetric matrix, so it's itself a symmetric matrix, and it's also the adjoint

  • By definition of adjoint, $DT=TD=0$, so the rows and coloumns of $D$ are eigenvectors for the eigenvalue $L$ of $S$

  • If $L$ was a simple eigenvector of $S$, and $v=[a,b,c]^T$ is an eigenvector, then $D$ can only have the form of $$D=c v v^T$$ with $c$ a constant different from zero.

  • The trace of $D$ is now $c(a^2+b^2+c^2)=c|v|^2$, so $$D^*=ww^T \qquad w=v/|v|$$

  • If $L$ isn't a simple eigenvalues, then $D=0$

Conclusion: your method works only for simple eigenvalues

  • $\begingroup$ Thanks Exodd, I think I'm happy with that. I realised the matrix of co-factors was a scalar multiple of the inverse (symmetric matrix), and guess you meant DT=TD=I. I shall work it through properly when I get time. $\endgroup$ – Pete_Bate Jun 15 '15 at 12:48
  • $\begingroup$ WHOOPS! Yes, DT=TD=0 not I. $\endgroup$ – Pete_Bate Jun 15 '15 at 12:56
  • $\begingroup$ Or does it? I'm just getting confused... especially as <return> finishes the comment on this site. For the eigenvector thing to work, yes, DT=0 but is that a property of the adjoint? I thought it was just a scalar multiple of the inverse so that would not be true? $\endgroup$ – Pete_Bate Jun 15 '15 at 12:58
  • $\begingroup$ when the matrx is invertible, then it is a multiple of the inverse. But T, in this case, is singular. In general, if $A$ is a matrix, and $B$ her adjoint, then $$AB=BA=det(A)I$$ If $A$ is singular, then $det(A)=0$ $\endgroup$ – Exodd Jun 15 '15 at 14:37
  • $\begingroup$ Yes, of course; T is singular by definition! Thank you very much, Exodd, for both answering the question and putting up with the "brain scramble". @Exodd $\endgroup$ – Pete_Bate Jun 16 '15 at 18:04

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