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Let's say I am given this equation: \begin{align} (M-k*N) \cdot u = 0 \end{align} for given positive symmetric real matrices $M$ and $N$ of Dimension n, and yet to be determined constant $k$ and vector $u$.

It seems to me that the solution to this equation automatically diagonalizes $M$ and $N$ simultaneously, and I am in doubt why this should be true, since I can choose those matrices abitrarily.

To solve the equation,M-k*N needs to have a determinant of 0, so I will have a characteristic polynomial to solve, which yields at maximum n different possible values $k_i$, for each I will find one or more possible $u_i$, with a maximum of n different $u$.

Another way to look at the same problem is the following: \begin{align} M \cdot u &= k*N \cdot u \\ N^{-1} \cdot M \cdot u &= k* u \end{align} So I need to find the eigenvalues $k_i$ and eigenvectors $u_i$ of $N^{-1} \cdot M$, and those will also give me the solutions to the initial problem. Especially, I know know that the solutions $u_i$ will all be orthogonal - for different eigenvalues they will be orthogonal anyway, and if the eigenspace to a certain eigenvalue is more than 1 dimensional, I can choose two orthogonal representatives.

But now comes the catch: One can show, that the $\{u_i \}_{i=1, N}$ are not only orthogonal, but also M-orthogonal and N-orthogonal:

Take the two solutions, $u_i$ and $u_j$ with differing eigenvalues: \begin{align} u_j^T \cdot M \cdot u_i = k_i * u_j^T \cdot N \cdot u_i \\ \end{align} But also \begin{align} u_i^T \cdot M \cdot u_j = k_j * u_i^T \cdot N \cdot u_j \end{align} The second equality can be transposed, and because $M$ and $N$ are symmetric it follows, \begin{align} u_j^T \cdot M \cdot u_i = k_j * u_j^T \cdot N \cdot u_i \end{align} Subtracting this from the first equality yields: \begin{align} (k_j - k_i) * u_j^T \cdot N \cdot u_i = 0 \end{align} That means, when $k_i$ and $k_j$ are not the same, then $u_i$ and $u_j$ are neccessarily N-orthogonal. So in the case that we have degeneracy here, and we have one eigenvector $u_i$ per eigenvalue $k_i$, then the complete set of all the $u_i$ is N-orthogonal.

But from before, we also know that the $u_i$ are orthogonal, because they are solutions to the eigenvalue problem of $N^{-1} \cdot M$. If the vectors $u_i$ are all mutually orthogonal, and N-orthogonal, they are also eigenvectors of $N$ (See this question.

If $u_i$ is supposed to solve the initial problem, and is an eigenvector of $N$, it must also be an eigenvector of $M$.

So in the end, the set of ${u_i}$ simultaneously consists of eigenvectors of $M$ and $N$, and $N^{-1}\cdot M$ - I can't believe that this is true in general, but I also am not able to find my mistake on this topic.

Quantumwhisp
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1 Answers1

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Your mistake was to think that $M$ and $N$ symmetrical imply $N^{-1}M$ symmetrical. It is not always the case unless $M$ and $N$ commute. Indeed, $$ \ ^t(N^{-1}M) = \ ^tM\ ^tN^{-1} = MN^{-1}, $$ which is not $N^{-1}M$ unless $M$ and $N$ (therefore $M$ and $N^{-1}$) commute. In general, $M$ and $N$ are simultaneoulsy diagonalisable (i.e. diagonal in the same basis) if and only if they are diagonalisable and they commute (it generalises to a higher number of matrices).

Since $N^{-1}M$ is a priori not symetrical, the $u_i$ can not be taken orthogonal.

Cactus
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  • I don't know if I can follow: The solutions of the eigenvalue problem of $N^{-1}\cdot M$ are (if you have different eigenvalues) orthogonal, no matter wether $N^{-1} \cdot M$ is symmetrical. Symmetricity of $N^{-1}\cdot M$ is indeed not guaranteed - but it would only ensure that $N^{-1} \cdot M$'s eigenvalue problem actually has solutions. – Quantumwhisp May 23 '23 at 08:26
  • No, first of all, orthogonality depends on your scalar product but eigenvectors do not so a statement like "in any case, eignevectors are orthogonal" is non-sense. Take for example $M = \begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix}$ with the canonical scalar product. The eigenvectors are $(1,0)$ and $(1,-1)$, which are not orthogonal. – Cactus May 23 '23 at 08:45
  • Of course I'm talking about the canonical scalar, so I don't see non-sense from that side - still I see my mistake - Distinct Eigenvalues only ensure orthogonality if the matrix is indeed symmetric. – Quantumwhisp May 23 '23 at 09:01