Normal modes of oscillation: proving a matrix formula

Question

When dealing with normal modes in physics we often encounter the equations of motion:

$$ M \ddot{x} = -K x $$

Where $M$ and $K$ are (symmetric) positive definite matrices and $x$ is a vector of positions. We solve this by making the ansatz $x=e^{-i\omega t} q$ to get

$$ M^{-1}Kq = \omega^2 q $$

Because $M$ and $K$ are positive definite we have that $M^{-1}K$ is diagonalizable (see Why is this matrix product diagonalizable?) and has positive eigenvalues (The product of two positive definite matrices has real and positive eigenvalues?). We can write

$$ M^{-1} K Q = Q\Omega^2 $$

Where the columns of $Q$ have eigenvectors of $M^{-1}K$ and $\Omega^2$ is diagonal with positive eigenvalues (corresponding to eigenvector columns of $Q$) along the diagonal. We have $$ KQ = MQ \Omega^2 $$

so that $$ Q^T K Q = Q^T M Q \Omega^2 = \mathcal{M} \Omega^2 $$

Where I've defined $\mathcal{M} = Q^T M Q$. Transposing, and using the fact that $K$, $M$, and $\Omega$ are all symmetric, we get

$$ Q^T K Q = \Omega^2 \mathcal{M} $$

So we see $$ \mathcal{M} \Omega^2 = \Omega^2 \mathcal{M} $$

It is a theorem that if $\mathcal{M}$ commutes with diagonal matrix $\Omega^2$ with distinct values that $\mathcal{M}$ must then be diagonal (If $X$ is diagonal with distinct diagonal entries and $XY = YX$ then $Y$ is also diagonal matrix.).

Summary: Above I have proven that, for positive definite $M$ and $K$, if $M^{-1}K$ has distinct eigenvalues, then there are matrices $Q$ and $\Omega^2$ which satisfy $$ Q^TKQ = Q^T M Q \Omega^2 $$ AND $$ Q^T M Q $$ is diagonal.

Questions:

1. Is my proof above correct?
2. Does the theorem still hold if $M^{-1}K$ has a degeneracy? My guess is that the answer is yes, and that the proof is something like: If $M^{-1}K$ has a degeneracy then there is a freedom to select the columns of the matrix $Q$. This freedom can be used to ensure $Q^T M Q$ is diagonal. But how exactly does this proof look?
3. I believe the same result should hold (except $\Omega^2$ may now have negative entries) if $K$ is merely symmetric and not positive definite so long as we also assume $M^{-1}K$ is diagonalizable. Is this correct?

Answer 1

Consider $M$ positive-definite and $K$ symmetric. We seek matrices $Q$ and $\Omega^2$ which satisfy

• $Q$ is invertible
• $\Omega^2$ is diagonal
• $Q^T K Q = Q^T M Q \Omega^2$
• $Q^T M Q$ is diagonal

This is possible.

Consider $$ M^{-1} K = M^{-1/2} M^{-1/2}K M^{-1/2}M^{+1/2} $$ We see $M^{-1}K$ is similar to $M^{-1/2}K M^{-1/2}$ which is symmetric and thus has real eigenvalues and a complete eigenbasis, so $M^{-1}K$ shares those real eigenvalues and also has a complete eigenbasis. Furthermore, if $K$ is positive-definite, it can be shown that these eigenvalues are positive.

This means there is an invertible matrix $Q$ whose columns are eigenvectors of $M^{-1}K$ and a matrix $\Omega^2$ which is diagonal and whose diagonal elements are the corresponding eigenvalues of $M^{-1}K$ such that \begin{align} M^{-1}KQ =& Q\Omega^2\\ Q^TKQ=& Q^TMQ\Omega^2\\ Q^TKQ =& \mu \Omega^2 \end{align} Where I've defined $\mu = Q^T M Q$. Note that $\mu$ is symmetric because $M$ is symmetric. Because $K$, $\mu$, and $\Omega^2$ are symmetric we can see $$ \mu\Omega^2 = \Omega^2 \mu $$ so that $$ \mu_{ij} \omega_j^2 = \omega_i^2 \mu_{ij} $$ where $\omega_i^2$ are the diagonal elements of $\Omega^2$ and using the fact that $\Omega^2$ is diagonal. Suppose $i\not= j$. If $\omega_i \not= \omega_j$ we must have $\mu_{ij}= 0$.

Now suppose $i\not=j$ but $\omega_i = \omega_j$, that is $i$ and $j$ correspond to eigenvectors that share an eigenvalue $\omega=\omega_i=\omega_j$. Suppose the multiplicity of the eigenvalue $i$ is $d$. This means there is a $d$-dimensional subspace of vectors $q$ which satsify $$ M^{-1}K q = \omega^2q $$ We can see from above that $$ \mu_{ij} = q_i^T M q_j $$ Because $M$ is positive-definite we can see that $\mu_{ij}$ is an inner-product on this space. We can apply a Gram-Schmidt procedure to this space to find $d$ vectors which are orthogonal with respect to the non-trivial $M$ metric. In other words, we can find vectors $q$ such that $\mu_{ij}=0$ for $i\not = j$. $Q$ should be chosen so that vectors $q$ in degenerate sub-spaces satisfy this condition.

• $Q$ is invertible because $M^{-1}K$ admits a complete set of eigenvectors due to $K$ being symmetric and $M$ being positive definite.
• $\Omega^2$ is diagonal with the eigenvalues of $M^{-1}K$.
• We see above that $Q^TKQ = Q^TMQ\Omega^2 = \mu\Omega^2$
• $Q^TMQ = \mu$ is diagonal as proven above

My answers to the questions in the OP:

1. Yes, the proof in the post is correct (as far as I can tell), but it can be generalized.
2. Yes, the theorem holds even if $M^{1}K$ has degeneracy. In that case a Gram-Schmidt procedure can be used to find eigenvectors which are orthogonal with respect to the non-trivial metric $M$.
3. The theorem does hold if $K$ is only symmetric. But we need not assume $M^{-1}K$ is diagonalizable. We can prove it is diagonalizable from the fact that $M$ is positive definite and $K$ is symmetric. If $K$ is symmetric then the eigenvalues of $\Omega^2$ may be positive, negative, or zero. If $K$ is positive definite then the eigenvalues of $\Omega^2$ are positive.