TITLE: Diagonalization via the Toda flow
QUESTION [10 upvotes]: According to some almost indecipherable notes of a graduate Linear Algebra class, a symmetric matrix $A\in\mathbb R^{n\times n}$ can be diagonalised via the Toda flow. More specifically, if $X=X(t)\in\mathbb R^{n\times n}$ is the solution of the initial value problem
($n^2\times n^2$)
\begin{equation}
\frac{dX}{dt} \,=\, B(X)X-XB(X), \quad X(0)\,=\, A,
\end{equation}
where $B(X)=X_+-(X_+)^T=-B(X)^T$, and $X_+$ is the strict
upper part of $X$:
$$
X_+ \,=\, \left(\!
\begin{array}{cccccccc}
0 & x_{12} & x_{13} & x_{14} &\cdots & x_{1n}\\
   &  0      & x_{23} & x_{24} &\cdots & x_{2n}\\
   &          &  0 & x_{34} & \cdots & x_{3n}  \\
   &&&\ddots&& \vdots \\
 & & & &0&x_{n-1,n} \\
& & &&& 0
\end{array}
\!\right),
$$
then
$$
\lim_{t\to\infty} X(t)=\varLambda,
$$
where $\varLambda$ is a diagonal matrix containing the spectrum of $A$.
Is there any reference for this fact?
Update. This is due to Moser: 
J. MOSER, Finitely many mass points on the line under the influence of an exponential 
potential-an integrable system, in “Dynamic Systems Theory and Applications” 
(J. Moser, Ed.), pp. 467-497, Springer-Verlag, New York/Berlin, 1975. 
I was wondering if its proof is accessible anywhere.

REPLY [17 votes]: $\def\Tr{\mathrm{Tr}}$This proof is short enough that I thought I'd just write it out.
On a skim, this looks like the same proof that Christian Remling pointed you to, and which Deift-Li-Tomei say is the same as the proof of Moser.
Disclaimer: all signs in this argument have at best a 55% chance of being right.
First of all, let $C$ be any continuous function at all from $n \times n$ matrices to $n \times n$ matrices and define $Y(t)$ by the ODE
$$\frac{dY}{dt} = C(Y) Y - Y C(Y).$$
Then
$$\frac{d \Tr(Y^m)}{dt} = \Tr \left( \frac{dY}{dt} Y^{m-1} + Y \frac{dY}{dt} Y^{m-2} + \cdots + Y^{m-1} \frac{dY}{dt} \right)$$
$$=\Tr \left(  C(Y) Y^m - Y C(Y) Y^{m-1} + Y C(Y) Y^{m-1} - Y^2 C(Y) Y^{m-2} + \cdots + Y^{m-1} C(Y) Y  - Y^m C(Y) \right) = \Tr\left( C(Y) Y^m - Y^m C(Y) \right)=0.$$
So $\Tr(Y^m)$ is constant and all the $Y(t)$'s have the same spectrum.
Also, if $Y$ is symmetric and $C(Y)$ is skew-symmetric, then $C(Y) Y - Y C(Y)$ is symmetric, so symmetric matrices stay symmetric. 
Now, we specialize to the case of Toda flow. Let $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$ be the spectrum of $X$. A quick computation shows that
$$\frac{d X_{kk}}{dt} = - \sum_{i< k} X_{ik}^2 + \sum_{i>k} X_{ik}^2.$$
So
$$\frac{d (X_{11}+X_{22} + \cdots + X_{kk})}{dt} = \sum_{i \leq k,\ j > k} X_{ij}^2.$$
So all the quantities $X_{11}+X_{22} + \cdots + X_{kk}$ are increasing.
Since $X$ is symmetric we have $X_{ii} \leq \lambda_1$ (an inequality of Schur), so $X_{11} + \cdots + X_{kk}$ is bounded above and we conclude that $\lim_{t \to \infty} X_{11} + \cdots + X_{kk}$ exists. As a result, $\lim_{t \to \infty} X_{kk}$ exists, call it $\mu_k$.
Also, we see that $\lim_{t \to \infty} \sum_{i \leq k,\ j > k} X_{ij}^2 =0$ and we thus deduce that $\lim_{t \to \infty} X_{ij} =0$ for each $i \neq j$. So $\lim_{t \to \infty} X$ is a diagonal matrix, with diagonal entries $\mu_i$, and the same spectrum as $X$. So the $\mu$'s are a permutation of the $\lambda$'s.
Finally, we want to know in what order the $\lambda$'s occur. We can't answer this in general: all the diagonal matrices are fixed points of the flow. However, I claim that $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_n$ is the only stable fixed point. Proof: If $\mu_i < \mu_{i+1}$, then a tiny perturbation in direction $e_{i,i+1} + e_{i+1, i}$ is magnified, where $e_{i,j}$ is the matrix whose unique nonzero entry is a $1$ in position $(i,j)$. So almost all matrices flow to $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_n$.

On Toda flow and Morse flow The exact same proof works if 
$$B(X)_{ij} = c_{ij} X_{ij}$$
for any skew symmetric matrix $c$ with positive entries above the diagonal. In another answer, I work out that the Morse flow for the function $\psi(X) = \sum a_i X_{ii}$ is given by this equation with $c_{ij} = a_i - a_j$. (The metric on the set of matrices with fixed spectrum is induced by the $SO(n)$ action, and the inner product on $\mathfrak{so}(n)$ is the standard one.) So Toda flow would be Morse flow if we could arrange that $a_i -a_j = 1$ for all $i<j$. This is possible for tridiagonal matrices (a very cool lemma is that Toda flow preserves the property of having $X_{ij} = 0$ for $|i-j|>k$), but not in general. Still, I can imagine a fake history where Toda flow was discovered by writing down Morse flow for $\psi$ and then noticing that it still worked for any $c_{ij}$.