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Solve $\dot{Q}=-AQ$, $Q(0)=I$. Since $A^T=-A$, differentiation gives $Q^{T}Q=I$. To see why the Lax equation preserves the spectrum, direct differentiation givesThus $L(t)=Q(t)L(0)Q(t{)}^T$ is an orthogonal similarity, so all eigenvalues and all coefficients of its characteristic polynomial are first integrals.

A $2n$-dimensional Hamiltonian system is Liouville integrable when it has $n$ functionally independent first integrals, including the Hamiltonian, whose pairwise Poisson brackets vanish, on an open dense regular region.

For the three-particle periodic Toda lattice, take $p_i={\dot{q}}_i$. The proposed Hamiltonian haswhich are exactly the equations of motion. Differentiating the new variables givesThe transformation is not canonical. For examplerather than canonical coordinate brackets. Moreover $a_1{a}_2{a}_3=1/8$, so the map forgets the common translation of all the $q_i$ and is not an invertible coordinate change on the six-dimensional phase space.

Multiplication of the displayed matrices gives diagonal entries $2(a_i^2-a_{i-1}^2)$ and off-diagonal entries $(b_{i+1}-b_i)a_i$ in $LA-AL$. Therefore they form the required Lax pair. The three characteristic coefficients areThey are independent first integrals, as permitted without proof. Equivalently one may takeIndeed $C_1=-P/2$, $C_2=P^2/8-H/4$ and $C_3=1/4-J/8$. Their mutual Poisson commutation also follows directly: $P$ generates the common translation, and $\{H,J\}=0$ because $J$ is conserved.

A real square matrix $M$ is an orthogonal matrix when $M^{T}M=I$. The planar rotationobeys $R^{T}R=I$ by the Pythagorean trigonometric identity.

For $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, the off-diagonal entry of $RA{R}^T$ is $\frac{1}{2}(a-c)\sin 2\theta +b\cos 2\theta$. ThusIf $(a-c,b)\ne (0,0)$, equivalently $\theta =\frac{1}{2}atan2(-2b,a-c)+k\pi /2$; if $A$ is scalar, every angle works.

Because $RA{R}^T$ is an orthogonal similarity, its diagonal entries are the eigenvaluesThe matrix $C=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ has eigenvalues $3,-1$ along $(1,1{)}^T,(1,-1{)}^T$. Raising its orthogonal diagonalization to the even power gives

For symmetric $A$, apply at stage $k$ the same reflection on both sides, choosing it to zero entries below the first subdiagonal in column $k$. Orthogonal similarity preserves symmetry, so the corresponding row entries vanish too and previous zeros remain. After finitely many stages $QA{Q}^T$ is symmetric tridiagonal. Constructing each reflector uses only arithmetic and one square root.

Starting with $A_0=A$, suppose columns $1,\dots ,j-1$ have already been reduced to tridiagonal form. Let $x$ be the part of column $j$ in rows $j+1,\dots ,n$. A Householder reflection $P_j$ on these coordinates can map $x$ to $\pm \Vert x\Vert e_1$. Extend it by the identity on the first $j$ coordinates and call the resulting orthogonal matrix $U_j$.

The updatezeros every entry below row $j+1$ in column $j$. Because it is an orthogonal similarity, it preserves eigenvalues; because the same transformation is applied on both sides, it preserves symmetry and zeros the corresponding row entries without disturbing earlier columns. After $n-2$ such steps,is symmetric and tridiagonal and has the same eigenvalues as $A$. Every reflector is obtained from finitely many matrix entries using finitely many arithmetic operations and a square root, so this is a finite construction.