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A real skew-symmetric matrix is an orthogonal matrix exactly when its square is minus the identity matrix. Indeed, $B^{\mathsf{T}}B=-B^2$. Such a matrix can exist only in even dimension: taking determinants of $B^2=-I$ would otherwise give $(\det B{)}^2=-1$. In two dimensions every example isThese are rotation matrices through opposite right angles. Their orthogonal similarity transforms preserve eigenvalues, but do not in general produce a diagonal matrix.

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.

A QR decomposition is a factorization $A=QR$ with $Q^{T}Q=I$ and $R$ upper triangular. An orthogonal transformation preserves the Euclidean norm becauseFor the basic QR algorithm, start with $A_0=A$ and repeatedly formIn a convergent diagonal form, the diagonal entries give the eigenvalues; practical symmetric-matrix implementations use shifts and deflation to obtain and isolate such convergence. The basic unshifted iteration alone need not converge diagonally for every symmetric input.

At every step, however,Thus each iterate is orthogonally similar to its predecessor. Similar matrices have the same characteristic polynomial and eigenvalues, including multiplicities, so induction proves that every iterate has exactly the original eigenvalues. Orthogonal similarity also preserves symmetry.

A QR decomposition is $A=QR$, with $Q$ orthogonal and $R$ upper triangular. Orthogonality gives $Q^{\mathsf{T}}Q=I$, and therefore $\Vert Qx{\Vert }^2=x^{\mathsf{T}}{Q}^{\mathsf{T}}Qx=x^{\mathsf{T}}x=\Vert x{\Vert }^2$, proving invariance of the Euclidean norm.

The unshifted QR algorithm starts with $A_0=A$. At step $k$, factor $A_k=Q_k{R}_k$ and set $A_{k+1}=R_k{Q}_k$. ThenThus consecutive matrices are related by orthogonal similarity, so they have the same characteristic polynomial and eigenvalues. Induction proves every iterate has the original eigenvalues. Symmetry is also preserved. When the iteration converges to diagonal form, its diagonal entries are the eigenvalues; practical implementations use shifts and deflation to improve convergence.

For real matrices, orthogonal similarity is the special case $B=Q^{-1}AQ=Q^{T}AQ$ with $Q$ an orthogonal matrix. Then a symmetric matrix remains symmetric, because $B^T=Q^T{A}^{T}Q=B$. The expression $QA{Q}^T$ describes the reverse conversion and has the same property. An orthonormal basis retains the ordinary dot product formula for its coordinate vectors.

An arbitrary change of basis need not preserve entrywise symmetry. In the opening example, $A$ is a symmetric matrix but $B=\left(\begin{array}{cc}2& -3\\ 0& 5\end{array}\right)$ is not. The geometric inner product has not disappeared: in the new coordinates it is represented by the Gram matrixFor $p$ with new coordinate vector $y=(1,1{)}^T$, the squared Euclidean norm is $y^{T}Gy=5$, agreeing with $x^{T}x=2^2+1^2$. The naive coordinate sum $y^{T}y=2$ would use the wrong inner product.

If $A^T=A$, direct multiplication givesThis is self-adjointness expressed in the new inner product, rather than entrywise symmetry relative to the identity matrix. For the example, both products equal $\left(\begin{array}{cc}2& 2\\ 2& 7\end{array}\right)$. Thus arbitrary matrix similarity preserves the linear operator, while its geometric interpretation requires transforming the Gram matrix as well. Matrix congruence supplies the corresponding rule for a bilinear form and should be distinguished from matrix similarity.