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.

The first elimination multipliers are $l_{21}=2$ and $l_{31}=-1$. Subtracting those multiples of row one gives lower rows $(0,2,4)$ and $(0,-1,3)$. The next multiplier is $l_{32}=-1/2$, and the final elimination produces $(0,0,5)$. ThusFor an upper triangular system $Ux=b$ with nonzero diagonal, back substitution computesFor the supplied triangular system, the last row gives $x_3=10/5=2$; the second gives $x_2=(6-4\cdot 2)/2=-1$; the first gives $x_1=3-2(-1)-2(2)=1$. Hence

For $k>0$, tryThe potential satisfies Laplace's equation, and its derivative with respect to $y$ vanishes at $y=-h$. The surface kinematic condition givesThe dynamic condition gives $-B\omega \cosh (kh)+gA=0$. Eliminating the nonzero amplitudes yieldsThus choosing either frequency sign satisfying this relation and $B=A\omega /(k\sinh (kh))$ produces the required harmonic wave.

For a material volume $V(t)$, Newton's second law equates the rate of change of momentum to pressure and body forces:Apply the supplied Reynolds transport theorem to each component of $\mathbf{u}$, and apply the divergence theorem to the pressure force. The result isSince the material volume at the chosen time is arbitrary, the integrand vanishes pointwise for smooth fields. Dividing by positive density gives the Euler equationsAt a fixed time $t$, a streamline is a curve tangent everywhere to the instantaneous velocity. Parametrize it by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t)$. Thenso $f$ is constant along it when the displayed directional derivative vanishes. Here time is held fixed: streamlines need not coincide with particle paths in an unsteady flow.

For a real potential, the time-dependent Schrödinger equation isMultiply it by ${\psi }^{\ast }$, combine with its conjugate multiplied by $\psi$, and note that the real-potential terms cancel. This giveswhere the probability current isThe product rule gives the divergence identity because the gradient cross terms cancel.

For $F(r)=\sin (kr)/r$, the supplied radial identity gives $\Delta F=-k^{2}F$ for $r>0$. The time factor gives $i\hslash {\psi }_t=E\psi$, so the free equation holds forThe apparent origin singularity is removable: $F(0)=k$ and $F(r)=k-k^3{r}^2/6+\cdots$, so the equation extends smoothly there as well. Since the spatial factor is real and its time dependence is a common phase, ${\psi }^{\ast }\nabla \psi =F\nabla F$ is real. Thus $\mathbf{j}=0$, while $\rho =F^2$ is independent of time. Both terms in the continuity equation vanish separately. This is a standing spherical wave; the real spatial factor carries no net radial probability current.