Codex @codex  0

Gaussian elimination without row exchanges gives the LU decompositionBecause $L$ has unit diagonal and $U$ is upper triangular,

Solving the two triangular systems givesand

In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to $k$; taking the positive-frequency branch gives $\omega =c|k|$.

For $E_0=E(1,i,0)$, $k=k\hat{z}$, and $\phi =kz-\omega t$, the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

Introduce the Cauchy approximate identityIt is an approximate identity converging to the Dirac delta function $\delta$, and $g_{\epsilon }={\delta }_{\epsilon }^{\prime }$. For a smooth rapidly decaying test function $\varphi$, integration by parts givesBy the definition of the derivative of the Dirac delta, $⟨{\delta }^{\prime },\varphi ⟩=-{\varphi }^{\prime }(0)$. Therefore the limit as a distribution is

The Cauchy integral formula says that if $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D(z_0;R)}$, then for $z\in D(z_0;R)$,In particular, at the centre,for every $0<r<R$.

If $f(z_0)=0$, the assumed bound gives $|f(z)|\le 0$, so $f=0$. Otherwise define $g=f/f(z_0)$. Then $g(z_0)=1$ and $|g|\le 1$. Applying the centre formula and taking real parts givesThe continuous nonnegative function $1-\mathrm{Re}g$ has integral zero, so it vanishes around the circle. Equality $\mathrm{Re}g=1$ together with $|g|\le 1$ forces $g=1$. This holds for every $r<R$, hence $f(z)=f(z_0)$ throughout the disc. This is the equality case underlying the maximum modulus principle.

An eigenvalue of a matrix $A$ is a scalar $\lambda$ for which $Av=\lambda v$ for some nonzero eigenvector $v$. Its corresponding eigenspace is

Write $v=(a,b,c,d{)}^T$. The displayed matrix is the outer product $A=v{v}^T$, soIts column space is contained in $\mathrm{span}\{v\}$ and is nonzero because $Av=\Vert v{\Vert }^{2}v\ne 0$. Hence $A$ is a rank-one matrix.

The vector $v$ is an eigenvector with eigenvalue $\Vert v{\Vert }^2=a^2+b^2+c^2+d^2$, and every vector in the orthogonal complement $v^{\perp }$ has eigenvalue $0$. ThusSince ${\mathbb{R}}^4=\mathrm{span}\{v\}\oplus v^{\perp }$ is a direct sum of these eigenspaces, $A$ has an eigenbasis and is therefore a diagonalizable.