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The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

In vacuum, curl Faraday's law and use $\nabla \cdot E=0$ and the Ampère-Maxwell equation to obtainThe plane wave solves this when $\omega =c|k|$ and $k\cdot E_0=0$. Faraday's law gives

For the stated wave, propagation is along $+x$ and polarization is $(0,1,1)/\sqrt{2}$. ThusThe Poynting vector is electromagnetic energy flux, and its average is the wave intensity.

Separation with Dirichlet walls givesFor $a<b<c$, the ground state $(1,1,1)$ has a nondegenerate energy. The cheapest excitation changes the quantum number in the longest direction, so the first excited state $(1,1,2)$ also has a nondegenerate energy.

If $a<b=c$, the ground-state energy remains nondegenerate, but $(1,2,1)$ and $(1,1,2)$ have equal first-excited energy, giving degeneracy two.

The local maximum modulus principle says that if $|f|$ has a local maximum at an interior point of a connected domain, then $f$ is constant. On a small circle about the maximum, the mean-value property andforce equality everywhere. Equality in the triangle inequality makes the boundary values identical; Cauchy's formula then makes $f$ constant locally, and the identity theorem makes it constant on the domain.

Write $f=u+iv$. The hypothesis gives $u-v\le 0$. Thereforewith equality at zero. The local maximum modulus principle makes the exponential constant. Differentiation then gives $f^{\prime }=0$, so $f$ is constant; since $f(0)=0$, it is identically zero.

A contraction satisfies $d(Tx,Ty)\le qd(x,y)$ for some $0\le q<1$. Starting from $x_0$, define $x_{k+1}=T{x}_k$. The geometric bound on successive distances makes $(x_k)$ Cauchy, so completeness gives a limit $x$. Continuity of $T$ gives $Tx=x$. If $Ty=y$, then $d(x,y)\le qd(x,y)$, hence $x=y$. This is the contraction mapping theorem.

If $T^n$ is a contraction, it has a unique fixed point $x$. Since $T^n(Tx)=T(T^{n}x)=Tx$, the point $Tx$ is also fixed by $T^n$, so uniqueness gives $Tx=x$. Every fixed point of $T$ is fixed by $T^n$, so it too must equal $x$. Thus $T$ has exactly one fixed point.

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.