For $\mathrm{Re}p>0$, the complex exponential $e^{-pT}$ tends to zero, since its modulus is $e^{-T\mathrm{Re}p}$. Therefore direct integration givesThe condition on the real part also gives absolute convergence of this improper integral.

For the Laplace transform of a derivative, first apply integration by parts on a finite interval:Under the usual transform hypotheses, including sufficient regularity for integration by parts, existence of the derivative's Laplace transform, and $e^{-pT}x(T)\to 0$, passing to the limit givesThese conditions hold, for example, for a continuously differentiable $x$ with $x$ and $\dot{x}$ of exponential order, in a sufficiently far right half-plane. Existence of $\bar{x}(p)$ alone does not establish all of these derivative-transform hypotheses; the finite-interval identity displays exactly which boundary term must disappear. The polynomial solution in the last part satisfies them for every $\mathrm{Re}p>0$.

A holomorphic function on the open set $\Omega$ is complex differentiable at every point of $\Omega$: the limitexists independently of the direction of approach. Write $f(x+iy)=u(x,y)+iv(x,y)$. The Cauchy-Riemann equations areTo prove necessity, approach first with real $h$, obtaining $f^{\prime }(z)=u_x+i{v}_x$. Approach instead with $h=it$ for real $t$, obtainingEquality of real and imaginary parts yields the displayed Cauchy-Riemann equations.

A sufficient condition is that $u$ and $v$ have continuous first partial derivatives throughout $\Omega$ and satisfy the Cauchy-Riemann equations there. Indeed, real differentiability gives, for $h=a+ib$,The Cauchy-Riemann equations imply $f_y=i{f}_x$, so this is $f_{x}h+o(|h|)$. Dividing by $h$ proves complex differentiability and $f^{\prime }(z)=f_x(z)$ at every point. Merely having first partial derivatives and the equations at a point, without a suitable differentiability hypothesis, would not justify this argument.

Put $F={\mathbb{R}}^n\setminus \Omega$. This is a closed set. First assume $F$ is nonempty. The Heine-Borel theorem makes the nonempty bounded closed set $K$ a compact set. By the preceding parts, $x\mapsto d(x,F)$ is a continuous function and is strictly positive at every $x\in K$, because $K\subset \Omega$.

The extreme value theorem gives $x_{\ast }\in K$ attaining its minimum. ThereforeThe equality with the distance between two sets follows by taking the two infima successively. Crucially, compactness supplies a point attaining the minimum; positivity at each point alone would not make an arbitrary infimum positive. This is the principle that a compact subset of an open set has positive distance from its complement.

If $\Omega ={\mathbb{R}}^n$, its complement is empty. With the usual extended-real convention $\inf \varnothing =+\infty$, the stated inequality holds in that case too.

A subset $\Omega \subseteq {\mathbb{R}}^n$ is an open set if every $x\in \Omega$ admits a radius $r>0$ such that the open ballis contained in $\Omega$. A subset $F\subseteq {\mathbb{R}}^n$ is a closed set if its complement ${\mathbb{R}}^n\setminus F$ is an open set. Equivalently, $F$ contains the limit of every convergent sequence of its points.

Set $X=0$ to define a surjective ring homomorphismIts kernel of a ring homomorphism is exactly $(X)$: the constant coefficient in $X$ must vanish, so every remaining monomial is divisible by $X$. The first isomorphism theorem for rings therefore givesThe polynomial ring $\mathbb{Q}[Y]$ is an integral domain, because the product of two nonzero polynomials has nonzero leading coefficient. It is not a field: $Y$ has no polynomial multiplicative inverse, since a nonzero product $YQ(Y)$ has positive degree of a polynomial. Consequently $(X)$ is prime and is not maximal. Concretely, $(X)⊊(X,Y)⊊\mathbb{Q}[X,Y]$ is a strict chain of ideals.

If the quotient ring $R/I$ is an integral domain, then $I\ne R$, because an integral domain has $1\ne 0$. For $ab\in I$,There are no nonzero zero divisors in an integral domain, so $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$, proving $I$ is a prime ideal.

If $R/I$ is a field, again $I$ is proper. Let $J$ be an ideal properly containing $I$, and choose $a\in J\setminus I$. The nonzero element $a+I$ has a multiplicative inverse $b+I$, so $ab-1\in I\subset J$. Since $ab\in J$, subtraction gives $1\in J$, hence $J=R$. Therefore $I$ is a maximal ideal. These arguments establish the relevant prime ideal quotient criterion and maximal ideal quotient criterion.

A prime ideal is a proper ideal $I\ne R$ such that $ab\in I$ implies $a\in I$ or $b\in I$. A maximal ideal is a proper ideal $I\ne R$ such that there is no ideal strictly between $I$ and $R$.

Every ideal of the integers has the form $(n)=n\mathbb{Z}$ for a unique nonnegative integer $n$. Its prime ideals and maximal ideals are respectivelyThe zero ideal is a prime ideal because the integers form an integral domain. For a prime number $p$, divisibility of $ab$ by $p$ forces divisibility of one factor, so $(p)$ is a prime ideal. Also any ideal $(d)$ containing $(p)$ has $d\mid p$, so it equals $(p)$ or $(1)$; thus $(p)$ is a maximal ideal.

If $n>1$ is composite, write $n=ab$ with $1<a,b<n$. Then $ab\in (n)$ while neither factor belongs to $(n)$, so $(n)$ is not a prime ideal. A prime number $p\mid n$ gives $(n)⊊(p)⊊\mathbb{Z}$, so $(n)$ is not a maximal ideal either. The zero ideal is not maximal since $(0)⊊(2)⊊\mathbb{Z}$, and $(1)=\mathbb{Z}$ is not proper.

A diagonalizable matrix over the field $F$ admits an invertible matrix $P$ with entries in $F$ such thatEquivalently, the vector space $F^n$ has a basis of eigenvectors of $A$. Indeed, $AP=PD$ says that each column of $P$ is an eigenvector with the corresponding diagonal entry as its eigenvalue. Requiring the basis and eigenvalues to lie in $F$ makes the choice of field essential.

For a linear functional $\lambda \in V^{\prime }$, the definition of the dual map givesThus the kernel of a linear map is precisely the annihilator of a vector subspaceWrite $n=\dim U$, $m=\dim V$ and $r=\dim \mathrm{im}T$. The rank-nullity theorem, the dimension formula for an annihilator of a vector subspace, and $\dim V^{\prime }=m$ giveApplying the same dimension formula in $U$ givesFinally, if $\alpha =T^{\prime }\lambda$ and $u\in \ker T$, then $\alpha (u)=\lambda (Tu)=\lambda (0)=0$. Therefore $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^0$. These are finite-dimensional vector subspaces of equal dimension, so inclusion is equality:This is the dual image and kernel annihilator identity.

The annihilator of a vector subspace consists of the linear functionals that vanish on every element of that vector subspace:For the dimension identity, choose a basis $w_1,\dots ,w_k$ of $W$ and use basis extension to obtain $w_1,\dots ,w_n$ for $V$. A linear functional is determined by its values on these $n$ basis vectors. Belonging to the annihilator of a vector subspace sets its first $k$ values to zero and leaves its last $n-k$ values arbitrary. Equivalently the last $n-k$ elements of the dual basis form a basis of $W^0$.

The dual space consists of the linear functionals on the original vector space, with addition and scalar multiplication defined pointwise:The dual map acts by function composition with $T$:For $u\in U$, its value is $(T^{\prime }\lambda )(u)=\lambda (Tu)$. The composite is a linear functional because both $T$ and $\lambda$ are linear maps. Moreover $T^{\prime }(a\lambda +b\mu )=a{T}^{\prime }\lambda +b{T}^{\prime }\mu$, so the dual map is itself a linear map. Its direction is reversed because a measurement on the output of $T$ becomes a measurement on its input.

A pole of order $m\ge 1$ at $a$ meansEquivalently the lowest nonzero power in its Laurent series is $(z-a{)}^{-m}$. In particular, $(z-a{)}^{m}f(z)$ tends to a finite nonzero limit.