The minimal polynomial of $T$ is the monic nonzero polynomial of smallest degree annihilating $T$. Such a polynomial exists, since sufficiently many powers of $T$ are linearly dependent in the finite-dimensional space $\mathrm{End}(V)$. If two monic annihilators had the same least degree, their difference would be an annihilator of smaller degree and hence zero. Thus it is unique. Division with remainder also shows that every annihilating polynomial is a multiple of it.

The primary decomposition theorem states that if $m={\prod }_{j=1}^r{f}_j^{e_j}$ with distinct monic irreducible polynomials, thenThese subspaces are $T$-invariant, and the restriction to $V_j$ has minimal polynomial $f_j^{e_j}$.

Let $r=\gcd (q,m)$, chosen monic. The Bézout's identity for polynomials gives $r=aq+bm$. If $q(T)v=0$, then $r(T)v=a(T)q(T)v+b(T)m(T)v=0$. Conversely, $q=rh$, so $r(T)v=0$ implies $q(T)v=0$. ThereforeThe following Roman scopes correspond to the three labelled conditions, with their implications proved from the indicated condition.

The dual basis is $E^{\ast }=\{e_1^{\ast },\dots ,e_n^{\ast }\}$, where $e_i^{\ast }(e_j)={\delta }_{ij}$ and each coordinate linear functional extends linearly. Thus $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$. These functionals are a basis of $V^{\ast }$ because a linear functional is uniquely determined by its values on $E$.

For a linear subspace $U$ of dimension $k$, extend a basis $u_1,\dots ,u_k$ to one of $V$. In its dual basis, the annihilator of a vector subspace $U^{\circ }$ is spanned by the last $n-k$ coordinate functionals. Hence $\boxed{\dim U+\dim U^{\circ }=\dim V}$.

The transpose in the source is the algebraic dual map $T^{\ast }:V^{\ast }\to V^{\ast }$ defined by $T^{\ast }f=f\circ T$. Composition of linear maps is linear, so $T^{\ast }f$ is a linear functional; and $T^{\ast }(af+bg)=a{T}^{\ast }f+b{T}^{\ast }g$, proving linearity in its input. If $T(e_j)={\sum }_i{a}_{ij}{e}_i$, thenThe coefficient of $e_j^{\ast }$ in $T^{\ast }{e}_i^{\ast }$ is consequently $a_{ij}$, so $\boxed{[T^{\ast }{]}_{E^{\ast }}=([T{]}_E{)}^{\mathsf{T}}}$. No inner product is involved.

A linear functional $f$ annihilates $\mathrm{Im}T$ exactly when $f(Tv)=0$ for all $v$, equivalently $T^{\ast }f=0$. Thus $(\mathrm{Im}T{)}^{\circ }=\ker T^{\ast }$. If $v\in \ker T$, then $(T^{\ast }f)(v)=f(Tv)=0$, giving $\mathrm{Im}{T}^{\ast }\subseteq (\ker T{)}^{\circ }$. The first equality and the dimension formula give $\dim \ker T^{\ast }=n-\dim \mathrm{Im}T$. By the rank-nullity theorem,The inclusion is therefore an equality:

For a second-order linear partial differential equation with principal part $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}$, the discriminant is $B^2-AC$. Here $A=1$, $B=0$, $C=-x^2$, soThe equation is therefore a hyperbolic partial differential equation throughout its domain. The lower-order term $-u_x/x$ does not affect this classification.

The characteristic curves satisfyIntegrating gives the two families $y\mp x^2/2=\text{constant}$. Choose coordinates constant along these families:Their Jacobian determinant is $8x\ne 0$, so they define an invertible change of variables on $x>0$, with image $\xi +\eta >0$. Write $u(x,y)=U(\xi ,\eta )$. The chain rule givesBoth the pure second partial derivatives and the first partial derivatives cancel in the original equation, leavingThis is the full canonical form of a second-order partial differential equation, including the effect of the lower-order term. Integrating shows $U_{\xi }$ is a function of $\xi$ alone, hencewith arbitrary twice differentiable functions $f,g$. The representation is global on $x>0$: for each fixed $\xi$, the allowed $\eta >-\xi$ form an interval, so $U_{\xi }$ is independent of $\eta$ throughout that interval; after subtracting its antiderivative, the same reasoning in the other coordinate leaves only $g(\eta )$. These are quadratic characteristic coordinates for a variable-speed wave equation.

For the data on $x=1$, set $s=1-2y$, so the other argument is $2-s$. The two initial conditions becomeDifferentiating the first equation with respect to $s$ also gives $f^{\prime }(s)-g^{\prime }(2-s)=0$. Solving these two linear equations yields $f^{\prime }(s)=g^{\prime }(2-s)=1/2$. Since $s$ ranges over all real numbers,ConsequentlyIndeed $u_{xx}=2$, $u_{yy}=0$, $u_x=2x$, so the equation and both initial conditions hold directly.

Near $x=0$, separation of variables gives $dy/y=dx/(1-x)$, so $\log |y|=-\log (1-x)+C$. The initial condition selectsDirect substitution verifies both the ordinary differential equation and the initial condition. The endpoint $x=1$ is excluded because the coefficient and this solution are singular there; the displayed interval is the maximal connected real interval containing the initial point.

For $|x|\le h<1$ and any two admissible values $y,\tilde{y}$,This is a Lipschitz condition in the dependent variable, uniform in $x$, with Lipschitz constant $L=(1-h{)}^{-1}$. Also $|f(x,y)|\le (1+k)/(1-h)$ on the prescribed rectangle, so the usual local hypotheses of the Picard-Lindelöf theorem hold.

Integrating from the initial point gives the integral equationConversely, differentiating this equation using the fundamental theorem of calculus recovers the initial value problem. Start Picard iteration with the constant initial approximation $y_0(x)=1$ and setThe first three approximations areTo prove the general expression by mathematical induction, put $\ell (x)=\log (1-x)$. The initial approximation gives the formula for $n=0$. If it holds for $n$, then the substitution rulegivesThus the formula follows at every stage.

For every fixed real $x<1$, these are partial sums of the exponential series evaluated at the finite real number $-\log (1-x)$. ThereforeThis range is larger than $|x|<1$: the expansion is in powers of $\log (1-x)$, not a Taylor series in $x$. For example it converges at $x=-2$. It also converges uniformly on every compact interval contained in $(-\infty ,1)$, because $|\log (1-x)|$ is bounded there and the exponential series supplies a uniform summable bound. At $x=1$ the approximations are undefined, while for $x>1$ the real integral from zero crosses the singularity. This is Picard iteration for a simple-pole linear differential equation.

A conformal map preserves oriented angles between intersecting smooth curves at each point, with a nonzero local scale factor. A sufficient criterion is that $f$ be a holomorphic function on the domain and that $f^{\prime }(z)\ne 0$ everywhere there. Its first-order action then multiplies every tangent direction by the same nonzero complex number, combining a rotation and a dilation. A conformal bijection additionally requires injectivity and surjectivity onto the target.

For $z=x+iy$ in the given half-strip,Thus $0<|w|<1$ and $0<\arg w<\pi$. Conversely every such $w$ has a unique argument in $(0,\pi )$ and unique radius in $(0,1)$, giving $x=\arg w$, $y=-\log |w|$. HenceThe complex exponential is injective on $S$, and its derivative $i{e}^{iz}$ never vanishes.

On $D_{+}$, use the Joukowski map with a sign change,Sinceit maps into the upper half-plane. To prove injectivity, $J(w_1)=J(w_2)$ givesBoth moduli are below one, so $w_1{w}_2\ne 1$ and $w_1=w_2$.

For surjectivity, take any $\zeta$ in the upper half-plane and solve $w^2+\zeta w+1=0$. The two roots have product one. Neither can lie on the unit circle, since there $-(w+w^{-1})$ is real. Exactly one root therefore has modulus below one. For this root, the displayed imaginary-part identity forces $\mathrm{Im}w>0$, so it belongs to $D_{+}$. Finally $J^{\prime }(w)=w^{-2}-1$ has no zero in $D_{+}$. This proves the Joukowski map from the upper half-disc to the upper half-plane is a conformal bijection.

Composing gives the required conformal mapIts derivative $2\sin z$ is nonzero on $S$, since the zeros of the complex sine are the real integer multiples of $\pi$.