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Use ordinary matrix addition and multiplication on $R=K[A]$. Sums and products of polynomials in $A$ remain such polynomials, and they commute because powers of the same matrix commute. The identity is $A^0=I$. Evaluation $\varphi (\sum c_i{X}^i)=\sum c_i{A}^i$ is a surjective unital algebra homomorphism over $K$. It is unique among maps fixing the scalar matrices and sending $X$ to $A$.

The scalar condition is needed for the word “unique”: as an unrestricted ring-homomorphism assertion over an arbitrary field, prescribing only the image of $X$ does not suffice. For example, over $\mathbb{C}$ with a real scalar matrix $A$, conjugating every coefficient gives another unital ring homomorphism with the same image of $X$. The intended polynomial-evaluation map is the $K$-algebra map just defined.

The kernel of a ring homomorphism is an ideal of the principal ideal domain $K[X]$. It is nonzero because finitely many powers of $A$ must become linearly dependent in the finite-dimensional matrix space. Its unique monic generator is precisely the minimal polynomial $m$: division by $m$ shows that a polynomial killing $A$ has zero remainder, by minimality. Hence $\ker \varphi =(m)$ and $R\cong K[X]/(m)$.

If $\gcd (f,m)=1$, Bézout's identity gives $uf+vm=1$; evaluation gives $u(A)f(A)=I$. Conversely an inverse in $R$ has form $g(A)$, so $fg-1\in (m)$, which forces $\gcd (f,m)=1$. Thus invertibility in this ring is exactly coprimality with the minimal polynomial.

The orthogonal complement is $W^{\perp }=\{v\in V:⟨v,w⟩=0\text{ for every }w\in W\}$. Put $Pv={\sum }_i⟨v,e_i⟩e_i$, so $T=I-P$. Orthonormality gives $P^2=P$, and therefore $T^2=T$. Also $⟨Tv,e_j⟩=0$ for every $j$, so $Tv\in W^{\perp }$. Every $v=Pv+Tv$ is thus a sum from $W$ and $W^{\perp }$, and their intersection is zero, since a vector orthogonal to itself has squared norm zero. This proves $V=W\oplus W^{\perp }$. The decomposition is unique, so its two projection maps depend only on $W$, not on the chosen orthonormal basis.

Finite dimensionality of $W$ cannot simply be dropped. Take the real Hilbert space $V={\ell }^2(\mathbb{N};\mathbb{R})$ and $W=c_{00}$, the finitely supported sequences. This is an infinite-dimensional proper subspace. A vector orthogonal to every coordinate unit vector is zero, so $W^{\perp }=\{0\}$. Yet, for example, $(2^{-j}{)}_{j\ge 1}$ belongs to $V\setminus W$. Hence $W\oplus W^{\perp }=W\ne V$. The failure occurs even though $V$ itself is complete; the proper dense subspace $W$ is not closed.

Use the characteristic parameter $t$ and writeDirect differentiation givesA characteristic beginning at $(s,0,0)$ therefore satisfies $5x+4y+3u=5s$ and $x^2+y^2+u^2=s^2$. Eliminating $s$ givesIts discriminant as a quadratic in $u$ is $100(9x^2+40xy)$. Requiring the root to be zero when $y=0$ selectsThe selected branch is differentiable in each of the two regions where the radical is strictly positive, and substitution into the differentiated quadratic identity verifies the partial differential equation there. On the initial line away from the origin the characteristic vector has nonzero $y$ component $-3s$, so these are the locally unique branches determined by that line.

The maximal real-valued continuation of these branches has the printed domainAt $x=0$ the continuous limiting value is $u(0,y)=3y/4$. The two bounding lines are $x=0$ and $y=-9x/40$, and the included sectors are the one above the sloping line on the right and the one below it on the left.

As in the fold boundary of a characteristic graph, the closed-domain statement concerns the formula's real values. A classical solution requires first derivatives on an open set. Along the nonzero boundary points $9x^2+40xy=0$, the radical's transverse derivative is singular, so the formula is generally not a classical solution there; its two classical domains are the interiors of the shaded sectors. The origin is characteristic for the initial data and does not supply a differentiable neighborhood solution. This qualification preserves the source's stated closed set while distinguishing it from the domain of a classical solution.

Differentiate the Wronskian:It is therefore constant. Linear dependence plainly makes it zero. Conversely, if it is zero, the two initial-data vectors $(u(0),u^{\prime }(0))$ and $(v(0),v^{\prime }(0))$ are linearly dependent. Some constants $A,B$, not both zero, make $Au+Bv$ and its derivative zero at zero. Uniqueness for a linear second-order equation with continuous coefficients then makes $Au+Bv$ identically zero. Thus $\boxed{w=0⟺u,v\text{ are linearly dependent}}$.

For the prescribed integral put $K(x,s)=[u(x)v(s)-u(s)v(x)]/w$. Then $K(x,x)=0$, $K_x(x,x)=-1$, and $K_{xx}(x,s)=-q(x)K(x,s)$. Differentiation under the integral, using continuity of the coefficients and forcing, givesThis proves the required inhomogeneous equation, with the sign of the forcing as printed.

For the explicit forcing take $u=\cosh x$, $v=\sinh x$, so $w=1$ and $K(x,s)=-\sinh (x-s)$. Since $2/(e^s+e^{-s})=\mathrm{sech}s$, the integral isIndeed expand $\sinh (x-s)=\sinh x\cosh s-\cosh x\sinh s$ and integrate $1$ and $\tanh s$. Adding the entire homogeneous solution givesFinally multiply the other equation by $\sin x$ and integrate from zero to $\pi$. Integration by parts givesThus $\boxed{{\int }_0^{\pi }f(s)\sin sds=0}$ is necessary. It is also sufficient: the solution with $y(0)=0$ is $B\sin x-{\int }_0^x\sin (x-s)f(s)ds$, and the integral condition gives its value zero at $\pi$.

The identity theorem for holomorphic functions says that a holomorphic function on a connected open set which has distinct zeros accumulating at a point of the set is identically zero. Equivalently, two holomorphic functions agreeing on such a set of points agree everywhere in that connected open set.

To prove it, expand at an accumulation point $a$. If the first nonzero Taylor coefficient had index $m$, the function would factor as $(z-a{)}^{m}h(z)$ with $h(a)\ne 0$. Continuity would make $h$ nonzero near $a$, so no other zeros could approach $a$, a contradiction. Thus all coefficients vanish and the function vanishes on a neighborhood of $a$. Let $E$ be the set of points having such a zero neighborhood. It is nonempty and open. If a point in the domain is a limit point of $E$, it is either already in $E$ or is an accumulation point of distinct zeros, and the same Taylor argument puts it in $E$. Thus $E$ is relatively closed as well. Connectedness forces $E$ to be the whole domain. Apply this result to the difference of two functions for the equivalent formulation.

An isolated singularity at $a$ means that the function is holomorphic throughout some punctured disc $0<|z-a|<r$. Its residue is the Laurent coefficient $c_{-1}$, equivalently $(2\pi i{)}^{-1}{\int }_{|z-a|=\rho }f(z)dz$. The singularity is removable if all negative Laurent coefficients vanish, a pole of order $k$ if only finitely many are nonzero and the lowest power is $-k$, and essential if infinitely many negative coefficients are nonzero.

For the final contour integral, expand $e^{tz}{e}^{t/z}$. On the circle these exponential series converge absolutely and uniformly for $t$ in every bounded set, so their product can be integrated term by term. Inonly $j-k-2=-1$ contributes. Set $k=m$, $j=m+1$. ThusThe resulting power series converges for every complex $t$ by the ratio test. The coefficients are independent of the positive radius $R$.