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Put $p(t)={\prod }_{j=1}^m(t-{\mu }_j)$ and define the Lagrange interpolation polynomialsTheir denominators are nonzero and ${\ell }_j({\mu }_i)={\delta }_{ij}$. The polynomial ${\sum }_j{\ell }_j-1$ has degree at most $m-1$ and vanishes at all $m$ distinct scalars, so it is zero. Suppose condition (a) holds, and put $E_j={\ell }_j(T)$. Since $(t-{\mu }_j){\ell }_j(t)$ is a constant multiple of $p(t)$,Thus each $E_{j}v$ lies in $V_j$, and every vector is a sum of such vectors. If ${\sum }_j{v}_j=0$ with $v_j\in V_j$, applying $E_i$ gives $v_i=0$, because ${\ell }_i(T)v_j={\ell }_i({\mu }_j)v_j$. Hence the sum is direct, proving (b). Conversely, on each $V_j$ the factor $T-{\mu }_{j}I$ vanishes; the factors commute, so $p(T)$ vanishes on every summand and therefore on $V$. This proves the equivalence without invoking primary decomposition.

If $ST=TS$ and $v\in V_j$, then $(T-{\mu }_{j}I)Sv=S(T-{\mu }_{j}I)v=0$, so $S[V_j]\subseteq V_j$. Conversely, if each summand is preserved, then on $V_j$ both $ST$ and $TS$ act as ${\mu }_{j}S$. They agree on all summands and hence on $V$. Therefore

For a nonzero principal part the discriminant $\Delta =b^2-ac$ classifies the equation: it is hyperbolic for $\Delta >0$, parabolic for $\Delta =0$, and elliptic for $\Delta <0$. These are the three definitions requested by the inline roman labels.

In the hyperbolic case the characteristic equation $a(dy{)}^2-2bdxdy+c(dx{)}^2=0$ gives two distinct real tangent directions. Locally, with suitably smooth coefficients, solve the corresponding characteristic ordinary differential equations to obtain two independent families of curves. Label their level sets $\varphi =\text{constant}$ and $\psi =\text{constant}$. Their gradients satisfyand the coordinate Jacobian is nonzero. The chain rule therefore eliminates $u_{\varphi \varphi }$ and $u_{\psi \psi }$. The remaining mixed coefficient $2[a{\varphi }_x{\psi }_x+b({\varphi }_x{\psi }_y+{\varphi }_y{\psi }_x)+c{\varphi }_y{\psi }_y]$ is nonzero: otherwise the invertible coordinate change would send the nondegenerate principal quadratic form to zero. Divide by this coefficient and absorb all first-order terms into $g$, giving $u_{\varphi \psi }=g(\varphi ,\psi ,u,u_{\varphi },u_{\psi })$. This canonical reduction is local; a global chart is not automatic.

For the specific equation $a=y,b=0,c=-1$, so it is hyperbolic exactly on $y>0$, elliptic for $y<0$, and degenerate parabolic on $y=0$. For $\varphi =3x-2y^{3/2}$ and $\psi =3x+2y^{3/2}$, the Jacobian is $18\sqrt{y}\ne 0$. Direct differentiation givesSince $\psi -\varphi =4y^{3/2}$, the canonical form isThe characteristic families are $3x-2y^{3/2}=c$ and $3x+2y^{3/2}=c$, with slopes $dx/dy=\pm \sqrt{y}$. They are distinct for $y>0$ and meet the degeneracy line with coincident vertical tangent directions.

PutLeibniz differentiation gives $z^{\prime }=-{\int }_0^x\cos (x-s)f(s)ds$ and $z^{\prime \prime }=-f(x)+{\int }_0^x\sin (x-s)f(s)ds=-f-z$. Thus $z$ satisfies both zero initial conditions and the inhomogeneous equation. Every solution with $y(0)=0$ is $y=A\sin x+z(x)$.

On $[0,1]$, the second endpoint determines $A=-z(1)/\sin 1$ uniquely, since $\sin 1\ne 0$. Collecting the integrals, and using the sine addition formula for $s<x$, givesThis continuous kernel vanishes at the endpoints, satisfies $G_{xx}+G=0$ off $x=s$, and has derivative jump $-1$, as required for $-y^{\prime \prime }-y=f$. The preceding construction proves existence, and a homogeneous solution with both endpoints zero must vanish, proving uniqueness.

On $[0,\pi ]$, instead, the second boundary value becomesThus solvability is equivalent to ${\int }_0^{\pi }f(s)\sin sds=0$. When it holds, all solutions areThere is one free homogeneous mode because the endpoint length is resonant.

On $[-a,a]$ let $F=\Vert f{\Vert }_{\infty }$ and bound $|K|$ by $M$ on the compact integration region. Write $Th(x)={\int }_0^{x}K(x,s)h(s)ds$. The iteration gives $y_n={\sum }_{j=0}^n{T}^{j}f$. Induction, integrating absolute values also when $x<0$, givesThe exponential series is summable uniformly for $|x|\le a$. The Weierstrass M-test therefore proves uniform convergence to a continuous function $y$. Also $\Vert T(y_n-y){\Vert }_{\infty }\le Ma\Vert y_n-y{\Vert }_{\infty }\to 0$, so passing to the limit yields $y=f+Ty$.

For the indicated kernel, twice differentiating its integral givesConversely integrating the differential equation twice, with these initial conditions, gives the integral equation. This is valid also for negative $x$, either by oriented integration or by differentiating the double-integral identity.

For every finite $a$ the continuous kernel $(x-s)s$ meets the preceding hypotheses. Solutions on overlapping intervals agree: if $d=Td$ is their difference, iteration gives $\Vert d{\Vert }_{\infty }\le \Vert d{\Vert }_{\infty }(Ma{)}^j/j!$ for every $j$, forcing $d=0$. The solutions therefore patch to one solution defined on all of $\mathbb{R}$.

For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, a singular point at zero is regular when $xp(x)$ and $x^{2}q(x)$ extend analytically to zero. Hereso these products are analytic at zero. The equation remains singular there because $q$ has a pole.

Substitute $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$. The lowest power gives the indicial equation $\sigma (\sigma -1+\lambda )=0$. Comparing the next powers givesFor $0<\lambda <1$ the denominators do not vanish. With $\sigma =0$, $a_1=-a_0/\lambda$ and $a_2=0$, soWith $\sigma =1-\lambda$, the recurrence becomesConsequently $y_2=a_0{x}^{1-\lambda }(1-x{)}^{\lambda }$ on $0<x<1$, with the binomial series for its last factor. The nonzero coefficient ratios tend to one, giving radius one for that power-series factor. The factor $x^{1-\lambda }$ specifies the Frobenius branch; the different leading powers prove independence.

When $\lambda =1$, both indicial roots become zero and the analytic Frobenius solution in both cases is a multiple of $1-x$. To obtain the missing solution use reduction of order. Away from zero and one, $p=1/x$, henceThus the real general solution on any interval avoiding the singular points isIts Wronskian with $1-x$ is $1/x\ne 0$, proving independence. For complex solutions choose a logarithm branch on a suitable simply connected region.

Expanding gives $(I-E{)}^2-(I-E)=E^2-E$, so either operator is idempotent exactly when the other is. For an idempotent $E$,If $w\in \ker E\cap \mathrm{Im}E$, write $w=Eu$; then $0=Ew=E^{2}u=Eu=w$. Hence $V=\ker E\oplus \mathrm{Im}E$.

Next,If the right side vanishes, left multiplication by $E_1$ gives $E_1{E}_2+E_1{E}_2{E}_1=0$, while right multiplication gives $E_1{E}_2{E}_1+E_2{E}_1=0$. Subtraction shows $E_1{E}_2=E_2{E}_1$, and their original sum now gives $2E_1{E}_2=0$. Over the real or complex numbers this implies $E_1{E}_2=E_2{E}_1=0$.

For three idempotents summing to $I$, each pair sum equals the complementary idempotent $I-E_k$. Thus every product of distinct operators vanishes. Every vector equals $E_{1}v+E_{2}v+E_{3}v$. If $v_1+v_2+v_3=0$ with $v_i\in \mathrm{Im}{E}_i$, applying $E_j$ gives $v_j=0$. ThereforeTake a basis of each summand and unite the three bases. A vector in the $j$th basis has eigenvalue one for $E_j$ and zero for the other two, so this is the requested common eigenbasis. Zero summands simply contribute no basis vectors.

For a nonnegative integer $n$, put $z=e^{i\theta }$ on the unit circle. Since $\cos \theta =(z+z^{-1})/2$ and $d\theta =dz/(iz)$,The residue at zero is the coefficient of $z^{2n}$ in $(1+z^2{)}^{2n}$, namely $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. The residue theorem givesThe contour integrand has no other poles. The formula also includes $n=0$, giving $2\pi$.

Integrate $F(z)=e^{iz}/(z^2+1)$ around the real segment from $-R$ to $R$ and the counterclockwise upper semicircle, with $R>1$. The only enclosed pole is $i$, whose residue is $e^{-1}/(2i)$. On the semicircle, $|e^{iz}|\le 1$ and $|z^2+1|\ge R^2-1$, so its integral has magnitude at most $\pi R/(R^2-1)\to 0$.

The residue theorem thus gives ${\int }_{-\infty }^{\infty }{e}^{ix}/(x^2+1)dx=\pi e^{-1}$. The line integral converges absolutely, so taking its real part is justified:

The inverse image $I={\theta }^{-1}(J)$ is an ideal, by preservation of differences and multiplication. It is proper since $\theta (1)=1\notin J$. If $ab\in I$, then $\theta (a)\theta (b)\in J$; primality of $J$ gives $\theta (a)\in J$ or $\theta (b)\in J$, hence $a\in I$ or $b\in I$. Therefore the inverse image is prime.

For maximality, take the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ and $J=(0)$. This is a maximal ideal of the field $\mathbb{Q}$, but its inverse image is $(0)$ in $\mathbb{Z}$, strictly contained in the proper ideal $2\mathbb{Z}$, so it is not maximal. Preservation of identity matters for proper inverse images: without it the zero homomorphism could pull a proper ideal back to the whole ring.

Use the standard definition that a prime ideal is proper, in addition to the product condition. The quotient operations areIts zero and identity are $I$ and $1+I$, which differ because $I$ is proper. If $(r+I)(s+I)=I$, then $rs\in I$, so primality gives $r\in I$ or $s\in I$. Thus one factor is zero. The commutative quotient ring has no nonzero zero divisors and is therefore an integral domain.

The printed definition omits properness. If that omission were read literally, $I=R$ would satisfy its product condition, but $R/I$ would be the zero ring, which is not an integral domain under the standard convention $1\ne 0$. Properness is the necessary qualification, retained explicitly in this proof.