A finite-dimensional representation of a group $G$ on a vector space $V$ is a group homomorphism $D:G\to \mathrm{GL}(V)$. Thus each $D(X)$ is invertible andHere the group identity and the identity linear map are different objects, represented by corresponding symbols. Applying the multiplication property to $X{X}^{-1}=X^{-1}X=I$ givesBy the defining property of a matrix inverse,If the definition only states the multiplication law together with invertibility, $D(I{)}^2=D(I)$ already forces $D(I)=I_V$ after multiplication by its inverse.

Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

The relevant weighted inner product is $⟨u,v{⟩}_w={\int }_0^{1}w(x)u(x)v(x)dx$. The normalized eigenfunctions of the Sturm-Liouville problem have the orthonormality propertyHere ${\delta }_{nm}$ is the Kronecker delta: the integral is one for $n=m$ and zero otherwise. The weight is $w$, rather than $p$ or the unweighted measure.

Define $h=f-y^{\prime }{f}_{y^{\prime }}$. For a density without explicit $x$ dependence, the chain rule givesThus the Euler-Lagrange equation implies the Beltrami identityWhere $y^{\prime }\ne 0$ this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full Euler-Lagrange equation, so such candidate paths must still be checked in the original equation.

Insert the Frobenius solution and its differentiated series into the equation. Collecting the terms before shifting an index givesThe coefficient of the lowest power $x^{\sigma -1}$ is $a_0\sigma (\sigma +\gamma -1)$. Because $a_0\ne 0$, the indicial equation givesFor the coefficient of $x^{n+\sigma -1}$ with $n\ge 1$, shift the index in the second sum. The hypergeometric Frobenius recurrence isWhenever the left coefficient is nonzero, division yieldsFor noninteger $\gamma$, the two independent local solutions can therefore be identified aswith a consistent local branch for the fractional power. The hypergeometric function coefficients follow by iterating the recurrence using rising factorials.

There is a parameter qualification to the PDF's statement for all real constants: a zero denominator cannot be divided through. At an undetermined coefficient at Frobenius resonance, use the undivided coefficient equation to test compatibility. For example, $\gamma =0,\sigma =0,n=1$ requires $0=\alpha \beta a_0$, so a pure exponent-zero series with $a_0\ne 0$ is impossible if $\alpha \beta \ne 0$. When $\gamma =1$ the two indicial roots coincide, and a second independent solution generally needs a logarithmic term. The indicial roots and undivided relation remain the valid coefficient conclusions in these exceptional cases.

The origin is an ordinary point of the Hermite differential equation, so put $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$. Substitution and matching powers givesThus the linear recurrence relation isEven and odd coefficients evolve independently from arbitrary $a_0,a_1$. The general power series solution iswith an empty product equal to one. Its first terms areBoth series converge at every finite $x$, since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero polynomial of degree $N$, the highest coefficient in the differential equation requires $\lambda =2N$. Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. HenceHere $H_N$ is a physicists' Hermite polynomial. Two explicit choices areDirect differentiation verifies both. These are polynomial solutions up to an arbitrary overall factor. Without a polynomial requirement or additional boundary conditions, the equation also has nonpolynomial solutions for other real $\lambda$; discreteness here comes from termination.

Differentiate the prescribed imaginary part with respect to $y$:The Cauchy-Riemann equation $u_x=v_y$ integrates toIts $y$ derivative is $u_y=-e^x[(x+1)\sin y+y\cos y]+F^{\prime }(y)$. On the other hand, $-v_x=-e^x[(x+1)\sin y+y\cos y]$. The second Cauchy-Riemann equation therefore requires $F^{\prime }(y)=0$. Combining the real and imaginary parts,Thus the most general answer on a connected domain isA real constant preserves the given imaginary part. Conversely, the difference of any two such analytic functions has zero imaginary part, and its Cauchy-Riemann derivatives force it to be constant. This proves completeness, rather than only exhibiting one possible harmonic conjugate.

The Cauchy-Riemann equations give $u_x=v_y$ and $u_y=-v_x$. ThereforeTaking the nonnegative square root yieldsThe two gradients are also orthogonal, since $\nabla u\cdot \nabla v=v_y{v}_x-v_x{v}_y=0$, although equality of their lengths alone is the requested result.

Since $C$ is unitary, $C^{-1}=C^{†}$. Hence $B=C^{-1}AC=C^{†}AC$ is Hermitian, and so is its positive integer power:It follows that $B^p$ has real eigenvalues. More explicitly, cancellation of adjacent $C{C}^{-1}$ factors gives the similarity transformationIf the real eigenvalues of $A$ are ${\lambda }_j$, the eigenvalues in question are thereforeFor even $p$ they are nonnegative; for odd $p$ their signs can be either. The similarity argument actually works for every invertible $C$, although the Hermiticity argument uses its unitarity.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.