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An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

For $m=1$, $Y=(1-{\eta }^2)/2$. Expanding both sides of the perturbation equation givesFor $y={\sum }_{r\ge 0}{a}_r{\eta }^r$, coefficient comparison givesThe even and odd series terminate exactly when $6\lambda =k(k+1)$ with the corresponding parity. Thus every positive integer $k$ gives the regular polynomial solution, a Legendre polynomial up to scale.

Write $p={\sum }_{j=0}^n{a}_j{x}^{n-j}{y}^j$. Equating the coefficient of $x^{n-j-2}{y}^j$ in ${\nabla }^{2}p$ givesThus all even coefficients are determined by $a_0$ and all odd coefficients by $a_1$, with no further constraints. These two choices give two independent harmonic homogeneous polynomials.

Green's theorem states, for a positively oriented simple boundary $C=\partial R$,The defining polynomial becomesso the region is the cardioid

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$. For each degree, the coefficient tuples for polynomials in $A[x]$ form a finite Cartesian power of the countable set $A$; a countable union over degrees is countable. In particular the algebraic numbers, being roots of countably many integer polynomials with finitely many roots each, are countable. Since $\mathbb{R}$ is uncountable, uncountably many real numbers are transcendental.

The set $Q$ is uncountable. Partition $N$ into four-element blocks and, independently on each block, choose either of two fixed-point-free permutations. Binary sequences then inject into $Q$.

Finally choose a line $L$ distinct from and nonparallel to every $L_i$, possible because only countably many directions are excluded. Each $L\cap L_i$ has at most one point, so the union covers only countably many points of the uncountable line $L$. It cannot cover the plane.

Choose an entire $F$ with real part $u$. The bound implies that $e^F$ has at most polynomial growth, so Cauchy's estimates make it a polynomial. It has no zeros, hence that polynomial is constant. Therefore $F$, and in particular $u$, is constant.

The first characteristic polynomial factors asThe root condition holds precisely for $0<\alpha <2$: in this range the two reciprocal roots are distinct and on the unit circle; at either endpoint a unit root is repeated, and outside it one root has modulus greater than one. Consistency is given, so these and only these values are convergent. The exceptional order-three value $\alpha =6$ is outside this interval; consequently every convergent case has order two.

If $f$ had a pole at one of the finitely many candidate points, choose a polynomial vanishing to sufficiently high order at all the other candidates. Hermite interpolation lets its remaining Taylor coefficients be chosen so that the residue of $p^{2}f$ at the selected pole is nonzero. A small contour around that pole would then have nonzero integral by the residue theorem, contradicting the hypothesis. Hence none of the candidate points is a pole and $f$ is entire.

If a prime divided every coefficient of $fg$, reducing modulo that prime would make the product of the two nonzero reduced primitive polynomials zero in a domain, impossible. Thus a product of primitive polynomials is primitive. Factoring the contents from arbitrary $f,g$ then givesfor a unit $u$.

The content $c(f)$ is a greatest common divisor of the coefficients, defined up to a unit. A polynomial is primitive when its content is a unit.

Regard the polynomial as an element of $\mathbb{C}[x][y]$:The prime $x-1$ divides every nonleading coefficient, its square does not divide the constant coefficient, and it does not divide the leading coefficient $x^3$. Eisenstein's criterion proves irreducibility over $\mathbb{C}(x)[y]$; the polynomial is primitive in $\mathbb{C}[x][y]$, so Gauss's lemma proves irreducibility there.

If the monic orthogonal polynomial $p_n$ had fewer than $n$ distinct sign-changing zeros in $(-1,1)$, multiply the factors corresponding to those zeros to form a polynomial $q$ of degree below $n$. Then $p_{n}q$ has one sign and is not identically zero, so its positive-weight integral cannot vanish, contradicting orthogonality. Thus all $n$ zeros are distinct and lie in the open interval.

The three-term recurrence theorem with the displayed coefficients identifies $P_n$ with the monic orthogonal polynomial $p_n$. Part (b) says its zeros are precisely the eigenvalues of the real symmetric Jacobi matrix $A_n$, and part (a) says those zeros are distinct and all belong to $(-1,1)$.

The minimal polynomial $m_{\varphi }$ is the unique monic polynomial of least positive degree with $m_{\varphi }(\varphi )=0$. Uniqueness follows because division of one monic annihilator by another would produce a lower-degree annihilator. Cayley–Hamilton says that the characteristic polynomial annihilates $\varphi$; division then shows