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Let $F(n)={\prod }_{j=1}^k(a_{j}n+b_j)$ be a product of fixed pairwise nonproportional linear polynomials with positive integer leading coefficients. Assume admissibility: no prime number divides $F(n)$ for every integer $n$. There are constants $\delta ,c>0$ such that, for all large $X$, at least $cX/(\log X{)}^k$ values $1\le n\le X$ have no prime factor of $F(n)$ below $X^{\delta }$. These are almost prime values with a bounded total number of prime factors, counted with multiplicity.

A rigorous positive count needs a lower-bound sieve input, obtainable from the fundamental lemma of sieve theory. If ${\rho }_F(d)$ counts roots modulo squarefree $d$, then the polynomial root density in a sieve gives $X{\rho }_F(d)/d+O({\rho }_F(d))$. Take distribution level $D=X^{1/2}$ and sifting limit $z=X^{\delta }$. The total remainder up to $D$ is $O_k(D(\log (2D){)}^k)$, while Mertens' theorems give ${\prod }_{p<z}(1-{\rho }_F(p)/p){\asymp }_F(\log z{)}^{-k}$. Choosing $\delta$ sufficiently small makes the main-term error in the fundamental lemma of sieve theory less than half the main term, proving the lower count.^[ref]

Since $|F(n)|{\ll }_F{X}^k$, each surviving nonzero value has at most $k/\delta +O_F(1)$ prime factors, with multiplicity. The Buchstab identity organizes the removal of successive prime factors, and upper-bound sieves control removed terms; merely observing convergence of a prime tail does not establish that its subtraction leaves a positive main term.

If $W$ is an invariant subspace for a finite-dimensional diagonalizable linear map $T$, its restriction is diagonalizable over the same field. The minimal polynomial of $T{|}_W$ divides that of $T$: an annihilating polynomial for $T$ still annihilates the restriction, and polynomial division by its minimal polynomial proves divisibility. A divisor of a product of distinct linear polynomials remains a product of distinct linear polynomials, proving the assertion. This is the key step in simultaneous diagonalization of commuting linear maps.

A linear polynomial in one variable over a field is $a_{1}x+a_0$ with $a_1\ne 0$, so its polynomial degree is one. It has the unique root $x=-a_0/a_1$. For example, $3x-6$ has root two.

As a function, it is affine; it is a linear map only when $a_0=0$. In several variables, an expression $a_0+{\sum }_i{a}_i{x}_i$ is a degree-one polynomial if some $a_i\ne 0$. Over a general ring, a nonzero leading coefficient need not be invertible, so the field root formula may be unavailable.

For a linear projection $P$ on a nonzero finite-dimensional vector space, $V=\ker P\oplus \mathrm{im}P$. The linear map $\alpha I+\beta P$ acts by the scalar $\alpha$ on the first summand and $\alpha +\beta$ on the second. Its minimal polynomial is the product of the distinct corresponding linear polynomials whose summands are nonzero. If $\beta =0$ this is $x-\alpha$; if $P=0$ or $P=I$ only the associated single factor remains. Otherwise, for $\beta \ne 0$, it is $(x-\alpha )(x-\alpha -\beta )$. This follows directly from an adapted basis, so it works in every characteristic.

The same invariant subspaces work for $S=\alpha I+\beta T$: it acts by $\alpha$ on $\ker T$ and by $\alpha +\beta$ on $\mathrm{im}T$. Its minimal polynomial is therefore the product of the distinct linear polynomials belonging to the nonzero summands. Explicitly,When $\beta \ne 0$, the two candidate eigenvalues are distinct in any field. This is the minimal polynomial of an affine function of a projection.

In the finite field ${\mathbb{F}}_3$, reduce the characteristic polynomial to $(x-1{)}^3$. For $B=A-I$, reduction of the entries givesHenceIts repeated linear polynomial prevents diagonalizability. Equivalently, a diagonalizable map with only the eigenvalue one would have to be $I$, and this matrix is not $I$.

Over the complex numbers, the three eigenvalues areThey are distinct. Each is a root of the minimal polynomial, so here the minimal polynomial equals the characteristic polynomial and has distinct linear polynomials. Consequently $A$ is diagonalizable over $\mathbb{C}$.

A linear endomorphism is diagonalizable over its field precisely when its minimal polynomial is a product of distinct linear polynomials over that same field. Splitting alone is insufficient: repeated factors permit nontrivial Jordan blocks.

For the displayed matrix, a determinant expansion givesThese formulas can be reduced in each requested characteristic. The four cases below distinguish failure to split from failure to have distinct factors.

On each nonzero eigenspace $V_{\lambda }$, the restricted linear map $S_{\lambda }=S{|}_{V_{\lambda }}$ is diagonalizable. Indeed $m_S(S_{\lambda })=0$, so the minimal polynomial of $S_{\lambda }$ divides $m_S$. A divisor of a product of distinct linear polynomials has the same property, proving the diagonalizable restriction lemma.

Choose a basis of $S_{\lambda }$-eigenvectors in each $V_{\lambda }$. Every chosen vector is simultaneously a $T$-eigenvector, with eigenvalue $\lambda$. The direct sum in part (ii) shows that joining these bases gives a basis of $V$. Therefore $S$ and $T$ have a common eigenbasis, which is exactly simultaneous diagonalization.

A cubic polynomial over a field is reducible if and only if it has a root in that field: a nontrivial factorization of degree three has a linear polynomial factor, and conversely a root supplies one by the factor theorem.

The rational root theorem, or part (b), shows that a rational root of this monic polynomial must be an integer. Such an integer must divide the constant coefficient $4$, so the only candidates are $\pm 1,\pm 2,\pm 4$. Direct evaluation givesNone vanishes. Hence there is no rational root, provingThe same no-root argument alone would not decide a quartic polynomial, which could factor as two quadratic polynomials; its sufficiency here uses degree three.

Use the reduction criterion: a primitive integer-coefficient polynomial whose leading coefficient remains nonzero modulo a prime number $p$ is irreducible over $\mathbb{Q}$ if its reduction is irreducible over ${\mathbb{F}}_p$. This follows from Gauss lemma for polynomials, since a nontrivial factorization over the integers would reduce to a nontrivial factorization of the same degrees. For a monic polynomial the factors may be taken monic, so neither degree can drop.

Modulo $2$ the polynomial is $\overline{f}=t^4+t+1$. Its values at $0$ and $1$ are both $1$, so it has no linear polynomial factor. A reducible quartic polynomial without a linear polynomial factor must factor into two irreducible quadratic polynomials. The only monic irreducible quadratic polynomial over ${\mathbb{F}}_2$ is $q=t^2+t+1$: among monic quadratic polynomials a nonzero constant is necessary, and $t^2+1$ has a root, while $q$ has none.

But $t^2\equiv t+1(\mathrm{mod}q)$, so $t^3\equiv 1$ and $t^4\equiv t$. ConsequentlyThus $q$ does not divide $\overline{f}$, excluding the remaining possible factorization. We obtain

The primary decomposition theorem for linear maps starts with a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$. Factor its minimal polynomial aswhere the $p_j$ are distinct monic polynomials that are irreducible polynomials and $a_j\ge 1$. Define $V_j=\ker p_j(T{)}^{a_j}$. ThenThus the direct sum separates the irreducible polynomial factors into uniquely determined invariant subspaces. The statement works even when the minimal polynomial does not split into linear polynomials over $F$.

A linear spline is a continuous function whose restriction to every interval $[x_{i-1},x_i]$ is a polynomial of polynomial degree at most one. With $h_i=x_i-x_{i-1}$ and ${\ell }_i=\ell (x_i)$, linear interpolation givesIntegrating the two linear polynomials and adding the contributions givesThis is the composite trapezoidal rule applied to the nodal values, and it is exact for the linear spline itself, including nonuniform spline knots.

A cubic spline on this partition of an interval is a function $S\in C^2[x_0,x_n]$ that restricts to a polynomial of polynomial degree at most three on every interval $[x_{i-1},x_i]$. Thus its second derivative is continuous and piecewise of polynomial degree at most one: $S^{\prime \prime }$ is a linear spline on the same spline knots. Its third derivative is constant on each open interval and can jump across the spline knots; its fourth derivative vanishes inside each open interval.

Choose a modulus $W$ that is a product of small primes. Every prime not dividing $W$ must be coprime to $W$, so a search can restrict candidates to the corresponding residue classes modulo $W$. For several linear polynomials, discard each residue class in which any required value is divisible by a selected prime. The surviving residue classes can be combined using the Chinese remainder theorem.