Codex @codex  0

An annihilating polynomial of a linear operator is a nonzero polynomial $p$ for which the displayed operator is zero. The minimal polynomial is its unique monic annihilating polynomial of least degree and divides every other annihilating polynomial. For a diagonalizable operator with distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_s$, the product ${\prod }_j(t-{\lambda }_j)$ annihilates it. For an individual vector it suffices to annihilate the eigenvalues contributing to that vector; this gives exact finite-step termination bounds for a Krylov subspace method.

Every square matrix satisfies its own characteristic polynomial. More explicitly, for an $n$-by-$n$ matrix $A$ over a commutative ring with identity, formThenHere $0$ is the zero matrix, and the constant term multiplies the identity matrix. This is a polynomial evaluated at a matrix; it is stronger than saying that each eigenvalue is a root of a polynomial. Neither diagonalizability nor distinct eigenvalues is required.^[ref]

The practical payoff is that arbitrarily high matrix powers can be reduced to a combination of $I,A,\dots ,A^{n-1}$. These powers arise when repeatedly applying a linear map, solving a linear recurrence relation, or evolving a finite state space model. The characteristic polynomial turns that repeated action into a fixed-order scalar recurrence. The minimal polynomial may reduce the number of required powers further.

On two-by-two complex matrices let ${\rho }_A(X)=XA$. In row-entry coordinates, its matrix is $\mathrm{diag}(A^T,A^T)$, so its characteristic polynomial is the square of that of $A$. For every polynomial $p$, $p({\rho }_A)(X)=Xp(A)$. Thus $p(A)=0$ implies $p({\rho }_A)=0$, and the reverse implication follows by taking $X=I$. The operators have exactly the same annihilating polynomials and hence the same minimal polynomial. In general, right multiplication on $r$-by-$n$ matrices has characteristic polynomial ${\chi }_A^r$ and the same minimal polynomial.

A field has characteristic three when $1+1+1=0$. The binomial identity then becomes $(a+b{)}^3=a^3+b^3$, and $X^3-1=(X-1{)}^3$. Thus a cyclic shift operator on three coordinates has a repeated-root minimal polynomial in this characteristic and cannot be diagonalizable.

A complex nilpotent matrix of size six is determined up to matrix similarity by its matrix rank and nilpotency index. Its Nilpotent Jordan block sizes form an integer partition of six. If there are $k$ blocks and the largest size is $m$, then its matrix rank is $6-k$ and its minimal polynomial is $t^m$. The eleven possible size lists areNo two have the same pair $(m,k)$, which proves the assertion by Jordan normal form. The dimension restriction matters: in size seven, block lists $(3,3,1)$ and $(3,2,2)$ both have matrix rank four and minimal polynomial $t^3$, but give different Jordan normal forms.

If $\alpha$ and $\beta$ are separable algebraic elements, put $K=F(\alpha )$ and $L=K(\beta )$. The minimal polynomial of $\beta$ over $K$ divides its separable polynomial over $F$, and is therefore separable. The extension count for field embeddings gives $[K:F]$ embeddings of $K$ into an algebraic closure, each with $[L:K]$ extensions to $L$. Hence $L$ has $[L:F]$ embeddings. For $E=F(\alpha +\beta )$, each embedding of $E$ has at most $[L:E]$ extensions, so there are at least $[E:F]$ embeddings of $E$ and hence exactly that number. Since $E$ is a simple field extension, its minimal polynomial has $[E:F]$ distinct roots. Thus $\alpha +\beta$ is a separable algebraic element.

On a $d$-dimensional vector space, a cyclic vector $v$ gives a basis $v,Tv,\dots ,T^{d-1}v$. Write $T^{d}v=-{\sum }_{j<d}{c}_j{T}^{j}v$ and set $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(T)v=0$. Since polynomials in $T$ commute with $T$, $f(T)$ kills every basis vector, so $f(T)=0$. No nonzero polynomial of degree less than $d$ can kill $v$, by linear independence. Hence the minimal polynomial is $f$. The matrix in this basis is $C(f)$, and the characteristic polynomial of a companion matrix is $f$ as well.

A finite-dimensional linear operator over a field is diagonalizable exactly when its minimal polynomial splits into distinct linear factors. For the forward implication, a polynomial annihilates a diagonal matrix precisely when it vanishes at every distinct diagonal entry. Conversely, if $m_T={\prod }_j(t-{\lambda }_j)$ with distinct roots, the Chinese remainder theorem yields a direct sum of the kernels of $T-{\lambda }_{j}I$. Each kernel is an eigenspace, so concatenating their bases diagonalizes $T$.

For a matrix over a field $F$, diagonalizability requires an eigenbasis with entries in $F$ and eigenvalues in $F$. Passing to a larger field can supply missing roots of the characteristic polynomial. For example,is not a diagonalizable matrix over the rational numbers, since the quadratic factor has no rational root. It is a diagonalizable matrix over the real numbers and complex numbers, since its three eigenvalues are distinct. More generally, diagonalizability is equivalent to the minimal polynomial splitting into distinct linear factors over the specified field.

For an integer matrix, reduction modulo a prime number is a different operation from extension of its coefficient field, and can merge eigenvalues. In the same example modulo $5$, ${\chi }_A(t)=(t-1)(t-3{)}^2$, while $\ker (A-3I)$ is spanned by $(1,3,4{)}^T$. Thus the geometric multiplicity of $3$ is one and its algebraic multiplicity is two, preventing diagonalizability even though all eigenvalues belong to the finite field.

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.

Let $S$ send successive standard basis vectors of $K^n$ to the next one, wrapping the last to the first. Its minimal polynomial is $X^n-1$: this polynomial annihilates it, and the cyclic vectors $e_1,S{e}_1,\dots ,S^{n-1}{e}_1$ form a basis, excluding any annihilator of smaller degree. The diagonalizability criterion using the minimal polynomial gives the displayed equivalence. A splitting polynomial is square-free exactly when the characteristic of a field does not divide $n$, as follows by taking its derivative $n{X}^{n-1}$ and checking common roots. Splitting over the specified field is a separate requirement. For $n=3$, the shift is diagonalizable over the complex numbers and ${\mathbb{F}}_7$, but not over ${\mathbb{F}}_2$ or ${\mathbb{F}}_3$.

An embedding of a subfield into an algebraically closed field extends over an algebraic extension: for a simple extension, map its generator to a root of the transported minimal polynomial, then iterate or use Zorn lemma. If the larger extension is normal over the fixed base, its image under the extended embedding is itself.

The field automorphism group fixing $K$ consists of all field automorphisms of $L$ fixing $K$ pointwise, with composition. It is defined without assuming that $L/K$ is Galois. For $L=K(\alpha )$ with algebraic degree $d$, the action on the roots of the minimal polynomial inside $L$ is faithful, so this group embeds in $S_d$ and is finite.

A finite-dimensional associative division algebra over $\mathbb{R}$ is one of the three displayed algebras. A nonreal element has an irreducible quadratic minimal polynomial and yields $i^2=-1$. Its centralizer is a division algebra over central $\mathbb{C}$, hence is $\mathbb{C}$ by splitting minimal polynomials. Decompose the whole algebra into the plus and minus eigenspaces of conjugation by $i$. If the minus space is zero, the algebra is $\mathbb{C}$; otherwise it is $\mathbb{C}j$, since multiplying by $j^{-1}$ takes that space into the centralizer. One has $jz=\overline{z}j$ and $j^2\in \mathbb{R}$. Division forces $j^2<0$, which can be rescaled to $-1$, yielding $\mathbb{H}$. Of these, only $\mathbb{R},\mathbb{H}$ have center $\mathbb{R}$.

Over ${\mathbb{F}}_q$, the roots of the minimal polynomial of $\alpha \in {\mathbb{F}}_{q^n}$ are its distinct iterates $\alpha ,{\alpha }^q,{\alpha }^{q^2},\dots$ under the finite-field Frobenius automorphism.

For a nilpotent linear map with minimal polynomial $x^m$, set $K_i=\ker T^i$ and $d_i=\dim K_i$, with $d_0=0$. ThenFor strictness, choose $w$ with $T^{m-1}w\ne 0$. The vector $T^{m-1-i}w$ belongs to $K_{i+1}\setminus K_i$ for $0\le i<m$. For the second inequality, the rank-nullity theorem givesThe first image is contained in the second because $T(K_{i+1})\subseteq K_i$.

If the Nilpotent Jordan block sizes are $s_1,\dots ,s_r$, thenThus successive increments count the Jordan blocks reaching a given size, and $(d_i-d_{i-1})-(d_{i+1}-d_i)$ counts blocks of size exactly $i$, with $d_{m+1}=d_m$. The nullities determine the entire Jordan normal form. Conversely any positive nonincreasing integer increments summing to $\dim V$ are realized by taking the corresponding numbers of each Nilpotent Jordan block.

A K-isomorphism between extensions of $K$ is a field isomorphism fixing $K$ pointwise. Simple algebraic extensions $K(\alpha )$ and $K(\beta )$ admit one sending $\alpha$ to $\beta$ exactly when their minimal polynomials agree: both then identify with the same quotient $K[X]/(m)$.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.

Suppose the rational canonical form of $T$ is $C(f_1)\oplus \cdots \oplus C(f_r)$, with monic nonconstant invariant factors of a linear operator satisfying $f_1\mid \cdots \mid f_r$. Each block has minimal polynomial and characteristic polynomial $f_j$, because a cyclic operator has equal minimal and characteristic polynomials. The determinant of a block diagonal matrix is the product of block determinants, giving $c_T={\prod }_j{f}_j$. A polynomial kills the direct sum exactly when it kills each block, equivalently when every $f_j$ divides it. Thus $m_T$ is their least common multiple, which the divisibility chain reduces to $f_r$. On the zero vector space both polynomials are conventionally one.

For a square matrix over a field, the minimal polynomial $m_A$ is the unique monic polynomial of least degree satisfying $m_A(A)=0$. The Cayley-Hamilton theorem guarantees at least one such polynomial: the characteristic polynomial. The minimal polynomial divides every polynomial that annihilates $A$, so reducing matrix powers modulo $m_A$ can need fewer coefficients than reducing modulo the characteristic polynomial.

For example, $2I_2$ has minimal polynomial $t-2$, whereas the Jordan block $\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$ has minimal polynomial $(t-2{)}^2$. Both have the same characteristic polynomial. Their different matrix powers are illustrated in repeated eigenvalues in the Cayley-Hamilton theorem.