Polynomial operators, Cayley–Hamilton and decomposition

Proof

Prerequisite proofs used below

Polynomial division and the root factor identity over a field

Let $F$ be any field, and use formal polynomials in $F[t]$ . The zero polynomial is a separate case whenever degree is used. For nonzero $f,g$ , the leading coefficient of $fg$ is the product of their nonzero leading coefficients, so $\deg(fg)=\deg f+\deg g$ ; in particular there are no zero divisors.

For every $f$ and every nonzero $g$ there are unique polynomials $q,r$ with $f=qg+r$ , where $r=0$ or $\deg r<\deg g$ . Start with residual $f$ and quotient zero. If a nonzero residual has degree $d\ge e=\deg g$ , subtract $(a/b)t^{d-e}g$ , where $a,b$ are its and $g$ 's leading coefficients, and add $(a/b)t^{d-e}$ to the quotient. The new residual is zero or has strictly smaller degree. Nonnegative integer degrees cannot strictly decrease indefinitely, so the process stops. If two answers exist, $g(q-q')=r'-r$ ; a nonzero left side has degree at least $e$ , whereas a nonzero right side has degree below $e$ . Thus both differences are zero.

For $\lambda\in F$ , the finite identity

t^k-\lambda^k=(t-\lambda)\sum_{j=0}^{k-1}t^{k-1-j}\lambda^j\quad(k\ge1)

gives $f(t)-f(\lambda)=(t-\lambda)h(t)$ . Consequently $f(\lambda)=0$ exactly when $t-\lambda$ divides $f$ . This does not assert that any root exists in $F$ .

Monic gcd, Euclid and Bezout, including the zero cases

For $f,g$ not both zero, their monic greatest common divisor is the monic polynomial dividing both and divisible by every common divisor. If $g=0,f\ne0$ , it is $d=f/\operatorname{lc}(f)$ , with $d=(1/\operatorname{lc}(f))f+0g$ ; if $f=0,g\ne0$ , interchange the roles.

Otherwise divide successively, $r_{i-1}=q_ir_i+r_{i+1}$ , until the next remainder is zero. Strictly decreasing nonzero remainder degrees make this finite. The equation $r_{i+1}=r_{i-1}-q_ir_i$ shows that a divisor of $r_{i-1},r_i$ divides $r_i,r_{i+1}$ , and the reverse equation shows the converse. The last nonzero remainder therefore has exactly the common divisors of the original pair. Backward use of the division equations shows it divides every previous remainder and both original inputs. Each remainder is also a polynomial linear combination of the original inputs, by forward induction. Normalize the final remainder by its nonzero leading coefficient. This yields the monic gcd $d=af+bg$ . Any two such monic gcds divide each other; equality of degrees and monic normalization force equality. In particular, relatively prime inputs have $af+bg=1$ .

Irreducible factors, primality and unique factorization

A nonconstant polynomial is irreducible if it is not a product of two positive-degree polynomials. If a monic irreducible $p$ does not divide $a$ , its monic gcd with $a$ is one: a nonconstant divisor of $p$ is associated to $p$ , and hence cannot divide $a$ . Bezout gives $up+va=1$ . If $p\mid ab$ , multiplication by $b$ now gives $p\mid b$ . Thus irreducibles are prime.

Every nonzero polynomial is its leading coefficient times a product of monic irreducibles, with the product empty for a nonzero constant. Prove this by degree induction: a reducible positive-degree input splits into two smaller positive degrees, to which induction applies. For uniqueness, a first irreducible in one factorization divides the product in another, hence one of its factors by the prime property. Monic irreducibles dividing one another are equal. Cancel this factor, using absence of zero divisors, and repeat; leading coefficients fix the final unit. This proves uniqueness up to factor order, including multiplicities. In particular, powers of distinct monic irreducibles are relatively prime: any nonconstant common divisor would have an irreducible factor, which by primality would be equal to both of the distinct factors. This supplies the precise polynomial algebra used in primary decomposition, over the original arbitrary field.

Leibniz determinant and alternation over a commutative ring

Let $R$ be a commutative ring with identity, $n\ge1$ , and $A\in M_n(R)$ . Define

\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)},\qquad \det[\ ]=1\quad(n=0).

The sign is $(-1)^{\operatorname{inv}(\sigma)}$ , interpreted in $R$ . Exchanging adjacent entries in a permutation changes its inversion count by one in parity. Sorting by adjacent inversions therefore gives its sign as the product of the signs of those exchanges. Applying an adjacent-exchange decomposition of $\tau$ to the positions of $\sigma$ proves $\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$ ; a transposition has sign $-1$ . These are integer parity facts before mapping the signs into $R$ .

The finite formula is multilinear in each row and column. If two rows are equal, pair each permutation term with the term obtained by exchanging the two assigned columns. The pair is distinct, has the same product and opposite signs; its sum is zero in every characteristic. Thus the determinant vanishes on repeated rows; the column statement is identical. This is a paired cancellation, not an inference from $2\det A=0$ . Exchanging rows or columns changes sign by the same permutation substitution. Also $\det I=1$ . Substituting $\sigma^{-1}$ in the sum gives $\det A^{\mathsf T}=\det A$ .

Here is the needed uniqueness principle, valid even with zero divisors. If $D$ is an alternating multilinear function of $n$ column vectors in $R^n$ , expand each input in the standard basis. Terms with a repeated basis vector vanish. The other index lists are permutations; swapping two slots negates $D$ , since expansion of $D(\ldots,u+v,\ldots,u+v,\ldots)=0$ gives $D(\ldots,u,\ldots,v,\ldots)+D(\ldots,v,\ldots,u,\ldots)=0$ . Hence the expansion equals $D(e_1,\ldots,e_n)$ times the Leibniz determinant of the input columns. No division or field assumption was used.

Multiplicativity, triangular blocks and the adjugate

For fixed $A$ , the function $D(v_1,\ldots,v_n)=\det(Av_1,\ldots,Av_n)$ is alternating multilinear, with $D(e_1,\ldots,e_n)=\det A$ . Applying the preceding uniqueness principle to the columns of $B$ proves

\det(AB)=\det A\det B.

For $\begin{pmatrix}B&C\\0&D\end{pmatrix}$ with square block sizes $p,q$ , a nonzero Leibniz term must assign all last $q$ rows to the last $q$ columns; otherwise one lower-left entry is zero. Those rows use all those columns, so the first $p$ rows use the first $p$ columns. The remaining permutations are exactly pairs of permutations inside the two blocks, with product sign and product monomial. Summing gives $\det B\det D$ , including an empty block using determinant one. This argument also covers terms that vanish for other reasons; it does not require absence of zero divisors.

Let $A_{ij}^{\widehat{\ } }$ delete row $i$ and column $j$ , and put $C_{ij}=(-1)^{i+j}\det A_{ij}^{\widehat{\ } }$ . Grouping the Leibniz sum by the value $\sigma(i)=j$ gives row cofactor expansion $\det A=\sum_j a_{ij}C_{ij}$ . The sign factor follows by moving row $i$ and column $j$ to the first positions, making $i+j-2$ adjacent exchanges; the remaining permutation is on the undeleted indices. Transpose gives the column expansion. Define $\operatorname{adj}(A)_{ji}=C_{ij}$ . The $(i,j)$ entry of $A\operatorname{adj}(A)$ is $\sum_k a_{ik}C_{jk}$ , the cofactor expansion of the matrix obtained by replacing row $j$ by row $i$ . It is $\det A$ if $i=j$ , and zero otherwise by repeated-row cancellation. The column argument gives the other product. Therefore

A\operatorname{adj}(A)=\operatorname{adj}(A)A=(\det A)I.

For $n=1$ , the empty minor has determinant one, so the adjugate is $[1]$ and the identity still holds. For $n=0$ , define the adjugate to be the empty matrix; the identity is the equality of the unique empty matrices, while the determinant remains one. All results in this section apply to $R=F[t]$ ; no division by a determinant has occurred.

A vector space is a set with addition and scalar multiplication satisfying the usual abelian-group and distributive axioms over a field $F$ . Columns are one model. Polynomials of degree at most $d$ , denoted $P_d(F)$ , and linear maps between spaces are others. The zero polynomial is included in $P_d(F)$ although its degree is not ordinarily defined.

A subspace contains zero and is closed under sums and scalar multiples. A list spans if every vector is a linear combination of its entries; it is independent if only the zero coefficients give zero. A basis is both. A finite spanning list can be reduced to a basis by deleting dependent entries; an independent list can be extended to a basis by adding vectors outside its span until the given finite spanning list is spanned.

Exchange lemma. An independent list of $r$ vectors in the span of $s$ vectors has $r\le s$ . To see this, express the first independent vector in the spanning list, use a nonzero coefficient to replace one spanning vector by it, and repeat. At each stage independence ensures the next independent vector has a nonzero coefficient on an as-yet unreplaced spanning vector. There must be at least $r$ such slots. Comparing two bases in both directions proves that their lengths agree; this common length is $\dim V$ .

Consequently, in dimension $n$ , any $n$ independent vectors form a basis, as do any $n$ spanning vectors. Counting a list is useful only after its independence or spanning has been justified.

A map $T:V\to W$ is linear if $T(\alpha v+\beta u)=\alpha Tv+\beta Tu$ . Its kernel and image are subspaces. Choose a basis $k_1,\ldots,k_r$ of $\ker T$ and extend to a basis with $v_1,\ldots,v_s$ . Then $Tv_1,\ldots,Tv_s$ form a basis of $\operatorname{im}T$ : spanning follows by expanding any input; a dependence among these images places the corresponding input combination in the kernel, and independence of the extended basis makes all coefficients zero. Thus

\dim V=\dim\ker T+\dim\operatorname{im}T.

This is rank–nullity. In equal finite dimensions, injectivity and surjectivity are equivalent, but the equivalence can fail in infinite-dimensional spaces.

The field bridge: singularity and characteristic polynomials

Now let $R=F$ be a field. If $\det A\ne0$ , the adjugate identity gives inverse $\operatorname{adj}(A)/\det A$ . If $A$ is invertible, multiplicativity gives $\det A\det A^{-1}=1$ , so $\det A\ne0$ . Rank–nullity, proved in C01, shows a finite square linear map has zero kernel exactly when it is invertible. Thus $\det A=0$ exactly when its kernel is nonzero.

For a fixed $A\in M_n(F)$ , apply the ring construction to $tI-A\in M_n(F[t])$ . Its determinant is monic of degree $n$ : the identity-permutation product contributes $t^n$ , whereas every other permutation uses an off-diagonal constant entry, so its product is either zero or has degree less than $n$ . Multiplicativity proves invariance under $A\mapsto P^{-1}AP$ with invertible constant $P$ ; the block-triangular formula proves restriction–quotient factorization. Finally every adjugate entry is zero or has degree at most $n-1$ , so their coefficient comparison is legitimate in the C02 Cayley–Hamilton proof. In dimension zero, the characteristic polynomial is the empty determinant one; evaluating it gives $I$ on the zero space, which is its zero operator, so Cayley–Hamilton is separate and immediate. This supplies the actual commutative-ring determinant prerequisites without extending any real-only determinant authority.

Throughout, $V$ is a nonzero finite-dimensional vector space over a field $F$ , and $T\in\operatorname{End}_F(V)$ . No algebraic closure is assumed unless stated.

1. An operator can satisfy an equation without being a number

For $p(t)=\sum_{j=0}^dc_jt^j$ , define $p(T)=\sum c_jT^j$ , with $T^0=I$ . Because powers of the same operator commute, $(pq)(T)=p(T)q(T)$ and evaluation is a ring homomorphism $F[t]\to\operatorname{End}(V)$ . This does not permit replacing variables by arbitrary noncommuting matrices in every matrix-coefficient identity.

A subspace $U$ is $T$ -invariant when $T(U)\subseteq U$ . Kernels and images of $p(T)$ are invariant because $T$ commutes with $p(T)$ . On an invariant subspace the restriction is an operator, and on $V/U$ the formula $v+U\mapsto Tv+U$ is well-defined. An adapted basis makes the full matrix block upper triangular. Its characteristic polynomial is the product of those of the restriction and quotient, since the determinant of a block triangular matrix is the product of diagonal-block determinants.

The characteristic polynomial is $\chi_T(t)=\det(tI-A)$ in any basis. Similarity preserves it because $tI-P^{-1}AP=P^{-1}(tI-A)P$ . It is monic of degree $n=\dim V$ . A scalar $\lambda\in F$ is an eigenvalue exactly when $\chi_T(\lambda)=0$ , because a square matrix is singular exactly when it has a nonzero kernel.

2. Cayley–Hamilton without an invalid substitution

Theorem. $\chi_T(T)=0$ .

Proof. Let $\chi_A(t)=\sum_{k=0}^nc_kt^k$ with $c_n=1$ . The adjugate identity over the commutative polynomial ring $F[t]$ gives

(tI-A)B(t)=\chi_A(t)I,\qquad B(t)=\operatorname{adj}(tI-A)=\sum_{k=0}^{n-1}B_kt^k.

Set $B_{-1}=B_n=0$ . Comparing coefficients yields $B_{k-1}-AB_k=c_kI$ for $0\le k\le n$ . Multiply the $k$ th identity on the left by $A^k$ and sum. The terms telescope:

\sum_{k=0}^nc_kA^k=\sum_{k=0}^n\left(A^kB_{k-1}-A^{k+1}B_k\right)=0.

This is $\chi_A(A)=0$ . It uses coefficient identities and associativity only, not an unexplained substitution into a matrix polynomial with potentially noncommuting coefficients. ∎

For $A=\begin{pmatrix}2&1\\0&3\end{pmatrix}$ , $\chi_A=t^2-5t+6$ , so $A^2=5A-6I$ . Repeated powers reduce to combinations of $A$ and $I$ . Since $A(A-5I)=-6I$ , in characteristic zero $A^{-1}=(5I-A)/6$ . If the field has characteristic two or three, that division is invalid and the particular matrix is not invertible; the general Cayley–Hamilton identity still holds.

3. Minimal polynomial: the shortest exact operator equation

Since Cayley–Hamilton supplies a nonzero annihilating polynomial, choose the monic one of least degree and call it $m_T$ . Polynomial division proves

p(T)=0\quad\Longleftrightarrow\quad m_T\mid p.

Indeed, write $p=qm_T+r$ with $\deg r<\deg m_T$ . If $p(T)=0$ , then $r(T)=0$ , and minimality forces $r=0$ . Uniqueness follows from the same argument. In particular $m_T\mid\chi_T$ . A constant nonzero polynomial cannot annihilate a nonzero space, so $m_T$ has positive degree.

A scalar $\lambda\in F$ is a root of $m_T$ exactly when it is an eigenvalue. If $Tv=\lambda v$ for $v\ne0$ , then $0=m_T(T)v=m_T(\lambda)v$ , so $m_T(\lambda)=0$ . Conversely if $m_T=(t-\lambda)q$ and $T-\lambda I$ were invertible, its inverse applied to $0=(T-\lambda I)q(T)$ would give $q(T)=0$ , contradicting minimal degree.

The characteristic polynomial records total algebraic multiplicities; the minimal polynomial records the strongest polynomial relation needed on any vector. For $\lambda I$ on an $n$ -dimensional space, $\chi=(t-\lambda)^n$ but $m=t-\lambda$ . For $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ , $(T-\lambda I)^2=0$ but $T-\lambda I\ne0$ , so $m=(t-\lambda)^2$ .

4. Coprime factors split the space

Two-factor decomposition. Suppose $f,g\in F[t]$ are relatively prime and $f(T)g(T)=0$ . Then

V=\ker f(T)\oplus\ker g(T).

Proof. Polynomial Euclid gives $af+bg=1$ . For any $v$ ,

v=a(T)f(T)v+b(T)g(T)v.

The first term lies in $\ker g(T)$ and the second in $\ker f(T)$ , because polynomial operators commute and $f(T)g(T)=0$ . If $v$ lies in both kernels, the same identity gives $v=0$ . ∎

Applying this repeatedly to the factorization

m_T=p_1^{r_1}\cdots p_s^{r_s}

into distinct monic irreducibles gives the primary decomposition

V=\bigoplus_{j=1}^s\ker p_j(T)^{r_j}.

Each summand is invariant. Over a field where $m_T$ splits into linear factors, these are generalized eigenspaces. Over $\mathbb R$ , an irreducible quadratic may remain and must not be relabelled as a real eigenspace. The real rotation matrix with $m=t^2+1$ illustrates this obstruction.

Diagonalization criterion. $T$ is diagonalizable over $F$ exactly when its minimal polynomial is a product of distinct linear factors in $F[t]$ .

If the factors are distinct linear factors, primary decomposition gives a direct sum of ordinary eigenspaces, whose bases combine to an eigenbasis. Conversely if an eigenbasis has distinct eigenvalues among $\lambda_1,\ldots,\lambda_s$ , the product $\prod(t-\lambda_j)$ kills every basis vector, so $m_T$ divides this square-free split polynomial. Every eigenvalue must be a root of $m_T$ , hence $m_T$ is exactly that product. Merely having no repeated roots in an extension field is not diagonalizability over the original field unless it splits there.

5. Polynomial reduction as a computational method

For any polynomial $p$ , its remainder modulo $m_T$ gives the same operator $p(T)$ and has degree below $\deg m_T$ . If $m_T(0)\ne0$ , Euclid gives $ut+vm_T=1$ , so $u(T)T=I$ : $T$ is invertible and its inverse is a polynomial in $T$ . Conversely if $m_T(0)=0$ , zero is an eigenvalue and $T$ is not invertible.

Worked example. Suppose $(T-I)(T-3I)=0$ over $\mathbb R$ . The projections

P_1=(3I-T)/2,\qquad P_3=(T-I)/2

satisfy $P_1+P_3=I$ , $P_1P_3=0$ and $P_j^2=P_j$ , by reducing their products modulo $(t-1)(t-3)$ . Their images are the corresponding eigenspaces; for each integer $k\ge0$ ,

T^k=P_1+3^kP_3.

For negative $k$ the same formula holds because both eigenvalues are nonzero and $T$ is invertible. The divisions by two require characteristic different from two; in characteristic two the two named scalars coincide and this decomposition is not valid.

6. Exercises

P01 — Foundation. Find characteristic and minimal polynomials of $A=\operatorname{diag}(2,2,5)$ over $\mathbb R$ .

P02 — Core. For $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ compute $A^k$ for all integers $k\ge0$ using polynomial reduction.

P03 — Core. An operator over $\mathbb R$ satisfies $T^2=I$ . Prove it is diagonalizable and give its two possible eigenspace projections.

P04 — Diagnosis. Explain why the conclusion of P03 can fail in characteristic two, with an explicit matrix.

P05 — Core. If $A^2-3A+2I=0$ over $\mathbb R$ , express $A^{-1}$ as a polynomial in $A$ .

P06 — Proof. If $S=P^{-1}TP$ , show $m_S=m_T$ .

P07 — Core. Determine whether the real rotation $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is diagonalizable over $\mathbb R$ and over $\mathbb C$ , using its minimal polynomial.

P08 — Extension. Suppose $T$ has minimal polynomial $(t-1)^2(t+2)$ over $\mathbb R$ . State its primary decomposition and determine whether it is diagonalizable.

P09 — Proof. If $T$ is diagonalizable and $U$ is invariant, prove that $T|_U$ is diagonalizable without assuming a selected eigenbasis of $V$ is adapted to $U$ .

7. Full solutions

P01. $\chi_A=(t-2)^2(t-5)$ . A polynomial kills the diagonal matrix exactly when it vanishes at both two and five, so $m_A=(t-2)(t-5)$ .

P02. Put $N=A-I$ , with $N^2=0$ . The binomial identity gives $A^k=(I+N)^k=I+kN=\begin{pmatrix}1&k\\0&1\end{pmatrix}$ , including $k=0$ .

P03. Its minimal polynomial divides $t^2-1=(t-1)(t+1)$ , a product of distinct linear factors over $\mathbb R$ . Thus it is diagonalizable. The projections are $(I+T)/2$ onto $E_1$ and $(I-T)/2$ onto $E_{-1}$ ; a zero eigenspace corresponds to a zero projection.

P04. In characteristic two, $(t-1)(t+1)=(t-1)^2$ . Take $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ . Then $A^2=I$ because $2=0$ , but $A\ne I$ and its only eigenvalue is one. Its eigenspace is one-dimensional, so it is not diagonalizable.

P05. Rearranging gives $A(3I-A)=2I$ , hence $A^{-1}=(3I-A)/2$ . Both left and right products agree because the factor is a polynomial in $A$ .

P06. For every polynomial, $p(S)=P^{-1}p(T)P$ . Thus the annihilating polynomials are identical, and their unique monic least-degree member is identical.

P07. $A^2=-I$ and $A$ is not scalar, so $m=t^2+1$ . It does not split over $\mathbb R$ , preventing real diagonalization. Over $\mathbb C$ , it factors as $(t-i)(t+i)$ with distinct roots, so the matrix is diagonalizable there.

P08. $V=\ker(T-I)^2\oplus\ker(T+2I)$ . The repeated factor in the minimal polynomial prevents diagonalizability. It cannot be removed merely because the two primary factors are coprime; the first summand may contain nontrivial generalized eigenvectors.

P09. Any polynomial killing $T$ also kills its restriction, so $m_{T|_U}$ divides $m_T$ when $U\ne0$ . A divisor of a product of distinct linear factors has the same property, hence the criterion gives diagonalizability on $U$ . The zero-dimensional restriction is trivially diagonalizable. This argument avoids the unjustified claim that an arbitrary chosen eigenbasis contains a basis of $U$ .

8. Summary

Cayley–Hamilton guarantees an operator equation; the minimal polynomial compresses it to the exact relation. Coprime factors give invariant direct sums, and square-free split factors characterize diagonalization. Repeated factors lead to nilpotent pieces and Jordan chains, developed with a full existence proof next.

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