Cayley–Hamilton and minimal-polynomial applications: diagnosis and complete practice

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.

Choose an ordered basis, write $A=[T]$ and $n=\dim V$, and use the monic convention $\chi_T(t)=\det(tI-A)$.

1. Why eigenvalues alone do not prove the general theorem

For a diagonalizable operator with eigenbasis $v_j$, polynomial evaluation satisfies $p(T)v_j=p(\lambda_j)v_j$. Since every eigenvalue is a root of $\chi_T$, the identity $\chi_T(T)=0$ follows immediately in that case. But a defective operator has no eigenbasis. Checking an operator on its existing eigenvectors may check only a proper subspace; it is not a proof on all of $V$.

The general Cayley–Hamilton theorem supplies a polynomial relation even in the defective case. It lets us reduce high powers, construct inverses when possible, and distinguish two operators having the same characteristic polynomial but different structure.

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.

2. A full general Cayley–Hamilton proof

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.

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. ∎

3. Minimal polynomials and the diagonalization criterion

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 $m=(t-\lambda)^2$.

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.

4. Power and inverse calculations

For $A=\begin{pmatrix}1&1\\0&2\end{pmatrix}$, Cayley–Hamilton gives $A^2-3A+2I=0$. Its eigenvalues are distinct, so the projections are $P_1=2I-A$ and $P_2=A-I$. They satisfy $P_1+P_2=I$ and project onto the two eigenspaces. Hence

$$A^k=P_1+2^kP_2=\begin{pmatrix}1&2^k-1\\0&2^k\end{pmatrix},\qquad k\ge0.$$

For the defective matrix $J=2I+N$ with $N^2=0$, the binomial identity gives instead $J^k=2^kI+k2^{k-1}N$ for $k\ge1$. The extra factor $k$ records the nilpotent part that eigenvalues alone miss.

An operator is invertible exactly when $m_T(0)\ne0$. If the constant term is nonzero, Euclid gives polynomials $u,v$ with $ut+vm_T=1$, hence $u(T)T=I$. Conversely if $m_T=tq$ and $T$ were invertible, $Tq(T)=0$ would imply $q(T)=0$, contradicting minimality. Thus a nonsingular operator's inverse is itself a polynomial in the operator.

5. Mixed graded exercises

P01 — Foundation. Over an arbitrary field $F$, let $A=\begin{pmatrix}2&1\\0&3\end{pmatrix}$, where integers denote their images in $F$. Compute $\chi_A$, verify Cayley–Hamilton by multiplication, decide exactly when $A$ is invertible, and express $A^{-1}$ using $A$ when it exists.

P02 — Core. Find $m_A$ and $\chi_A$ for $A=\operatorname{diag}(1,1,4)$ over $\mathbb R$.

P03 — Core. Show that $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ is not diagonalizable and derive its $k$th power for $k\ge0$.

P04 — Proof. If a real operator satisfies $T^2=4I$, prove that it is diagonalizable, allowing the possibility that one of its eigenspaces is zero.

P05 — Diagnosis. If two matrices have the same characteristic polynomial, must they have the same minimal polynomial? Give a counterexample.

P06 — Proof. Prove that $m_T$ is invariant under similarity.

P07 — Core. A real matrix has minimal polynomial $t^2+1$. Is it diagonalizable over $\mathbb R$? Over $\mathbb C$?

P08 — Proof. If $T$ is self-adjoint on a finite-dimensional real inner-product space, explain why its minimal polynomial has no repeated factor. State the theorem used.

P09 — Extension. Suppose a real operator satisfies $(T-I)(T-3I)=0$. Write explicit projections onto the two eigenspaces and use them to find $T^{-1}$.

P10 — Proof audit. A proof checks $\chi_T(T)v=0$ for every eigenvector and concludes $\chi_T(T)=0$. State a sufficient hypothesis that repairs this proof and identify why the general theorem needs more.

6. Full solutions

P01. In every characteristic, $\chi_A=t^2-5t+6$ and direct multiplication gives $A^2=\begin{pmatrix}4&5\\0&9\end{pmatrix}=5A-6I$. Since $\det A=6$, the matrix is invertible exactly when $\operatorname{char}F\notin\{2,3\}$. In that case $A(5I-A)=(5I-A)A=6I$, so $A^{-1}=(5I-A)/6=\begin{pmatrix}1/2&-1/6\\0&1/3\end{pmatrix}$. In characteristic two or three the determinant is zero and no inverse exists; division by six is invalid. The characteristic-polynomial identity itself remains valid in both singular cases.

P02. The characteristic polynomial is $(t-1)^2(t-4)$. The minimal polynomial is $(t-1)(t-4)$ because a polynomial kills the diagonal matrix precisely when it vanishes at both distinct diagonal values. The problem explicitly works over $\mathbb R$, so these two values are distinct. For the same written diagonal matrix over a general field, the conclusion has a boundary: in characteristic three, $4=1$, so $A=I_3$, $m_A=t-1$ and $\chi_A=(t-1)^3$. This does not change the stated real-field answer.

P03. The only eigenvalue is one and its eigenspace is $\operatorname{span}(e_1)$, so there is no two-vector eigenbasis. With $N=A-I$, $N^2=0$, hence $A^k=I+kN=\begin{pmatrix}1&k\\0&1\end{pmatrix}$, including $k=0$.

P04. The minimal polynomial divides $t^2-4=(t-2)(t+2)$, a product of distinct real linear factors, so the criterion applies. It could be only $t-2$ or only $t+2$, corresponding to a scalar operator; a missing eigenvalue does not invalidate diagonalization.

P05. $I_2$ and $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ both have characteristic polynomial $(t-1)^2$, but their minimal polynomials are $t-1$ and $(t-1)^2$ respectively.

P06. If $S=P^{-1}TP$, then $p(S)=P^{-1}p(T)P$ for every polynomial, so exactly the same polynomials kill $S$ and $T$. Their unique monic least-degree annihilators agree.

P07. Over $\mathbb R$, the polynomial does not split into real linear factors, preventing diagonalization. Over $\mathbb C$ it is $(t-i)(t+i)$ with distinct roots, so the matrix is diagonalizable. The polynomial is nonrepeated over the extension, but ground-field splitting is still essential.

P08. The real symmetric spectral theorem supplies an orthonormal eigenbasis. Evaluating the product of distinct eigenvalue factors on it shows that product annihilates $T$. The minimal polynomial divides the product and therefore has no repeated factor. This conclusion uses self-adjointness and the spectral theorem, not merely a claim that all eigenvalues are real.

P09. $P_1=(3I-T)/2$ and $P_3=(T-I)/2$ sum to $I$ and have images in $E_1,E_3$. The direct eigenspace sum gives $T^{-1}=P_1+\tfrac13P_3=(4I-T)/3$. Multiplying by $T$ and using $T^2=4T-3I$ verifies the inverse directly.

P10. Diagonalizability is sufficient because eigenvectors then include a full basis, and a linear operator vanishing on a basis vanishes everywhere. A defective operator may have eigenvectors spanning only a proper subspace. The coefficient proof in §2 covers all vectors without requiring an eigenbasis.

7. Summary

The general Cayley–Hamilton proof closes the gap left by checking only eigenvectors. Minimal polynomials distinguish the shortest operator relation from characteristic multiplicity, and their splitting and repeated factors decide diagonalizability. Polynomial reduction unifies inverse formulas, powers and eigenspace projections while keeping the ground field explicit.