Find the shortest polynomial relation an operator obeys

Try

Take A=diag(1,1,4). Its characteristic polynomial repeats the factor t−1, yet (A−I)(A−4I)=0 already. Evaluate the product on each coordinate: on either first coordinate the first factor kills the vector; on the last coordinate the second factor does. No nonzero linear polynomial can annihilate A because it would have to vanish at both 1 and 4. The shortest monic relation is therefore m_A(t)=(t−1)(t−4).

Predict

Predict what happens for B=[[1,1],[0,1]]. Here B−I is nonzero but (B−I)²=0. The repeated factor is now necessary. A repeated eigenvalue in the characteristic polynomial does not force a repeated factor in the minimal polynomial, but a repeated factor in the minimal polynomial signals a genuine obstruction to diagonalization when the polynomial splits. Distinguish these two statements with the examples rather than memorizing a slogan about repeated eigenvalues.

Experiment

Make a table for I, diag(1,4), and B, listing the first polynomial relation you can verify and why a lower degree cannot work. For B, compute (B−I)e₂=e₁ and (B−I)e₁=0. This two-step chain explains the square directly. For a general vector (a,b), calculate Bᵏ(a,b)=(a+kb,b); the factor k exposes behavior hidden by the eigenvalue list alone.

Understand

For a finite-dimensional V, the endomorphism space has finite dimension (dim V)². Therefore I,T,T²,… eventually become linearly dependent, giving a nonzero annihilating polynomial. Among them choose a monic polynomial of least degree, called m_T. It is unique: subtracting two monic candidates of the same least degree would give an annihilator of smaller degree. Polynomial division proves m_T divides every annihilator: if p=qm_T+r, evaluation gives r(T)=0, forcing r=0. Every eigenvalue λ is a root of m_T because applying m_T(T) to a λ-eigenvector gives m_T(λ)v. These facts do not yet rely on Cayley–Hamilton.

Apply

T is diagonalizable over F exactly when m_T is a product of distinct linear factors over F. One direction follows by evaluating a product over the distinct diagonal entries. For the other, write m_T as a product of t−λᵢ and form the Lagrange polynomials pᵢ(t)=∏ⱼ≠ᵢ(t−λⱼ)/(λᵢ−λⱼ). Their sum is one, and (t−λᵢ)pᵢ(t) is a multiple of m_T. Thus every v is the sum of pᵢ(T)v, each in the λᵢ-eigenspace. Distinct eigenspaces are independent, so a union of their bases is an eigenbasis. Check the construction numerically for eigenvalues 1 and 4.

The Lagrange construction also proves that its projections are pairwise disjoint: pᵢpⱼ is zero at every λ when i≠j, so the split square-free polynomial m_T divides that product. Likewise m_T divides pᵢ²−pᵢ. Evaluating gives PᵢPⱼ=0 and Pᵢ²=Pᵢ. These are algebraic projections, and without normality they need not be orthogonal projections. Test the distinction on a nonnormal matrix with two distinct eigenvalues.

Transfer: change the field, keep the relation

A field change can create eigenvectors without changing the matrix entries or the minimal polynomial. Use the real rotation J with columns (0,1) and (−1,0). Direct multiplication gives J²=−I. No nonzero polynomial of degree one annihilates J, because that would make J a scalar multiple of I. Thus its minimal polynomial is t²+1.

Over R this polynomial is square-free but does not split; J has no real eigenvalue. Over C it has the distinct roots i and −i. Check the candidate vectors explicitly:

$$J(1,-i)=i(1,-i),\qquad J(1,i)=-i(1,i).$$

They are linearly independent, so they form a complex eigenbasis. The characteristic and minimal polynomials have not become linear. Rather, the same polynomial now factors into linear pieces in the enlarged coefficient field.

Why the least degree does not secretly drop over C. For any real matrix T, suppose a complex polynomial p=u+iv annihilates T, where u and v have real coefficients. Then u(T)+iv(T)=0 implies u(T)=v(T)=0 entrywise. If p is monic, u has the same leading coefficient one and the same degree. Therefore a complex annihilator of smaller degree would give a real one of smaller degree. The real minimal polynomial consequently remains minimal over C. This proof is for real matrices with complex scalar extension; it does not assert arbitrary unrelated matrices have the same polynomial.

Predict, then inspect the certificate

Changed case Rotation over RSame matrix over C

J=[[0,−1],[1,0]]. Over R, is J diagonalizable?

Your prediction Choose a predictionYes: its minimal polynomial has no repeated factorsNo: its minimal polynomial does not splitNo: its minimal polynomial has a repeated factor Check reasoning Reset prediction

Two authored cases are practice, not a proof of the general theorem. Explain which hypothesis changed. Responses are temporary and do not change saved progress.

Changed task. Let T=2I+J. Find its real minimal polynomial, its complex eigenvalues and T¹⁰ without multiplying ten matrices. First express powers as aI+bJ by reducing modulo J²+I.

Feedback and a compact computation

The minimal polynomial is (t−2)²+1, because T−2I=J is non-scalar and squares to −I. Eigenvalues are 2±i. For powers, multiplication of pairs obeys (a,b)(c,d)=(ac−bd,ad+bc). Squaring gives T²=3I+4J, T⁴=−7I+24J, T⁸=−527I−336J, and T¹⁰=−237I−3116J. Verify the last multiplication using the pair rule. This is exact polynomial reduction, not a floating-point eigenvector computation.

Continue

The word “splits” matters: t²+1 is square-free over R but provides no real eigenbasis. Over C it factors into distinct linear terms. Minimal polynomials also support efficient evaluation of high powers, because every polynomial can be reduced modulo m_T. The next lesson proves that the characteristic polynomial always supplies an annihilating relation, then later lessons identify what repeated factors mean geometrically.

Review Linear Algebra I when a prerequisite needs practice. Continue to Turn polynomial identities into operator calculations.