Turn polynomial identities into operator calculations

Try

For A=[[2,1],[1,2]], trace is 4 and determinant is 3. Calculate A²=[[5,4],[4,5]] and verify A²−4A+3I=0 entry by entry. This relation lets you compute A³ without a third matrix multiplication: A³=4A²−3A=13A−12I. The result is [[14,13],[13,14]]. Compare it with direct multiplication once, so the polynomial shortcut is tied to a concrete calculation rather than an unexplained theorem.

Predict

Predict whether the identity A²−4A+3I=0 means each entry of A individually solves t²−4t+3=0. It does not: the entry 2 does not solve that scalar equation. Polynomial evaluation of an operator uses matrix powers and the identity matrix. Off-diagonal interactions are part of multiplication. This distinction is particularly important when moving between elementwise software operations and genuine matrix multiplication.

Experiment

Use the relation to build a small table: A⁰=I, A¹=A, A²=4A−3I, A³=13A−12I, A⁴=40A−39I. Derive each row from the previous one and check its trace using eigenvalues 1 and 3. Then rearrange the original relation to obtain A⁻¹=(4I−A)/3. Multiply A by this candidate to verify the inverse. Notice that the constant term 3 being nonzero is essential to this construction.

Understand

Cayley–Hamilton states χ_A(A)=0. Here is a coefficient proof that avoids careless substitution into matrix products. Write χ_A(t)=c₀+c₁t+⋯+cₙtⁿ with cₙ=1, and adj(tI−A)=B₀+B₁t+⋯+Bₙ₋₁tⁿ⁻¹. The adjugate identity gives −AB₀=c₀I, Bⱼ₋₁−ABⱼ=cⱼI for 1≤j≤n−1, and Bₙ₋₁=I. The adjugate commutes with tI−A, so comparison of coefficients gives ABⱼ=BⱼA. Multiply the jth equation by Aʲ and sum, including the first and final equations. The B terms telescope and leave c₀I+c₁A+⋯+Aⁿ=0. All operations are coefficient identities over the original field, so no diagonalizability assumption was used.

Apply

It follows that the minimal polynomial divides the characteristic polynomial. More generally, if two scalar polynomials differ by a multiple of m_T, their operator values agree. The map p↦p(T) respects addition and multiplication because powers of the same operator commute. Polynomial division therefore gives a finite computational calculus for T. For an invertible T, a nonzero constant coefficient in m_T produces a polynomial formula for T⁻¹; if the coefficient were zero, removing a factor t would contradict minimality after multiplying by T⁻¹. Explain why this argument works even for a defective matrix with no eigenbasis.

The coefficient calculation can be checked completely in dimension two. Write adj(tI−A)=B₀+tI. The t coefficient says B₀−A=c₁I, while the constant coefficient says −AB₀=c₀I. Substitute the first identity into the second to get A²+c₁A+c₀I=0. This is the same relation derived numerically above, now proved for every two-by-two matrix. Keeping this small proof alongside the general telescoping argument helps locate the precise role of matrix multiplication order.

Turn a scalar recurrence into a correctly indexed state

Let uₙ₊₂=s uₙ₊₁−p uₙ for n≥0, with two specified initial values. Define zₙ=(uₙ₊₁,uₙ). Then zₙ₊₁=Czₙ for C with rows (s,−p),(1,0), and z₀=(u₁,u₀). Induction gives zₙ=Cⁿz₀. Conversely, the second row copies the previous first coordinate, so this matrix evolution recovers exactly the scalar recurrence and its initial conditions. Swapping the state coordinates without changing C gives a different calculation.

The characteristic polynomial is t²−st+p. Cayley–Hamilton supplies C²−sC+pI=0, hence the recurrence for every initial vector. This is also C’s minimal polynomial: its nonzero lower-left entry makes C nonscalar, so no degree-one polynomial can annihilate it. A particular initial state or scalar observation can nonetheless remove a mode, giving a shorter recurrence for that sequence.

One matrix, different observed recurrence orders

Predict whether changing the initial pair removes a mode. The state is ordered (uₙ₊₁,uₙ), not (uₙ,uₙ₊₁). The companion matrix describes every initial pair; this particular scalar sequence may obey a shorter relation.

Characteristic roots 1 and 22 repeated0 repeated Initial u₀ -1012 Initial u₁ -1012 State index n 012345678910

Companion matrix rows (3, -2), (1, 0). Its minimal polynomial is (t−1)(t−2); this sequence's minimal shift polynomial is t−2, order 1. The latter is derived from the proved family, not guessed from this finite table.

n	Direct recurrence	Closed formula	Difference
0	1	1	0
1	2	2	0
2	4	4	0
3	8	8	0
4	16	16	0
5	32	32	0
6	64	64	0
7	128	128	0
8	256	256	0
9	512	512	0
10	1024	1024	0
11	2048	2048	0
12	4096	4096	0

The constant mode vanishes when u₁=2u₀; the 2ⁿ mode vanishes when u₁=u₀. The all-zero sequence has minimal polynomial1 and order0.

Reset

Thirteen terms and eleven state choices bound the work; all displayed values are exact bounded integers. A finite matching prefix alone cannot prove an all-index formula. State is temporary and awards no mastery.

Distinct roots, repeated roots and the zero boundary

For s=3,p=2, the roots are1 and2. Solving the initial equations gives uₙ=(2u₀−u₁)+(u₁−u₀)2ⁿ. Verify both initial values and substitute into the recurrence; induction then proves the formula for every n. Eigenvector decomposition of C gives the same two coefficients. If u₁=u₀ the exponential term disappears; if u₁=2u₀ the constant term disappears. The operator still has two modes even when the observed sequence has only one.

For s=4,p=4, write C=2I+N. The matrix N is nonzero and N²=0, so for n≥1 the binomial expansion gives Cⁿ=2ⁿI+n2ⁿ⁻¹N; C⁰=I separately. Therefore uₙ=u₀2ⁿ+(u₁−2u₀)n2ⁿ⁻¹ for n≥1, with u₀ at n=0. The factor n is the repeated-root contribution. It vanishes exactly when u₁=2u₀. This is not a failure of the recurrence; it is a special initial condition.

If both roots are zero, C²=0. The sequence is u₀,u₁,0,0,… . Treat this directly rather than substituting zero into an expression containing a negative root power. Nilpotent finite support is a legitimate solution, not an undefined numerical value. Initial conditions and the n=0 boundary must be checked before simplifying a closed form.

Operator annihilators need not be minimal sequence relations

For a sequence a, let E be the left shift, (Ea)ₙ=aₙ₊₁. A polynomial q(E) annihilates a when its linear combination of shifted values is zero for every n≥0. If aₙ=ℓ(Tⁿv), then q(T)=0 implies q(E)a=0 by applying ℓTⁿ to q(T)v. The converse need not hold: the initial state or measurement may miss part of T. The monic sequence annihilator of least degree divides every other one by polynomial division, just as for operators. The all-zero sequence has minimal polynomial1, of degree zero.

For the two distinct real modes above, if both coefficients are nonzero the first three terms satisfy u₀u₂−u₁²=(2u₀−u₁)(u₁−u₀)≠0. Any first-order relation uₙ₊₁=c uₙ would make this determinant zero, so the second-order recurrence is minimal. When one coefficient vanishes, the surviving mode gives t−1 or t−2. In the repeated-root case the same determinant is −(u₁−2u₀)², so a nonzero repeated-root term cannot obey a first-order homogeneous relation. For the zero-root case, u₁≠0 gives determinant−u₁² and minimal t²; if u₁=0 but u₀≠0 the minimal polynomial is t. These statements concern homogeneous constant-coefficient recurrences starting at n=0.

A finite matching table alone does not prove a recurrence for an unknown sequence: another sequence can share the entire displayed prefix and differ at the next index. Here the family and initial data are specified, and induction supplies the all-index justification. The table independently checks the implementation of that proved construction.

A changed sequence with a genuinely repeated mode

Consider aₙ=3+(n−1)2ⁿ. Its initial values are2,3,7. The polynomial (t−1)(t−2)²=t³−5t²+8t−4 annihilates it, giving aₙ₊₃=5aₙ₊₂−8aₙ₊₁+4aₙ. To prove minimality rather than guess from three values, note that q(E)(n2ⁿ)=2ⁿ(nq(2)+2q′(2)). Also q(E)1=q(1) and q(E)2ⁿ=2ⁿq(2). The sequences1,2ⁿ,n2ⁿ are independent: apply (E−2)² to a purported dependence to isolate the constant coefficient, then divide the remaining identity by2ⁿ and compare n=0,1. Hence any annihilator must satisfy q(1)=q(2)=q′(2)=0. It is divisible by the displayed cubic, proving the minimum order is three.

Before using a power formula, write the ordered state, initial vector and valid index range. Then compare a generic initial pair with one that cancels a mode. For preparation, revisit minimal polynomials and their divisibility property and Jordan chains.

Continue

Polynomial reduction is exact algebra. Floating-point evaluation can still amplify rounding error, especially for poorly conditioned coordinate systems or large powers; the theorem does not supply a numerical stability guarantee. The next lesson uses coprime factors of the minimal polynomial to separate invariant pieces. This is stronger information than a list of eigenvalues and provides the route to generalized eigenspaces and Jordan chains.

Review Linear Algebra I when a prerequisite needs practice. Continue to Separate an operator into its primary pieces.