Cayley Hamilton Theorem: Inverses, Powers and Minimal Polynomials

It lets you rewrite any polynomial in a square matrix A as a polynomial of degree less than n, where n is the size of A. The characteristic polynomial of A is p(x) = det(xI − A). The theorem states that p(A) = 0, the zero matrix, with the constant term multiplied by I. For a 2×2 matrix this reads A² − tr(A)·A + det(A)·I = 0, where tr(A) is the trace (the sum of the diagonal entries). The statement is in Hoffman and Kunze's Linear Algebra.

As of November 2026, the GATE 2027 Mathematics syllabus, organised by the Indian Institute of Technology (IIT) Madras, lists the minimal polynomial and the Cayley-Hamilton theorem. A syllabus document from the CSIR Human Resource Development Group lists "Eigenvalues and eigenvectors, Cayley-Hamilton theorem" and, separately, "canonical forms, diagonal forms, triangular forms, Jordan forms".

For a unit-by-unit reading of the CSIR NET syllabus, see our guide to the CSIR NET Mathematical Sciences syllabus.

They ask about the structure a polynomial relation forces on a matrix, not about arithmetic. We extracted the text of the GATE 2024 Mathematics paper, organised by the Indian Institute of Science (IISc) Bengaluru, and the GATE 2026 Mathematics paper, organised by IIT Guwahati. We scanned every question for matrix and eigenvalue wording and classified the hits by hand, placing a question in this family when its solution uses the characteristic polynomial, the minimal polynomial or the Jordan form. Three questions in 2024 and two in 2026 qualify:

That is 5 marks out of 100 in 2024 (1 + 2 + 2) and 3 marks in 2026 (1 + 2).

For scale, the GATE 2026 Statistical and Performance Report from IIT Guwahati shows that 2,559 of the 11,561 candidates who appeared in Mathematics qualified, with a General-category qualifying mark of 34.7 out of 100. A single 2-mark question is therefore worth about 6% of that cut-off (2 ÷ 34.7, our arithmetic).

A keyword scan can miss questions, and two papers is a small sample, so this is not a rule for later years. Linking these questions to the Cayley Hamilton theorem is our inference: the papers test the minimal polynomial and the Jordan form, and the theorem is the reason the minimal polynomial divides the characteristic polynomial.

For the full spread of marks, see our GATE Mathematics previous year papers analysis.

Pick the move from what the question gives you:

The next three sections work examples for every row of this table.

Rearrange p(A) = 0 so that A multiplies a polynomial in A, and read off the inverse. For A = [[4, 1], [2, 3]], tr(A) = 7 and det(A) = 10, so A² − 7A + 10I = 0. Multiply through by A⁻¹ to get A − 7I + 10A⁻¹ = 0. Then A⁻¹ = (7I − A)/10 = (1/10)·[[3, −1], [−2, 4]]. Multiplying back gives (1/10)·[[10, 0], [0, 10]] = I, which is your check.

For a 3×3 matrix, take D = [[1, 1, 0], [0, 1, 1], [1, 0, 1]]. Its trace is 3, the sum of its principal 2×2 minors is 3, and det D = 2, so p(x) = x³ − 3x² + 3x − 2. Hence D³ − 3D² + 3D − 2I = 0, which gives D⁻¹ = (D² − 3D + 3I)/2. With D² = [[1, 2, 1], [1, 1, 2], [2, 1, 1]], this gives D⁻¹ = (1/2)·[[1, −1, 1], [1, 1, −1], [−1, 1, 1]].

In our judgement the method pays off for 3×3 and for symbolic or structured matrices. For a plain 4×4 numeric matrix, row reduction is usually quicker.

Divide xⁿ by p(x) and use the remainder, which has degree below the size of A (ax + b for a 2×2 matrix). Because p(A) = 0, the quotient part vanishes and Aⁿ equals the remainder evaluated at A.

Distinct eigenvalues. For A = [[4, 1], [2, 3]] the eigenvalues are 2 and 5. Write xⁿ = q(x)(x − 2)(x − 5) + ax + b. Substituting x = 2 and x = 5 gives 2ⁿ = 2a + b and 5ⁿ = 5a + b. So a = (5ⁿ − 2ⁿ)/3 and b = (5·2ⁿ − 2·5ⁿ)/3, and Aⁿ = aA + bI. At n = 2, a = 7 and b = −10, which recovers A² = 7A − 10I.

A repeated eigenvalue. For B = [[2, 1], [0, 2]], p(x) = (x − 2)². Write xⁿ = q(x)(x − 2)² + ax + b. Substituting x = 2 gives 2ⁿ = 2a + b. The identity holds for every x, because both sides are polynomials. So you may differentiate it: nxⁿ⁻¹ = q′(x)(x − 2)² + 2q(x)(x − 2) + a, and at x = 2 this gives a = n·2ⁿ⁻¹. Then Bⁿ = n·2ⁿ⁻¹·B + (2ⁿ − n·2ⁿ)·I. For n = 5 this gives [[32, 80], [0, 32]].

A power that returns to I. If a matrix satisfies A² = A⁻¹, multiplying by A gives A³ = I. Then A¹⁰⁰ = (A³)³³·A = A, because 100 = 3·33 + 1. Reducing the exponent this way takes seconds.

The minimal polynomial of a matrix is the monic polynomial (leading coefficient 1) of least degree that kills it, meaning q(A) = 0. It divides every polynomial that kills the matrix, including the characteristic polynomial. So a relation like T² = T tells you the minimal polynomial divides x² − x = x(x − 1).

That has three consequences. The only possible eigenvalues are 0 and 1. The roots are distinct, so T is diagonalisable. And two such maps on the same space are similar exactly when they have the same rank, because the rank is the number of 1s on the diagonal. This is the reasoning behind the 2024 Q51 type of question.

The same pattern solves the 2026 Q28 type. A reflection through a plane in ℝ³ satisfies P² = I, so its eigenvalues are ±1. It fixes the plane (two eigenvalues equal to 1) and flips the normal direction (one eigenvalue equal to −1). So trace(P) = 1 and det(P) = −1, and 2·1 − 3·(−1) = 5.

The minimal polynomial and the characteristic polynomial can differ. Take C = diag(2, 2, 3). Its characteristic polynomial is (x − 2)²(x − 3), yet (C − 2I)(C − 3I) = 0, so its minimal polynomial is (x − 2)(x − 3).

A relation can also fail to pin things down. From A³ = I, the minimal polynomial divides x³ − 1 = (x − 1)(x² + x + 1). The matrix [[0, −1], [1, −1]] has minimal polynomial x² + x + 1 and still satisfies A³ = I.

The standard proof uses the adjugate. The adjugate adj(M) of a square matrix M is the transpose of its matrix of cofactors, and M·adj(M) = det(M)·I. Take M = xI − A with x a scalar, so (xI − A)·adj(xI − A) = p(x)·I. Every entry of the adjugate is a polynomial in x of degree at most n − 1, so adj(xI − A) = B₀ + B₁x + … + Bₙ₋₁xⁿ⁻¹, where the Bᵢ are matrices of plain numbers.

Expand the left side and match coefficients of each power of x, which gives one matrix equation per power. Multiply the equation for xᵏ on the left by Aᵏ and add them all: the left sides cancel in pairs to 0 and the right sides add to p(A). So p(A) = 0. The matching happens while x is still a scalar, which is why this works where the substitution in the first FAQ below does not.

Four mistakes are, in our view, the costliest:

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