Cayley-Hamilton Theorem
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Direct answer
Every square matrix satisfies its own characteristic equation. If the characteristic polynomial of A (from det(A - λI) = 0 or det(λI - A) = 0) is λ^n + c1 λ^(n-1) + ... + cn, then the matrix identity A^n + c1 A^(n-1) + ... + cn I = O holds — the theorem substitutes the matrix for the scalar variable, with the constant term landing on cn I. For a 2 × 2 matrix this reads A^2 - (tr A) A + (det A) I = O, and rearranging it produces the two workhorse results: A^2 = (tr A) A - (det A) I, and when det A ≠ 0, A^(-1) = [(tr A) I - A]/det A — the inverse found without adjoints or row reduction. Higher powers reduce through polynomial division: divide λ^k by the characteristic polynomial and substitute A into the remainder.
What you must remember
- The theorem: every square matrix obeys its own characteristic equation; the substitution replaces λ by A and the constant term by (constant) × I.
- 2 × 2 identity: A^2 - (tr A)A + (det A)I = O; from it, A^2 = (tr A)A - (det A)I and A^(-1) = ((tr A)I - A)/det A for det A ≠ 0.
- Characteristic polynomial conventions: det(A - λI) and det(λI - A) differ by (-1)^n; pick one and stay consistent, or the middle signs flip.
- Higher powers: to get A^k, divide x^k by the characteristic polynomial and evaluate the remainder at A — the quotient's contribution collapses by the theorem.
- Trace and determinant as fingerprints: eigenvalues satisfy Σλi = tr A and Πλi = det A; the theorem is the polynomial side of the same coin.
- 3 × 3 form: A^3 - (tr A)A^2 + (sum of principal 2 × 2 minors)A - (det A)I = O — the middle coefficient is the sum of the three principal minors.
- Singularity read instantly: a zero constant term in the characteristic polynomial means det A = 0, so the theorem directly reflects invertibility.
Inverting a matrix without adjoints
Let A have rows (1, 2) and (2, 1). The trace is 2 and the determinant is -3, so the Cayley-Hamilton identity reads A^2 - 2A - 3I = O. Verify by direct multiplication: A^2 has rows (5, 4) and (4, 5), and 2A + 3I has rows (5, 4) and (4, 5) — the identity holds. Now rearrange for the inverse: A^2 - 2A = 3I factors on the left as A(A - 2I) = 3I, so A^(-1) = (A - 2I)/3. Compute A - 2I: rows (-1, 2) and (2, -1), so A^(-1) has rows (-1/3, 2/3) and (2/3, -1/3). Confirm by multiplication: the first diagonal entry of A × A^(-1) is 1 × (-1/3) + 2 × (2/3) = -1/3 + 4/3 = 1, and the off-diagonal 1 × (2/3) + 2 × (-1/3) = 0, with the second row checking by symmetry. Note the equivalent closed form: A^(-1) = ((tr A)I - A)/det A = (2I - A)/(-3), the same matrix — and note where sign discipline matters, since the constant term (-3, not 3) must travel with the rearrangement. The whole inverse took three lines and no adjoint, which is precisely the exam appeal.
Where students slip
JEE Main asks the identity itself (compute A^2 - 5A + 7I for a matrix whose characteristic polynomial is known to be λ^2 - 5λ + 7 — the answer is O) or the quick inverse via ((tr A)I - A)/det A. Advanced wants A^k for larger k (say A^5) through remainder division, the 3 × 3 version with principal minors, or Cayley-Hamilton embedded in questions about eigenvalues. The recurring errors: substituting the constant term without the identity matrix — A^2 - 2A - 3 ≠ O, only A^2 - 2A - 3I = O — the single most marked slip in this chapter; mixing the two characteristic-polynomial conventions so middle signs flip; and assuming the theorem lets you divide by A when det A = 0 (the rearrangement for the inverse needs invertibility). A subtler point worth one line of defence: the theorem is a matrix identity, not a determinant statement — det(A - A·I) = 0 is vacuously true and proves nothing. Matrices form a named unit in both syllabi, and this theorem is its standard crown jewel question.
Frequently asked questions
What does the Cayley-Hamilton theorem state?
Every square matrix satisfies its own characteristic equation: substitute A for λ (and append I to the constant term) in det(λI - A) = 0, and the resulting matrix identity holds.
How is the inverse of a 2 × 2 matrix found using the theorem?
From A^2 - (tr A)A + (det A)I = O, rearrange to A^(-1) = ((tr A)I - A)/det A — valid whenever det A ≠ 0.
Why must the constant term be multiplied by I?
Because the other terms are matrices; only cI can be added to them, and the identity matrix plays the role of the scalar 1 in polynomial evaluation.
How do you compute high powers like A^5 for a 2 × 2 matrix?
Divide x^5 by the characteristic polynomial and substitute A into the remainder (of degree at most 1): the theorem collapses the quotient's contribution to zero.
Does Cayley-Hamilton work for singular matrices?
Yes — the identity holds for every square matrix; only the further step of dividing by A to extract an inverse requires det A ≠ 0.