In linear algebra, the

**Cayley–Hamilton theorem**(named after the mathematicians Arthur Cayley and William Hamilton) states that every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation.
More precisely:

If

*A*is a given*n*×*n*matrix and*I*is the_{n}*n*×*n*identity matrix, then the characteristic polynomial of*A*is defined as
where "det" is the determinant operation. Since the entries of the matrix are (linear or constant) polynomials in λ, the determinant is also a polynomial in λ. The Cayley–Hamilton theorem states that "substituting" the matrix

*A*for λ in this polynomial results in the zero matrix:
The powers of λ that have become powers of

*A*by the substitution should be computed by repeated matrix multiplication, and the constant term should be multiplied by the identity matrix (the zeroth power of*A*) so that it can be added to the other terms. The theorem allows*A*^{n}to be expressed as a linear combination of the lower matrix powers of*A*.
When the ring is a field, the Cayley–Hamilton theorem is equivalent to the statement that the minimal polynomial of a square matrix divides its characteristic polynomial.

**Example**

As a concrete example, let

- .

Its characteristic polynomial is given by

The Cayley–Hamilton theorem claims that, if we

*define**p*(*X*) =*X*^{2}− 5*X*− 2*I*_{2},

then

which one can verify easily.

### A direct algebraic proof

This proof uses just the kind of objects needed to formulate the Cayley–Hamilton theorem: matrices with polynomials as entries. The matrix

*t**I*_{n}−*A*whose determinant is the characteristic polynomial of*A*is such a matrix, and since polynomials form a commutative ring, it has an adjugate
Then according to the right hand fundamental relation of the adjugate one has

Since

*B*is also a matrix with polynomials in*t*as entries, one can for each*i*collect the coefficients of*t*^{i}in each entry to form a matrix*B*_{i}of numbers, such that one has
(the way the entries of

*B*are defined makes clear that no powers higher than*t*^{n − 1}occur). While this*looks*like a polynomial with matrices as coefficients, we shall not consider such a notion; it is just a way to write a matrix with polynomial entries as linear combination of constant matrices, and the coefficient*t*^{i}has been written to the left of the matrix to stress this point of view. Now one can expand the matrix product in our equation by bilinearity
Writing , one obtains an equality of two matrices with polynomial entries, written as linear combinations of constant matrices with powers of

*t*as coefficients. Such an equality can hold only if in any matrix position the entry that is multiplied by a given power*t*^{i}is the same on both sides; it follows that the constant matrices with coefficient*t*^{i}in both expressions must be equal. Writing these equations for*i*from*n*down to 0 one finds
We multiply the equation of the coefficients of

*t*^{i}from the left by*A*^{i}, and sum up; the left-hand sides form a telescoping sum and cancel completely, which results in the equation
This completes the proof.

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