Cayley- Hamilton Theorem: Every Matrix* A* is a root of its characterstic polynomial *∆(t).*

**Proof:** Let *A* be an arbitrary n- square matrix and let *∆(t)* be its characterstic polynomial, say,

*∆(t) = |tI-A| = t ^{n} +a_{n-1}t^{n-1}+ …+a_{1}t+a_{0} *

Now let *B(t)* denote the classical adjoint of the matrix *tI-A*. The elements of *B(t)* are cofactors of the matrix *tI-A*, and hence are polynomials in t of degree not exceeding *n-1*. Thus

*B(t) = B _{t-1}t^{n-1} + …+ B_{t}t+B_{0}*

**Where the B_{i} are n-square matrices over K which are independent of t. By the fundamental property of the clasical adjoint (tI-A) B(t) = |tI-A|I,** or, we have

*(tI-A)(B _{n-1}t^{n-1}+ …+B_{1}t + B_{0} = (t^{n}+a_{n-1}t^{n-1}+…+ a_{1}t+a_{0})I*

Removing the parenthesis and equating the corresponding powers of* t* yields

*B _{n-1} = I, B_{n-2}-AB_{n-1} = a_{n-1}I, …, B_{0}-AB_{1}=a_{1}I, -AB_{0} = a_{0}I*

#### Multiplying the above the equations by *A*^{n}, A^{n-1}, …, A, I, respectively, yields

^{n}, A

^{n-1}, …, A, I

*A ^{n}B_{n-1} = A_{n}I, A^{n-1}B_{n-2}-A^{n}B_{n-1} = a_{n-1}A^{n-1},…, AB_{0} – A^{2}B_{1} = a_{i}A, -AB_{0} = a_{0}I*

Adding the above matrix equations yields 0 on the left hand side and *∆(A)* on the right hand side, that is,

*0 = A ^{n} + a_{n-1}A^{n-1}+…+ a_{1}A+a_{0}I*

Therefore, *∆(A) = 0*, which is the Cayley- Hamilton theorem. (Proved)