MT 김신휘 Cayley-Hamilton Theorem - Revised정태일

Cayley-Hamilton Theorem

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

More precisely,[1][2][3][4] if A is a given n×n matrix and In  is the n×n identity matrix, then the characteristic polynomial of A is defined as

p(\lambda)=\det(\lambda I_n-A)~,

where "det" is the determinant operation. Since the entries of the matrix are (linear or constant) polynomials in λ, the determinant is also an n-th order polynomial in λ.

The Cayley–Hamilton theorem states that "substituting" the matrix A for λ in this polynomial results in the zero matrixp(A) = 0 

(Citation : Wikipedia - Cayley-Hamilton theorem,, 2013.11.25)


 의 eigenvalue들을   이라 하자.

Schur의 정리에 따라  인 unitary matrix  와 대각 성분이 인 upper triangular matrix  가 존재한다.

   since   is a polynomial.


Let  .

  의 첫 번째 column vector는  이다.

 이며  의   번째 column vector가  이라고 가정하자.

의  번째 column vector들을   이라고 하자.

의  번째 column vetor는 이며  번째 column vector도   이다.

(Citation : Fuzhen Zhang, 1999, Springer-Verlag, Matrix Theory, 70-71p, Theorem 3.8  and 정태일,, 2013.11.25)


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

(t I_n - A) \cdot B = \det(t I_n - A) I_n = p(t) I_n.

Since B is also a matrix with polynomials in t as entries, one can for each i collect the coefficients of ti in each entry to form a matrix B i of numbers, such that one has

B = \sum_{i = 0}^{n - 1} t^i B_i

(the way the entries of B are defined makes clear that no powers higher than tn−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 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

\begin{align}  p(t) I_n &= (t I_n - A) \cdot B \\  &=(t I_n - A) \cdot\sum_{i = 0}^{n - 1} t^i B_i  \\  &=\sum_{i = 0}^{n - 1} tI_n\cdot t^i B_i - \sum_{i = 0}^{n - 1} A\cdot t^i B_i \\  &=\sum_{i = 0}^{n - 1} t^{i + 1}  B_i- \sum_{i = 0}^{n - 1} t^i A\cdot B_i  \\  &=t^n B_{n - 1} + \sum_{i = 1}^{n - 1}  t^i(B_{i - 1} - A\cdot  B_i) - A \cdot B_0. \end{align}



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 ti is the same on both sides; it follows that the constant matrices with coefficient ti in both expressions must be equal. Writing these equations for i from n down to 0 one finds

B_{n - 1} = I_n, \qquad B_{i - 1} - A\cdot B_i = c_i I_n\quad \text{for }1 < i < n-1, \qquad -A B_0 = c_0 I_n.

We multiply the equation of the coefficients of ti from the left by Ai, and sum up; the left-hand sides form a telescoping sum and cancel completely, which results in the equation

 0=A^n+c_{n-1}A^{n-1}+\cdots+c_1A+c_0I_n= p(A).

This completes the proof.

(Citation : Wikipedia - Cayley-Hamilton theorem,, 2013.11.25)

MT Page 340 Problem 9.8

For the matrix , evaluate the matrix polynomial .

(Citation : Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 340p)


따라서 Cayley-Hamilton 정리에 의해이다.

따라서 이다.


Note : 

바로 sage를 이용해서 계산할 수도 있다.


Wikipedia - Cayley-Hamilton theorem,, 2013.11.25

Fuzhen Zhang, 1999, Springer-Verlag, Matrix Theory, 70-71p

Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 340p

정태일,, 2013.11.25

SKKU Matrix Theory Contents

Back to the Index Page