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

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, http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem, 2013.11.25)

Proof1)

$\small A$ 의 eigenvalue들을  $\small \lambda_{1}, \lambda_{2},\cdots, \lambda_{n}$ 이라 하자.

Schur의 정리에 따라 $\small U^{*}AU=T$ 인 unitary matrix  $\small U$와 대각 성분이 $\small \lambda_{1}, \lambda_{2},\cdots, \lambda_{n}$인 upper triangular matrix  $\small T$가 존재한다.

$p\left ( \lambda \right )=\left ( \lambda -\lambda _{1} \right )\left ( \lambda -\lambda _{2} \right )\cdots \left ( \lambda -\lambda _{n} \right )$

$\small \Rightarrow$ $\small p(A)=p(UTU^{*})=Up(T)U^{*}$  since  $\small p(\lambda)$ is a polynomial.

$p\left ( T \right )=\left ( T-\lambda _{1}I \right )\left ( T-\lambda _{2}I \right )\cdots \left ( T-\lambda _{n}I \right )$.

Let  $\small T:=[t_{ij}]_{n\times n}$.

$\small T-\lambda_{1}I$의 첫 번째 column vector는 $\small \rm \bold{0}$ 이다.

$\small k\in \left \{ 1,2,\cdots, n-1 \right \}$ 이며  $\small (T-\lambda_{1}I)(T-\lambda_{2}I)\cdots(T-\lambda_{k}I)$의   $\small 1,2,\cdots,k$번째 column vector가 $\small \rm \bold{0}$ 이라고 가정하자.

$\small (T-\lambda_{1}I)(T-\lambda_{2}I)\cdots(T-\lambda_{k}I)$의  $\small k+1,k+2,\cdots,n$번째 column vector들을  $\small \rm\bold c_{\it k+1},\rm\bold c_{\it k+2},\cdots,\rm\bold c_{\it n}$ 이라고 하자.

$\left ( T-\lambda _{1}I \right )\left ( T-\lambda _{2}I \right )\cdots \left ( T-\lambda _{k+1}I \right )=\begin{bmatrix} \mathbf{0} & \cdots & \mathbf{0} & \mathbf{c}_{k+1} & \cdots & \mathbf{c}_{n} \end{bmatrix}\begin{bmatrix} t_{11} & \cdots & t_{1k} & t_{1\left ( k+1 \right )} & \cdots &t_{1n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \\ 0 & \cdots & t_{kk} & t_{k\left ( k+1 \right )} & \cdots & t_{kn}\\ 0 & \cdots & 0 & 0 & \cdots & t_{\left ( k+1 \right )n}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 0 & \cdots & t_{nn} \end{bmatrix}$

$\small \Rightarrow (T-\lambda_{1}I)(T-\lambda_{2}I)\cdots(T-\lambda_{k+1}I)$의  $\small 1,2,\cdots,k$번째 column vetor는 $\small \rm \bold{0}$이며  $\small k+1$번째 column vector도  $\small \sum_{i=1}^{k}t_{i(k+1)}\rm\bold0=\rm\bold 0$ 이다.

$\therefore p\left ( T \right )=\mathbf{0}$

$\therefore p\left ( A \right )=\mathbf{0}$

(Citation : Fuzhen Zhang, 1999, Springer-Verlag, Matrix Theory, 70-71p, Theorem 3.8  and 정태일, http://math1.skku.ac.kr/home/pub/1530/, 2013.11.25)

Proof2)

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

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

(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

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 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

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

This completes the proof.

(Citation : Wikipedia - Cayley-Hamilton theorem, http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem, 2013.11.25)

MT Page 340 Problem 9.8

For the matrix $A=\begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{bmatrix}$, evaluate the matrix polynomial $A^{5}+3A^{4}+A^{3}-A^{2}+4A+6I$.

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

Sol)

따라서 Cayley-Hamilton 정리에 의해$p\left ( A \right )=A^{3}-5A^{2}+8A-4I=\mathbf{0}$이다.

$x^{5}+3x^{4}+x^{3}-x^{2}+4x+6-x^{2}p\left ( x \right )=8x^{4}-7x^{3}+3x^{2}+4x+6$

$x^{5}+3x^{4}+x^{3}-x^{2}+4x+6-\left ( x^{2}+8x \right )p\left ( x \right )=33x^{3}-61x^{2}+36x+6$

$x^{5}+3x^{4}+x^{3}-x^{2}+4x+6-\left ( x^{2}+8x+33 \right )p\left ( x \right )=104x^{2}-228x+138$

따라서 $A^{5}+3A^{4}+A^{3}-A^{2}+4A+6=104A^{2}-228A+138$이다.

Note :

바로 sage를 이용해서 계산할 수도 있다.

References

Wikipedia - Cayley-Hamilton theorem, http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem, 2013.11.25

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

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

정태일, http://math1.skku.ac.kr/home/pub/1530/, 2013.11.25

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