Cayley-Hamilton Theorem

Theorem

Every square ($n \times n$) matrix satisfies its own characteristic equation.

If the characteristic polynomial is $$p(\lambda) = \det(\lambda I - A) = \lambda^n + a_{n-1}\lambda^{n-1}+...+a_1\lambda + a_0$$ then from Cayley Hamilton theorem, $$p(A) = A^n + a_{n-1}A^{n-1} + a_{n-2}A^{n-2}+...+a_1 A + a_0 I = \Theta$$ where $\Theta$ is the zero matrix.

Any power of $A$ can be written as a linear combination of $I, A, A^2, ..., A^{n-1}$

Any function which has a convergent power series expansion can be expressed as a linear combination of $I, A, A^2, ..., A^{n-1}$ $$f(A) = \gamma_{n-1}A^{n-1} + \gamma_{n-2}A^{n-2} + ... + \gamma_1 A + \gamma_0 I = \sum_{j=0}^{n-1} \gamma_j A^j \triangleq g(A)$$ which is a finite polynomial.

Consider diagonalizable $A$ matrix, then, there exists a nonsingular matrix $T$ such that $T^{-1} A T = \Lambda$ where $\Lambda = \operatorname{diag}(\lambda_1...\lambda_n)$.

Then, $$f(\Lambda) = \sum_{j=0}^{n-1} \gamma_j \Lambda^j =g(\Lambda)$$

It follows that $$f(\lambda_i) = \sum_{j=0}^{n-1} \gamma_j \lambda_i^j = g(\lambda_i) \quad \text{for} \quad i = 1,2,...,n$$

By choosing $\gamma_0, \gamma_1,...\gamma_{n-1}$ to satisfy this equation we make $f(A) = g(A)$.

Theorem

For any two arbitrary polynomial functions $f(\lambda)$ and $g(\lambda)$ such that $$\frac{d^j f(\lambda_i)}{d\lambda^j} = \frac{d^j g(\lambda_i)}{d\lambda^j} \quad \text{for} \quad i=1,...,s; \quad j=0,...,n_{i-1}$$ (i.e. values of $f$ and $g$ on spectrum of $A$ are equal), then $f(A) = g(A)$.

This theorem is utilized in a special case above.

Application to finding matrix exponential

Matrix exponential can be found as follows,

#incomplete

References

1. https://crrl.poly.edu/6253/lectures/lect5.pdf
2. https://mathworld.wolfram.com/Cayley-HamiltonTheorem.html
3. P. E. Sarachik, Principles of Linear Systems, Cambridge Press, 1996, pp. 75-77.