5.md

Cayley-Hamilton Theorem

TOC

• Cayley-Hamilton Theorem
  • TOC
  • Explanation
    • 1. Characteristic Polynomial
    • 2. Characteristic Equation
    • 3. Cayley-Hamilton Theorem
    • 4. Implications
    • Example
  • Solution

Explanation

The Cayley-Hamilton theorem is a fundamental result in linear algebra that applies to square matrices. The theorem states that every square matrix satisfies its own characteristic equation.

Here's a step-by-step explanation of the concepts involved:

1. Characteristic Polynomial

Given an $n \times n$ square matrix $A$, you can find its characteristic polynomial by taking the determinant of $\lambda I - A$, where $\lambda$ is a scalar and $I$ is the identity matrix of the same size as $A$. The characteristic polynomial is given by:

$$ p(\lambda) = \det(\lambda I - A) $$

This polynomial is of degree $n$ and has coefficients that are functions of the entries of matrix $A$.

2. Characteristic Equation

The characteristic equation is obtained by setting the characteristic polynomial equal to zero:

$$ p(\lambda) = 0 $$

Solving this equation gives you the eigenvalues of the matrix $A$.

3. Cayley-Hamilton Theorem

The Cayley-Hamilton theorem tells us that if you substitute the matrix $A$ itself into its characteristic polynomial, the result is the zero matrix. In other words, if $p(\lambda)$ is the characteristic polynomial of $A$, then:

$$ p(A) = 0 $$

Here, $p(A)$ means that every occurrence of $\lambda$ in the polynomial is replaced by the matrix $A$, and the operations are carried out according to matrix algebra.

4. Implications

This theorem has several important implications:

• It provides a way to compute functions of matrices (like matrix exponentials) without directly dealing with potentially difficult or large-scale computations.
• It gives a direct method for verifying that a given list of numbers are indeed the eigenvalues of a matrix.
• It can be used to establish the minimal polynomial of a matrix, which is a polynomial of the least degree that the matrix satisfies.

Example

Consider a $2 \times 2$ matrix:

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

Its characteristic polynomial is:

$$ p(\lambda) = \det(\lambda I - A) = (\lambda - a)(\lambda - d) - bc $$

Expanding this, we get:

$$ p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc) $$

According to the Cayley-Hamilton theorem, $A$ should satisfy $p(A) = A^2 - (a + d)A + (ad - bc)I = 0$.

Would you like to see a concrete numerical example worked out, or do you have any specific questions about the theorem?

Solution

5. Prove that for any square matrix $A(n\times n)$ with eigenvalues ${\lambda_1, \lambda_2, ..., \lambda_n}$ the multiplication: $(A-\lambda_1I)(A-\lambda_2I)...(A-\lambda_nI)$ or produces the zero matrix? (12 pts.)

To prove that the multiplication $(A-\lambda_1I)(A-\lambda_2I)\ldots(A-\lambda_nI)$ produces the zero matrix for any square matrix $A$ of size $n \times n$ with eigenvalues ${\lambda_1, \lambda_2, \ldots, \lambda_n}$, we can use the Cayley-Hamilton theorem.

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. If we denote the characteristic polynomial of $A$ as $p(\lambda)$, then by the theorem, $p(A) = 0$, where $p(A)$ is obtained by substituting $A$ into the polynomial $p(\lambda)$.

For a matrix $A$ with eigenvalues ${\lambda_1, \lambda_2, \ldots, \lambda_n}$, the characteristic polynomial can be written as:

$$ p(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \ldots (\lambda - \lambda_n) $$

According to the Cayley-Hamilton theorem, by substituting $\lambda$ with $A$, we get:

$$ p(A) = (A - \lambda_1 I)(A - \lambda_2 I) \ldots (A - \lambda_n I) = 0 $$

This is the zero matrix. Thus, the multiplication of $(A - \lambda_1 I)(A - \lambda_2 I) \ldots (A - \lambda_n I)$ indeed produces the zero matrix, which is what we wanted to prove.

This proof relies on the fact that the polynomial we obtain by multiplying $(A - \lambda_i I)$ for each eigenvalue $\lambda_i$ is exactly the characteristic polynomial of $A$, and by the Cayley-Hamilton theorem, the substitution of $A$ into its characteristic polynomial yields the zero matrix.