To verify the Cayley-Hamilton theorem for a matrix $A$, we need to show that the matrix $A$ satisfies its own characteristic equation. The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation, which is given by:

$p_A(\lambda) = \det(A - \lambda I) = 0$

Where:

• $p_A(\lambda)$ is the characteristic polynomial of $A$,
• $\lambda$ is the eigenvalue,
• $I$ is the identity matrix of the same size as $A$.

Steps:

1. Write the matrix $A$: $A = \begin{bmatrix} 3 & 2 & 1 \\ 0 & 2 & 0 \\ 1 & 2 & 3 \end{bmatrix}$

2. Find the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$: $A - \lambda I = \begin{bmatrix} 3-\lambda & 2 & 1 \\ 0 & 2-\lambda & 0 \\ 1 & 2 & 3-\lambda \end{bmatrix}$

3. Compute the determinant of $A - \lambda I$: $\det(A - \lambda I) = \det\begin{bmatrix} 3-\lambda & 2 & 1 \\ 0 & 2-\lambda & 0 \\ 1 & 2 & 3-\lambda \end{bmatrix}$ This determinant can be computed via cofactor expansion.

4. Obtain the characteristic polynomial $p_A(\lambda)$.

5. Substitute the matrix $A$ into its own characteristic polynomial $p_A(A)$ to check if it equals the zero matrix $0$.

Let's compute the characteristic polynomial first and then verify the Cayley-Hamilton theorem.

I'll do the determinant calculation now. sympy as sp

Define the matrix A and the symbol for the eigenvalue lambda

lambda_ = sp.symbols('lambda') A = sp.Matrix([[3, 2, 1], [0, 2, 0], [1, 2, 3]])

Compute the characteristic polynomial p_A(lambda) = det(A - lambda*I)

I = sp.eye(3) # Identity matrix of size 3 char_poly = (A - lambda_ * I).det()

Simplify the characteristic polynomial

char_poly_simplified = sp.simplify(char_poly) char_poly_simplifiedThe characteristic polynomial of matrix $A$ is:

$p_A(\lambda) = -\lambda^3 + 8\lambda^2 - 20\lambda + 16$

Now, according to the Cayley-Hamilton theorem, the matrix $A$ must satisfy this polynomial. Therefore, we need to check if $p_A(A) = 0$. That is, we substitute $A$ into the characteristic polynomial and verify if it equals the zero matrix.

Let's compute $p_A(A)$. Substitute matrix A into the characteristic polynomial char_poly_at_A = char_poly_simplified.subs(lambda_, A)

Simplify the result to check if it is the zero matrix

char_poly_at_A_simplified = sp.simplify(char_poly_at_A) char_poly_at_A_simplifiedUpon substituting the matrix $A$ into the characteristic polynomial, we get:

$p_A(A) = 16 + \begin{bmatrix} -16 & 0 & 0 \\ 0 & -16 & 0 \\ 0 & 0 & -16 \end{bmatrix}$

This simplifies to:

$p_A(A) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

This is the zero matrix, which confirms that the matrix $A$ satisfies its own characteristic equation.

Thus, the Cayley-Hamilton theorem holds for this matrix.

Would you like further clarification on any step? Here are a few related questions for deeper exploration:

1. How do the eigenvalues of a matrix relate to the characteristic polynomial?
2. What is the general process for finding the eigenvalues of a matrix?
3. How can the Cayley-Hamilton theorem be used to find higher powers of matrices?
4. Can the Cayley-Hamilton theorem be applied to non-square matrices?
5. What is the connection between the minimal polynomial and the characteristic polynomial?

Tip: When using the Cayley-Hamilton theorem, remember that it provides a way to simplify matrix expressions by directly substituting the matrix into its own characteristic polynomial.