Characteristic Polynomial & Cayley-Hamilton Theorem for 2x2 Matrices

Concept Overview

This question tests the understanding of two fundamental concepts in linear algebra: the characteristic polynomial of a matrix and the Cayley-Hamilton theorem. The characteristic polynomial is derived from the determinant of $(A - \lambda I)$, where $A$ is the matrix, $\lambda$ is a scalar, and $I$ is the identity matrix. The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation, meaning if $p(\lambda)$ is the characteristic polynomial, then $p(A) = 0$.

Worked Solution

Let's consider a general 2x2 matrix $A$:

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

Step 1: Define the characteristic equation. The characteristic equation of a matrix $A$ is given by $|A - \lambda I| = 0$, where $\lambda$ is an eigenvalue and $I$ is the identity matrix of the same order as $A$. For a 2x2 matrix, $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

Step 2: Calculate $A - \lambda I$.

$$A - \lambda I = \begin{pmatrix} a & b \\ c & d \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix}$$

This step involves subtracting the scalar $\lambda$ from the diagonal elements of matrix $A$.

Step 3: Compute the determinant of $A - \lambda I$. The determinant of a 2x2 matrix $\begin{pmatrix} p & q \\ r & s \end{pmatrix}$ is $ps - qr$. Applying this to $A - \lambda I$:

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

This calculation expands the determinant expression.

Step 4: Expand and simplify to find the characteristic polynomial. Expanding the expression from Step 3:

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

The characteristic polynomial, $p(\lambda)$, is this expression set to zero:

$$p(\lambda) = \lambda^2 - (a+d)\lambda + (ad-bc) = 0$$

Notice that $(a+d)$ is the trace of matrix $A$ (sum of diagonal elements), denoted as $tr(A)$, and $(ad-bc)$ is the determinant of matrix $A$, denoted as $|A|$. So, the characteristic polynomial can be written as:

$$p(\lambda) = \lambda^2 - tr(A)\lambda + |A| = 0$$

Step 5: State the Cayley-Hamilton theorem. The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. This means if we replace $\lambda$ with the matrix $A$ in the characteristic polynomial equation, the result should be the zero matrix.

Step 6: Verify the Cayley-Hamilton theorem for matrix $A$. According to the theorem, $p(A)$ should be the zero matrix:

$$p(A) = A^2 - tr(A)A + |A|I = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

To verify this, we need to calculate $A^2$, $tr(A)$, $|A|$, and $I$. First, calculate $A^2$:

$$A^2 = A \cdot A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2 \end{pmatrix}$$

Next, identify $tr(A)$ and $|A|$:

$$tr(A) = a+d$$ $$|A| = ad-bc$$

Now, substitute these into the equation $A^2 - tr(A)A + |A|I$:

$$A^2 - (a+d)A + (ad-bc)I = \begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2 \end{pmatrix} - (a+d)\begin{pmatrix} a & b \\ c & d \end{pmatrix} + (ad-bc)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$= \begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2 \end{pmatrix} - \begin{pmatrix} a(a+d) & b(a+d) \\ c(a+d) & d(a+d) \end{pmatrix} + \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix}$$ $$= \begin{pmatrix} a^2 + bc - a^2 - ad & ab + bd - ab - bd \\ ca + dc - ca - cd & cb + d^2 - ad - d^2 + ad - bc \end{pmatrix}$$ $$= \begin{pmatrix} bc - ad & 0 \\ 0 & bc - ad \end{pmatrix} + \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix}$$ $$= \begin{pmatrix} (bc - ad) + (ad - bc) & 0 \\ 0 & (bc - ad) + (ad - bc) \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

Thus, the matrix $A$ satisfies its own characteristic equation, verifying the Cayley-Hamilton theorem.

Key Takeaways:

• The characteristic polynomial of a 2x2 matrix $A$ is given by $\lambda^2 - tr(A)\lambda + |A| = 0$.
• The Cayley-Hamilton theorem states that a matrix satisfies its own characteristic equation, i.e., $p(A) = 0$.
• For a 2x2 matrix $A$, this verification involves calculating $A^2$, $tr(A)$, $|A|$, and $I$, and showing that $A^2 - tr(A)A + |A|I$ equals the zero matrix.
• This theorem is crucial for finding matrix inverses and powers of matrices.

Answer: The characteristic polynomial is $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$, and the Cayley-Hamilton theorem is verified by showing $A^2 - (a+d)A + (ad-bc)I = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.