To diagonalize a $2 \times 2$ matrix, we need to follow these steps:

Given the matrix: $A = \begin{pmatrix} 4 & 2 \\ 3 & -1 \end{pmatrix}$

1. Find the eigenvalues by solving the characteristic equation:
   $\det(A - \lambda I) = 0$ where $I$ is the identity matrix, and $\lambda$ represents the eigenvalues.

2. Find the eigenvectors corresponding to each eigenvalue by solving $(A - \lambda I) \mathbf{v} = 0$.

3. Form the matrix $P$ using the eigenvectors as columns.

4. Construct the diagonal matrix $D$, where the diagonal elements are the eigenvalues.

Let's compute the eigenvalues first:

Step 1: Find the Eigenvalues

The characteristic equation is: $\det(A - \lambda I) = \det \begin{pmatrix} 4 - \lambda & 2 \\ 3 & -1 - \lambda \end{pmatrix} = 0$

Calculating the determinant: $(4 - \lambda)(-1 - \lambda) - 2 \cdot 3 = 0$

Expanding this: $(4 - \lambda)(-1 - \lambda) = -4 - 4\lambda + \lambda + \lambda^2 = \lambda^2 - 3\lambda - 4$

So: $\lambda^2 - 3\lambda - 4 - 6 = 0 \quad \Rightarrow \quad \lambda^2 - 3\lambda - 10 = 0$

Solving this quadratic equation: $\lambda = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2}$ $\lambda = \frac{3 \pm 7}{2}$ Thus, the eigenvalues are: $\lambda_1 = \frac{3 + 7}{2} = 5, \quad \lambda_2 = \frac{3 - 7}{2} = -2$

Step 2: Find the Eigenvectors

For $\lambda_1 = 5$, solve $(A - 5I) \mathbf{v} = 0$: $A - 5I = \begin{pmatrix} 4 - 5 & 2 \\ 3 & -1 - 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 3 & -6 \end{pmatrix}$ Solving $\begin{pmatrix} -1 & 2 \\ 3 & -6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, gives: $-1x + 2y = 0 \quad \Rightarrow \quad x = 2y$ Let $y = 1$, then $x = 2$. So an eigenvector corresponding to $\lambda_1 = 5$ is: $\mathbf{v_1} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$

For $\lambda_2 = -2$, solve $(A + 2I) \mathbf{v} = 0$: $A + 2I = \begin{pmatrix} 4 + 2 & 2 \\ 3 & -1 + 2 \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 3 & 1 \end{pmatrix}$ Solving $\begin{pmatrix} 6 & 2 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, gives: $6x + 2y = 0 \quad \Rightarrow \quad 3x + y = 0 \quad \Rightarrow \quad y = -3x$ Let $x = 1$, then $y = -3$. So an eigenvector corresponding to $\lambda_2 = -2$ is: $\mathbf{v_2} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$

Step 3: Form the Matrix $P$

The matrix $P$ is formed by using the eigenvectors as its columns: $P = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}$

Step 4: Construct the Diagonal Matrix $D$

The diagonal matrix $D$ is: $D = \begin{pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix}$

Thus, the matrix $A$ can be diagonalized as: $A = P D P^{-1}$

Would you like further details or have any questions?

Related Questions:

1. How do you compute the inverse of a matrix like $P$?
2. Why do we diagonalize matrices in linear algebra?
3. What are the applications of diagonalizing a matrix?
4. How would you diagonalize a $3 \times 3$ matrix?
5. What happens if a matrix does not have distinct eigenvalues?

Tip:

Diagonalizing a matrix simplifies many matrix computations, such as raising it to large powers.