The matrix given is $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$. To find the eigenvalues and eigenvectors of this matrix, we follow the steps below.

1. Eigenvalues

The eigenvalues $\lambda$ are found by solving the characteristic equation:

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

Where $I$ is the identity matrix and $\lambda$ is a scalar (the eigenvalue). Let's compute $A - \lambda I$:

$A - \lambda I = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix}$

Now, find the determinant of $A - \lambda I$:

$\det(A - \lambda I) = \det\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = (4 - \lambda)(3 - \lambda) - (1)(2)$

Expanding the determinant:

$(4 - \lambda)(3 - \lambda) - 2 = 12 - 7\lambda + \lambda^2 - 2 = \lambda^2 - 7\lambda + 10$

Set this equal to zero to find the eigenvalues:

$\lambda^2 - 7\lambda + 10 = 0$

Solve this quadratic equation using the quadratic formula:

$\lambda = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)} = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2}$

Thus, the eigenvalues are:

$\lambda_1 = \frac{7 + 3}{2} = 5, \quad \lambda_2 = \frac{7 - 3}{2} = 2$

2. Eigenvectors

To find the eigenvectors corresponding to each eigenvalue, we solve the equation:

$(A - \lambda I) \mathbf{v} = 0$

Eigenvector for $\lambda_1 = 5$:

Substitute $\lambda_1 = 5$ into $A - \lambda I$:

$A - 5I = \begin{pmatrix} 4 - 5 & 1 \\ 2 & 3 - 5 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}$

Now, solve $(A - 5I) \mathbf{v} = 0$, i.e.,

$\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

This gives the system of equations:

$- v_1 + v_2 = 0$ $2v_1 - 2v_2 = 0$

From the first equation, $v_1 = v_2$. Thus, the eigenvector corresponding to $\lambda_1 = 5$ is:

$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Eigenvector for $\lambda_2 = 2$:

Substitute $\lambda_2 = 2$ into $A - \lambda I$:

$A - 2I = \begin{pmatrix} 4 - 2 & 1 \\ 2 & 3 - 2 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}$

Now, solve $(A - 2I) \mathbf{v} = 0$, i.e.,

$\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

This gives the system of equations:

$2v_1 + v_2 = 0$ $2v_1 + v_2 = 0$

Both equations are the same, so we get $v_2 = -2v_1$. Thus, the eigenvector corresponding to $\lambda_2 = 2$ is:

$\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$

Summary:

• Eigenvalues: $\lambda_1 = 5$, $\lambda_2 = 2$
• Eigenvector for $\lambda_1 = 5$: $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
• Eigenvector for $\lambda_2 = 2$: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$

Would you like further details on any of these steps, or have any specific questions?

---

Here are five related questions:

1. How do you interpret eigenvalues and eigenvectors geometrically?
2. Can a matrix have complex eigenvalues?
3. What is the significance of the determinant when finding eigenvalues?
4. How do eigenvectors relate to the diagonalization of a matrix?
5. How are eigenvalues used in solving systems of differential equations?

Tip: The eigenvectors corresponding to distinct eigenvalues are always linearly independent!