Question

Given a 2x2 matrix: \(A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}\)

1. Derive the characteristic polynomial of matrix \(A\).
2. Find its eigenvalues.
3. Find the corresponding eigenvectors.
4. Show that \(Av = \lambda v\) for each eigenpair.
5. Verify your solution using Python (without using numpy.linalg.eig).

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Solution

1. Characteristic Polynomial

To find the eigenvalues, we solve the characteristic equation:

\[\det(A - \lambda I) = 0\]

Where \(I\) is the identity matrix. We compute:

\[A - \lambda I = \begin{bmatrix} 4 - \lambda & 2 \\ 1 & 3 - \lambda \end{bmatrix}\]

The determinant is:

\[(4 - \lambda)(3 - \lambda) - (2)(1) = 0\]

Expanding this:

\[(4 - \lambda)(3 - \lambda) - 2 = (12 - 4\lambda - 3\lambda + \lambda^2) - 2 = \lambda^2 - 7\lambda + 10 = 0\]

2. Eigenvalues

Solving the quadratic equation:

\[\lambda^2 - 7\lambda + 10 = 0\]

We use the quadratic formula:

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

Thus, the eigenvalues are:

\[\lambda_1 = 5, \quad \lambda_2 = 2\]

3. Eigenvectors

For \(\lambda_1 = 5\):

We solve:

\[(A - 5I)v = 0 \Rightarrow \begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0\]

From the equations:

\[-1x + 2y = 0 \Rightarrow x = 2y\]

So an eigenvector is:

\[v_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\]

For \(\lambda_2 = 2\):

We solve:

\[(A - 2I)v = 0 \Rightarrow \begin{bmatrix} 2 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0\]

From the equations:

\[2x + 2y = 0 \Rightarrow x = -y\]

So an eigenvector is:

\[v_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\]

4. Verification

We verify \(Av = \lambda v\):

For \(\lambda = 5\) and \(v = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\):

\[A v = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} = 5 \cdot \begin{bmatrix} 2 \\ 1 \end{bmatrix}\]

For \(\lambda = 2\) and \(v = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\):

\[A v = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \end{bmatrix} = 2 \cdot \begin{bmatrix} -1 \\ 1 \end{bmatrix}\]

5. Python Code (Without np.linalg.eig)

def find_eigen(A):
    a, b = A[0]
    c, d = A[1]

    # Characteristic polynomial coefficients: λ² - (a+d)λ + (ad - bc)
    trace = a + d
    det = a*d - b*c

    discriminant = trace**2 - 4 * det
    sqrt_disc = discriminant**(1/2)

    lambda1 = (trace + sqrt_disc) / 2
    lambda2 = (trace - sqrt_disc) / 2

    def eigenvector(matrix, eigenvalue):
        a, b = matrix[0][0] - eigenvalue, matrix[0][1]
        c, d = matrix[1][0], matrix[1][1] - eigenvalue

        if abs(a) > abs(c):
            return [b, -a]
        else:
            return [d, -c]

    v1 = eigenvector(A, lambda1)
    v2 = eigenvector(A, lambda2)

    return (lambda1, v1), (lambda2, v2)

A = [[4, 2], [1, 3]]
eig1, eig2 = find_eigen(A)

print("Eigenvalue 1:", eig1[0], "Eigenvector:", eig1[1])
print("Eigenvalue 2:", eig2[0], "Eigenvector:", eig2[1])