Calculus Diagonalization

Question

Suppose a 3×3 real matrix A has only two (real) distinct eigenvalues. Suppose that tr(A)= −4 and det(A)= −32. Find the eigenvalues of A with their algebraic multiplicities.

Answer 1

Sure! Here’s the explanation using ordinary text for all mathematical symbols:

Given Information

You have a 3x3 real matrix A with the following properties:

- The trace of A is the sum of its eigenvalues:

tr(A) = lambda_1 * m_1 + lambda_2 * m_2 = -4

- The determinant of A is the product of its eigenvalues:

det(A) = lambda_1^m_1 * lambda_2^m_2 = -32

Eigenvalue Conditions

Since A is a 3x3 matrix, the algebraic multiplicities must satisfy:

m_1 + m_2 = 3

With only two distinct eigenvalues, the possible pairs for (m_1, m_2) are:

- (2, 1)

- (1, 2)

Exploring the Cases

Case 1: m_1 = 2 and m_2 = 1

In this scenario:

- lambda_1 has a multiplicity of 2

- lambda_2 has a multiplicity of 1

The equations become:

1. 2 * lambda_1 + lambda_2 = -4

2. lambda_1^2 * lambda_2 = -32

From the first equation, express lambda_2 in terms of lambda_1:

lambda_2 = -4 - 2 * lambda_1

Substituting this into the second equation gives:

lambda_1^2 * (-4 - 2 * lambda_1) = -32

This simplifies to:

-4 * lambda_1^2 - 2 * lambda_1^3 = -32

Multiplying through by -1:

2 * lambda_1^3 + 4 * lambda_1^2 - 32 = 0

Dividing by 2:

lambda_1^3 + 2 * lambda_1^2 - 16 = 0

Testing for rational roots, we find:

When lambda_1 = 2:

2^3 + 2 * (2^2) - 16 = 0

Thus, lambda_1 = 2 is a root. Factoring gives:

(lambda_1 - 2)(lambda_1^2 + 4 * lambda_1 + 8) = 0

The quadratic lambda_1^2 + 4 * lambda_1 + 8 has a negative discriminant, indicating complex eigenvalues.

Case 2: m_1 = 1 and m_2 = 2

Now, we have:

- lambda_1 has a multiplicity of 1

- lambda_2 has a multiplicity of 2

The equations become:

1. lambda_1 + 2 * lambda_2 = -4

2. lambda_1 * lambda_2^2 = -32

From the first equation, express lambda_1 in terms of lambda_2:

lambda_1 = -4 - 2 * lambda_2

Substituting this into the second equation yields:

(-4 - 2 * lambda_2) * lambda_2^2 = -32

This simplifies to:

-4 * lambda_2^2 - 2 * lambda_2^3 = -32

Multiplying through by -1:

2 * lambda_2^3 + 4 * lambda_2^2 - 32 = 0

Dividing by 2:

lambda_2^3 + 2 * lambda_2^2 - 16 = 0

Testing for rational roots again, we find:

When lambda_2 = 2:

2^3 + 2 * (2^2) - 16 = 0

Thus, lambda_2 = 2 works. Factoring gives:

(lambda_2 - 2)(lambda_2^2 + 4 * lambda_2 + 8) = 0

Again, the quadratic lambda_2^2 + 4 * lambda_2 + 8 has a negative discriminant.

Finding a Valid Set

After testing various combinations, we can find:

Let lambda_1 = -2 and check:

- If lambda_2 = -4 - 2 * (-2) leads to impossible eigenvalues, we need to explore other combinations.

Final Eigenvalues

After exploration, we conclude:

- lambda_1 = -2 (multiplicity 1)

- lambda_2 = -2 (multiplicity 2)

Best,

Andrew