Need help with linear algebra problem involving diagonalization.

Question

1. Let A be the matrix given below, in which a and b are real numbers.

A = 

[1 0 0 b]

[0 a a 0]

[0 0 a 0]

[0 0 0 b]

(a) Find all values of a and b for which A is diagonalizable.

(b) For any values for which it is diagonalizable, diagonalize it (meaning find the invertible P and diagonal D such that A = PDP^−1 ).

Some notes about the problem. • In order to give a complete answer to part (a), you must both show that A is diagonalizable for the values you chose and show it is not diagonalizable for the other values. • Take care to organize your work clearly and explain, in detail and with words, how and why you’re choosing the cases you’re choosing, and how you’re coming to your conclusions in each case. Your submission for this problem should be the polished final product of a larger amount of rough work. Answers which are a disorganized collage of unexplained computations will get low scores, even if those computations appear to be correct.

Answer 1

Hey Alex, here is a step by step solution:

Let A be the matrix:

A =

[1 0 0 b]

[0 a a 0]

[0 0 a 0]

[0 0 0 b]

Part (a): Conditions for Diagonalizability

A matrix is diagonalizable if it has a complete set of linearly independent eigenvectors. We will find the eigenvalues and their corresponding eigenvectors to determine the conditions on parameters a and b

Step 1: Finding the Eigenvalues

The characteristic polynomial can be found using the determinant of (A - lambda * I):

A - lambda * I =

[1 - lambda 0 0 b]

[0 a - lambda a 0]

[0 0 a - lambda 0]

[0 0 0 b - lambda]

The determinant is calculated as: det(A - lambda * I) = (1 - lambda) * det

[ a - lambda a]

[ 0 a - lambda]

Calculating the 2x2 determinant: det

[ a - lambda a]

[ 0 a - lambda]

= (a - lambda)^2

Thus, the characteristic polynomial becomes: det(A - lambda * I) = (1 - lambda) * (a - lambda)^2 * (b - lambda)

The eigenvalues are:

1. lambda_1 = 1
2. lambda_2 = a (with algebraic multiplicity 2)
3. lambda_3 = b

Step 2: Conditions for Diagonalizability

1. Distinct Eigenvalues: If all eigenvalues are distinct, then A is diagonalizable. This occurs when a is not equal to 1 and a is not equal to b.
2. Repeated Eigenvalues: When a is repeated (multiplicity 2), we need to check the geometric multiplicity. For A to be diagonalizable when a has multiplicity 2, the dimension of the eigenspace associated with a must equal 2.

Case Analysis:

1. Case 1: a is not equal to 1 and a is not equal to b
2. Diagonalizable: All eigenvalues are distinct.
3. Case 2: a is equal to b and a is not equal to 1
4. Not Diagonalizable: Eigenvalue a has multiplicity 2 but does not provide 2 linearly independent eigenvectors.
5. Case 3: a is equal to 1 and a is not equal to b
6. Diagonalizable: Eigenvalues are distinct.
7. Case 4: a is equal to b and a is equal to 1
8. Not Diagonalizable: Eigenvalue a has multiplicity 3 (not enough independent eigenvectors).

Summary of Diagonalizability Conditions

1. Diagonalizable:
2. a is not equal to 1 and a is not equal to b
3. a is equal to 1 and a is not equal to b
4. Not Diagonalizable:
5. a is equal to b and a is not equal to 1
6. a is equal to b and a is equal to 1

I could not answer part B due to character limit in this post.

Best,

Andrew