Explanation:

a) A is similar to B:
If A and B have the same minimal polynomial, it means that they have the same characteristic polynomial, and therefore the same eigenvalues. This implies that A and B have the same eigenvalue multiplicities.

Since A and B have the same eigenvalues and eigenvalue multiplicities, it follows that they have the same Jordan canonical form. The Jordan canonical form is unique up to the order of the Jordan blocks, so A and B must be similar.

b) A is diagonalizable if B is diagonalizable:
If B is diagonalizable, it means that it can be written as B = PDP^-1, where D is a diagonal matrix and P is an invertible matrix. Since A and B have the same minimal polynomial, they have the same characteristic polynomial, and therefore the same eigenvalues.

If A is diagonalizable, it means that it can be written as A = QDQ^-1, where D is a diagonal matrix and Q is an invertible matrix. Since A and B have the same eigenvalues, it follows that D is the same diagonal matrix for both A and B.

Therefore, if B is diagonalizable, A can also be written as A = PDQ^-1, where D is the same diagonal matrix as in the diagonalization of B. This implies that A is also diagonalizable.

c) A - B is singular:
Since A and B have the same minimal polynomial, they have the same characteristic polynomial, and therefore the same eigenvalues. Let λ be an eigenvalue of A and B. Then A - B has eigenvalues given by λ - λ = 0.

If A - B has eigenvalue 0, it means that A - B is not invertible, and therefore it is singular.

d) A and B commute:
To show that A and B commute, we need to show that AB = BA. Since A and B have the same minimal polynomial, they have the same characteristic polynomial, and therefore the same eigenvalues.

Let λ be an eigenvalue of A and B. Then we have:

ABv = BA v = λ Bv

where v is an eigenvector corresponding to the eigenvalue λ. This implies that ABv and BA v are scalar multiples of Bv. Since A and B have the same eigenvalues, ABv and BA v are scalar multiples of v as well.

Therefore, ABv = BA v for all eigenvalues λ and corresponding eigenvectors v, which implies that A and B commute.

Therefore, the correct answer is option 'B' - A is diagonalizable if B is diagonalizable.