(Here α and β are replaced by A and B, and T stands for transposition)

First of all, I said that ab is linearly independent, so A and B must not be zero vectors, that is, ab^T must not be a zero matrix.

A

The rank of f is the rank of ab T, because AB T = (B 1A T, b2a^T, b3a^T), obviously the latter two terms can be changed to 0 through the first term, that is, the rank is 1. (Of course, it can also be written as (a1b t, a2b^T, A3B T) T, and the conclusion is the same. )

B

Since A is right, the standard form should have only one variable, that is, f = z 2.

C

If it is positive definite, the eigenvalue must be greater than 0, rank = matrix order =3. We know that the rank of f is 1, and the eigenvalue must be 0, so it must not be positive definite (except positive definite, it is obviously not positive definite, there is only one eigenvalue, and the rest are 0). If the unique non-zero eigenvalue is positive, it is semi-positive, and if it is negative, it is semi-negative. The unique eigenvalue cannot be 0, because if it is 0, it is inconsistent with the linearity mentioned at the beginning)

D

Let a = ab t, | ab t+ba t | = | a+a t |

Then f = x tax, let g = x t (ba t) x = x t (ab t) tx = x ta tx, and both f and g are functional expressions. Obviously, f=g, so f+g=2f, simply multiplying by a constant will not change the rank of the matrix, so the rank of f+g = 1, that is, the rank of A+A is T = 1. Obviously, A+A T is not full rank, and the determinant must be 0.