Let a be a real matrix of order n and a ^ t be a transpose matrix of order A. It is proved that R (a) = R (a ^ TA) Even give 100 points for the answer

Let a be a real matrix of order n and a ^ t be a transpose matrix of order A. It is proved that R (a) = R (a ^ TA) Even give 100 points for the answer

We use this property: if a and B are matrices of order n, then there must be
The generalization theorem of R (AB) ≤ min {R (a), R (b)} was mentioned in Gao Dai of Peking University
Then R (a) = R (AE) = R (a * a ^ t * a) ≤ R (a ^ t * a) ≤ R (a)
(this step is to use the inequality of the above theorem to expand and shrink, using such a mathematical idea: to prove a = B, just prove a ≥ B and a ≤ b)
In other words, we obtain that R (a) ≤ R (a ^ t * a) ≤ R (a)
R (a) = R (a ^ t * a)

Given that the rank of M × n matrix A is n-1, α 1 and α 2 are two different solutions of the homogeneous linear equation system AX = 0, K is an arbitrary constant, then the general solution of the equation system AX = 0 is ()
A. kα1B. kα2C. k（α1+α2）D. k（α1-α2）

From the rank of M × n matrix A is n-1, we know that the fundamental solution system of AX = 0 contains only one solution vector. Therefore, in order to form the fundamental solution system, the solution vector must be a non-zero vector. We know that α 1 and α 2 are two different solutions of the homogeneous linear equations AX = 0. α 1 - α 2 must be a non-zero solution of AX = 0. The general solution of AX = 0 can be expressed as K (...)

Let a be a square matrix of order n. if there is a nonzero matrix B such that ab = 0, we prove that | a | = 0

If R (a) = n, then a is reversible. A ^ (- 1) [AB] = a ^ (- 1) * 0 = 0, and a ^ (- 1) [AB] = B, therefore, B = 0. It is contradictory with B not equal to 0