Proving that $a^tax = a^tb$ has a common solution

This is a very interesting inference question, combined with a lot of interesting conclusions.

First we need to make it clear that ATb a^tb is a vector, so the conclusion to be proven is: the non-homogeneous equations have solutions.

Therefore, the problem is turned into verification:

R (ATA) =r (ata| ATB) R (a^ta) = R (a^ta| A^TB)

So how to think about proving this proposition.

We thought that if the ATb A^TB could be represented by a column vector of at a^t, then the question could be simplified.

Because we have special proof: ata,a is equal rank. A^ta,a are equal-rank. Therefore, it is possible to consider the At,ata a^t,a^ta column block.
at= (α1,α2,α3,..., αn) a^t = (\alpha_1,\alpha_2,\alpha_3,..., \alpha_n)
B=ata= (β1,β2,β3,..., βn) b= A^ta = (\beta_1,\beta_2,\beta_3,..., \beta_n)

Then: b= (α1,α2,α3,..., αn) a= (