Let a be mxn real matrix and prove rank (ATA) = rank (a) emergency

As long as it is proved that the equations a'ax = 0 and Ax = 0 have the same solution (a '= at)
If x is the solution of AX = 0, then obviously x is also the solution of a'ax = 0
If x is the solution of a'ax = 0
Then x'a'ax = x'0 = 0
(Ax)'(Ax)=0
||Ax||=0
The norm of ax is 0 if and only if AX = 0
So x is the solution of AX = 0

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