Let A = .


Then   =  *  = .


If  = , then 

a^2 + bc = 0     (1)   as the element (i,j) = (1,1) of the squared matrix;

ab + bd  = 1     (2)   as the element (i,j) = (1,2) of the squared matrix;  

ac + cd  = 0     (3)   as the element (i,j) = (2,1) of the squared matrix;  

bc + d^2 = 0     (4)   as the element (i,j) = (2,2) of the squared matrix.  


Equation (3) implies that at least one of the two options takes place:

             1) c = 0   OR   2) a = -d.


Below I consider every of these two options and derive a contradiction from each of them.


    1)       if c = 0, then  from (1) a^2 = 0; hence, a = 0.

       Also, if c = 0, then  from (4) d^2 = 0; hence, d = 0.

             Then (2) becomes 0 = 1, CONTRADICTION.


    2)  If a = -d, then again (2) becomes  0 = 1,  CONTRADICTION.


These contradictions prove the statement.