Let #A=((0,1),(0,1))#. Let T linier operator to#R^(2x2)#, and #T(X)=AX-XA#, #AAX##inR^(2x2)#. Determine #rank(T)# ?
1 Answer
We have:
# bb(A) = ( (0, 1), (0, 1) ) \ \ \ # ; and;# \ \ ul(bb(T))(bb(X)) = bb(AX) - bb(XA) \ \ AA bb(X) in RR^(2xx2)#
Consider a generic element
# bb(X) = ( (a_11, a_12), (a_21, a_22) ) #
Then consider the effect of the linear operator
# ul(bb(T))(bb(X)) = bb(AX) - bb(XA) #
# \ \ \ \ \ \ \ \ \ = ( (0, 1), (0, 1) )( (a_11, a_12), (a_21, a_22) ) - ( (a_11, a_12), (a_21, a_22) )( (0, 1), (0, 1) ) #
# \ \ \ \ \ \ \ \ \ = ( (a_21, a_22), (a_21, a_22) ) - ( (0, a_11+a_12), (0, a_21+a_22) ) #
# \ \ \ \ \ \ \ \ \ = ( (a_21, a_22-a_11-a_12), (a_21, a_22-a_21-a_22) ) #
# \ \ \ \ \ \ \ \ \ = ( (a_21, a_22-a_11-a_12), (a_21, -a_21) ) #
The vectors:
# ( (a_21), (a_21) ) \ \ \ # ; and;# ( (a_22-a_11-a_12), (-a_21) ) #
Are linearly independent, and therefore:
# rank(ul(bb(T))) = 2#
even though