Given #A=((0,1),(0,1))#. Let say #T# linear operator on #R^(2x2)# with #T(X)=AX-XA#, #AAX in R^(2x2)#. How to determine #rank(T)# ?

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Answer 1 · 2017-04-11T17:38:19

See below.

There is an universal way to determine it.

Calling #T(x)=Theta=(t_(11),t_(12),t_(21),t_(22))#

and

#Xi=(x_(11),x_(12),x_(21),x_(22))# we have

#T(X)=AX-XA iff ((t_(11)=x_(21)),(t_(12)=-x_(11)-x_(12)+x_(22)),(t_(21)=x_(21)),(t_(22)=-x_(21)))#

or

#Theta=M Xi# with #M =((0, 0, 1, 0),(-1, -1, 0, 1),(0, 0, 1, 0),(0, 0, -1, 0))#

but

#"rank"(M) = 2#

so

#"rank"(T(X))=2#