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

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

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