A question about the rank of a linear operator

[sanctifier]

Let T is a linear transformation from a vector space V to V itself. The dimension of V, denoted by dim(V), is finite.

If the rank of T, denoted by rk(T), is equivalent to the rank of TT, i.e., rk(T)=rk(TT)

why is the intersection of image of T(denoted by im(T)) and the kernel of T(denoted by ker(T)) is the zero vector, i.e., im(T)$$\cap$$ker(T)={0}?

Thanks for any help.

 

[Cantab Morgan]

Think first of the special case where T is a projection. Then the image of T and the image of TT are clearly the same. In such a case, if the kernel of T intersected the image, then T wouldn't be a projection.

 

[sanctifier]

You are right.

Thanks!