Let T is a linear transformation from a vector space V to V itself. The dimension of V, denoted by dim(V), is finite.(adsbygoogle = window.adsbygoogle || []).push({});

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)[tex]\cap[/tex]ker(T)={0}?

Thanks for any help.

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# A question about the rank of a linear operator

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