if V is a finite dimensional over F and T belong to A(V)
T is singulare iff there exists v not equal to zero...such that vT=0
please prove the reverse part....that is if vT=0 then T is singular
Thanks
Weird definition...but never minds: if $\displaystyle T$ were invertible then there'd exist $\displaystyle S\in A(V)$ s.t. $\displaystyle ST =TS=I$ , with I = the identity
(operator or matrix, it never minds), and then $\displaystyle 0=Tv\Longrightarrow 0 = Sv = STv=Iv= v\Longrightarrow v=0$
Tonio