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pages/Math429/Math429_L13.md

@@ -26,8 +26,6 @@ Suppose $T$ is injective, then $null\ T={0}$, i.e $dim(null\ T)=0$, since $dim\
 
 If $T$ is surjective, then $dim\ range\ T=dim\ W$ but then $dim\ V=dim\ null\ T+dim\ W\implies dim\ null\ T=0$, so $T$ is injective, $T\ surjective\implies T\ injective$.
 
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 #### Theorem 3.68
 
 Suppose $V,W$ finite dimensional $dim\ V=dim\ W$, then for $T\in\mathscr{L}(V,W)$ and $S\in \mathscr{L}(W,V)$, then $ST=I\implies TS=I$