Let $T_1,T_2 : R^5 \to R^3$ be linear transformations s.t rank($T_1$)=3 and nullity ($T_2$)=3 . Let $T_3:R^3 \to R^3 $be linear transformation s.t $ T_3(T_1)=T_2.$ Then find rank of $T_3$
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By the rank-nullity theorem we have
$$\operatorname{rank}(T_2)=\dim \Bbb R^5-\operatorname{nullity}(T_2)=2$$ so
$$2=\operatorname{rank}(T_2)=\operatorname{rank}(T_3(T_1))=\dim(T_3(T_1(\Bbb R^5))=\dim(T_3(\Bbb R^3))=\operatorname{rank}(T_3)$$
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Ya mine thoughts were similiar. Thanks for confirming it – ketan Jan 15 '15 at 20:08
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I am not getting how $\dim(T_3(T_1(\Bbb R^5))=\dim(T_3(\Bbb R^3))$? – math student Apr 06 '20 at 06:17