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Let $T_1,T_2 : \mathbb{R}^5 \to \mathbb{R}^3$ be linear transformations such that $\text{rank }(T_1) = 3$ and $\text{nullity } (T_2) = 3$ . Let $T_3 : \mathbb{R}^3 \to \mathbb{R}^3$ be a L.T. such that $T_3 \circ T_1 = T_2$ . Find the rank of $T_3$ .

My approach :- $\text{ Rank} (T_3) = \text{dim(range set of } T_3) = \text{ dim}(T_2)$ . But how do I find the dimension of $T_2$ ?

Rushabh Mehta
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Beginner in Maths
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3 Answers3

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It is not clear what you mean by "$\dim(T_2)$" in your approach.

Hint: Because $T_1$ is surjective (onto), we know that $\operatorname{range}(T_3\circ T_1) = \operatorname{range}(T_3)$ (why is $T_1$ surjective? Why are the ranges equal as a consequence?) From there, $$ \operatorname{rank}(T_2) = \operatorname{rank}(T_3\circ T_1) = \dim\operatorname{range}(T_3\circ T_1) = \dim\operatorname{range}(T_3) = \operatorname{rank}(T_3). $$

Ben Grossmann
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By the rank-nullity theorem:

$$5= rank(T_2)+nullity (T_2).$$

Fred
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Using the fact that $\text{rank } (T_3 \circ T_1) \le \min \{ \text{rank } T_3 , \text{rank } T_1 \}$ (see proof here ), we have $2= \text{rank } T_2 = \text{rank } (T_3 \circ T_1) \le \min \{ \text{rank } T_3 , 3 \}$. So $\text{rank } (T_3)$ is either $2$ or $3$. If $\text{rank } (T_3) =3$ then $T_3$ is invertible and hence it follows that $\text{rank } {T_1} = \text{rank }{T_2}$ which is not possible. Hence, $\text{rank } (T_3) =2$.

ashK
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