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Let $V$ and $W$ be two non-trivial vector spaces over ${\displaystyle \mathbb {F}}$ and let $T: V\rightarrow W$ be a linear transformation.

Prove that $T$ is onto iff $T^t$ is one-to-one.

Now the only thing I know is that when T is onto if $\operatorname{range}(T) = W$ and it is $1:1$ when $\mathrm{Nullity} = 0$.

I don't understand how to even start this question. If someone could give a hint at least, that would help me get started. (A similar question is already available on Stack Exchange but I could not understand the solution it had)

Bernard
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rand1
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1 Answers1

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  • Suppose $T$ is not surjective

We can write $$W = Im(T) \oplus W',\quad W' \neq 0$$ now define $\varphi \in W^{\star}$, s.t. $$ \begin{cases} \varphi\left.\right|_{Im(T)} = 0\\ \varphi\left.\right|_{W'} \neq 0 \end{cases} $$ we have $T^t(\varphi) = 0$, but $\varphi\neq 0$, so $T^t$ is not injective.

  • suppose $T$ is surjective: let $\varphi\in W^{\star}$, s.t. $T^t(\varphi) = 0$

so for all $v\in V$ $\varphi(T(v)) = 0$, but $T$ is surjective so $\varphi = 0$, i.e $T^t$ is injective.

user388615
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