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Prove that if $T:V \rightarrow U$ is an invertible linear transformation then the inverse $T^{-1}: U \rightarrow V$ is also linear

So, $T$ is linear and thus $T(v_1+v_2) = T(v_1)+T(v_2)$ for all $v_1,v_2 \in V$. Also, $T$ is invertible and so we can write $T^{-1}T(v)=v$ for all $v \in V$. I'm a bit confused trying to put this all together, help appreciated!!

FireFenix777
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Since the inverse is defined on $U$ I'm assuming $T$ is bijective. So take $u_1, u_2 \in U$. Then $u_1 = T(v_1)$ and $u_2 = T(v_2)$ for some $v_1,v_2\in V$. Now

$$T^{-1}(u_1 + u_2) = T^{-1}\left(T(v_1)+T(v_2)\right) = T^{-1}\left(T(v_1+v_2)\right)= v_1+v_2 = T^{-1}(u_1) + T^{-1}(u_2)$$

as required.

Enforce
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$T^{-1}(v+w) \overset{?}{=} T^{-1}(v)+T^{-1}(w)$

Let's apply $T$ on both side we get $w+v = T(T^{-1}(v)+T^{-1}(w)) \overset{T \hspace{0.1cm} \text{linear}}{=} T(T^{-1}(v))+T(T^{-1}(w))$ true.

jacopoburelli
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    Ok, and how does this prove that $T^{-1}$ is linear? – Trevor Gunn Jan 06 '21 at 19:01
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    Proof by contradiction works because a true statement cannot imply a false statement. "Proof by noncontradiction" is not a thing because both true and false statements can imply true statements. – Trevor Gunn Jan 06 '21 at 19:18
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    @TrevorGunn I agree with you, but $T$ is bijective in particular surjective so $\exists u \in V : v = T(u)$ – jacopoburelli Jan 06 '21 at 19:49