# Proof of an invertible linear transformation being one-to-one

I am confused of the following:

Suppose that there is an linear transformation $T : V \rightarrow W$ that is invertible. To prove that $T$ is one-to-one and onto, the proof that I saw says the following:

Let $\mathbb{v}$ be in the kernel of $T$. Then, $T^{-1}(T(\mathbb{v})) = T^{-1}(\mathbb{0}) \rightarrow I(\mathbb{v}) =0$

I am not sure how one can reach the conclusion of $I(\mathbb{v}) =0$.

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Well, you have everything there. For $\mathbf{v} \in \ker(T)$, notice $T^{-1}(T(\mathbf{v}))=T^{-1}(0)=0$ as $T^{-1}$ is linear. But since $T^{-1} \circ T=I$, we have that, $I(\mathbf{v})=\mathbf{v}=0$. Hope this helps. –  user21436 Aug 28 '12 at 7:56
A proper proof should not use that $T$ is linear, since it is irrelevant for the property of being bijective. –  Marc van Leeuwen Aug 28 '12 at 9:54

Suppose $T(v) = 0$. Then because $T$ is invertible we can apply $T^{-1}$ that is also a linear transformation to both sides to get that
$$T^{-1}T(v) = T^{-1}(0).$$
However the condition that $T$ having an inverse is exactly the same thing as saying that $T^{-1}T = I = T(T^{-1})$. In particular, the first part of this equality tell us that
$$T^{-1}T(v) = (T^{-1}T)(v) = I(v).$$
However I said earlier that $T^{-1}$ was linear so that $T^{-1}(0) = 0$. It follows that $I(v) = 0$.