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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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    A proper proof should _not_ use that $T$ is linear, since it is irrelevant for the property of being bijective.2012-08-28

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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 $I(v) = 0$.