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I need to show that a linear transformation $T \in L(W,W)$ is invertible if and only if $T$ is bijective.

I'm not sure how to go about this, can someone give me an idea where to start? I know is injective if $T_u = T_w$ aka null $T = {0}$, surjective if range $T = W$, and invertible if there is an inverse that give you the identity

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The key point is to check that if $T$ is bijective, then $T^{-1}$ is linear, hence $T$ is invertible. The other direction is easy. –  1015 Jan 29 '13 at 2:35
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More generally, if $A$ and $B$ are any sets, and $f:A\to B$ is any function, then $f$ is invertible if and only if $f$ is a bijection. –  Gerry Myerson Jan 29 '13 at 2:38
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@GerryMyerson, how is that more general? You described the relation between bijective functions and iso's in $\bf Set$, whereas OP's context is vector spaces. Your statement isn't true in all categories, see Thomas Rot's answer here: math.stackexchange.com/questions/54232/… –  alancalvitti Jan 29 '13 at 2:54
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@alan, it's more general in that all vector spaces are sets, whereas not all sets are vector spaces. –  Gerry Myerson Jan 29 '13 at 3:08
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Vector spaces are sets with additional structure (a lot of additional structure: groups, rings, fields, finally vectors). This is reflected in the behavior of the arrows. –  alancalvitti Jan 29 '13 at 3:17

1 Answer 1

From a category perspective, the natural definition for $T$ to be invertible here is that there exists $S\in L(W,W)$ such that $S\circ T=T\circ S=I$.

As usual, $S\circ T=I$ implies that $T$ is injective, and $T\circ S=I$ entails that $T$ is surjective. So $T$ invertible implies it is bijective.

Now assume that $T$ is bijective. Then we have $S=T^{-1}$ that fits the definition above, provided we can prove that it belongs to $L(W,W)$.

Let us write that $T$ is linear, namely $$T(\lambda x+\mu y)=\lambda T(x)+\mu T(y)$$ for all $x,y\in W$, and $\lambda,\mu\in\mathbb{R}$. This is true in particular for $x=T^{-1}(x')$ and $y=T^{-1}(y')$ for all $x',y'\in W$. So $$ T(\lambda T^{-1}(x')+\mu T^{-1}(y'))=\lambda TT^{-1}(x')+\mu TT^{-1}(y')=\lambda x'+\mu y'. $$ It only remains to apply $T^{-1}$ to the LHS and the RHS of the above. This proves the linearity of $T^{-1}$.

So yes, bijectivity is equivalent to invertibility in $L(W,W)$.

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