invertible if and only if bijective

Question

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

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. — Julien, Jan 29 '13 at 02:35
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 02:38
That depends on your textbook definition of an invertible linear map. During my undergraduate days, "invertible" only means the existence of an inverse, so it is essentially a synonym of "bijective". That $T^{-1}$ is linear is only a biproduct, not part of the definition of invertibility of $T$. Yet, as the comment of julien points out, some people may call a linear map $T$ invertible if $T$ has an inverse map that is linear (but I believe that such definition is uncommon). In this case the point is to prove the linearity of $T^{-1}$ given that $T$ is bijective. — user1551, Jan 29 '13 at 02:46
@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: http://math.stackexchange.com/questions/54232/difference-between-bijection-and-isomorphism — alancalvitti, Jan 29 '13 at 02:54
@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 03:08
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 03:17
@user1551 Such a definition is actually very common. See here: http://mathworld.wolfram.com/InvertibleLinearMap.html Not to mention categories. — Julien, Jan 29 '13 at 04:44
@GerryMyerson, I don't agree with you, though I could have explained it better. Restrictions on the arrows in a category must be taken into account (they are also elements of sets). In $\bf Top$, whose objects are also sets with additional structure, the iso's are restricted to have a continuous inverse. Counterexample: $e^{2 \pi i t} : [0,1) \to S^1$, even though it's a bijection in $\bf Set$ — alancalvitti, Jan 29 '13 at 15:51
@alan, we're not in Top. If we were doing topology, I would say a homeomorphism must have a continuous inverse, but a bijection doesn't even have to be continuous. I wouldn't use the word, bijection, and expect people to think I meant "homeomorphism". I wouldn't use "bijection" in linear algebra and expect people to think I meant "isomorphism". "Bijection" means one-one and onto, nothing more. — Gerry Myerson, Jan 29 '13 at 23:28
@Gerry, neither are we in $\bf Set$: The OP's context is $\bf Vect(k)$ yet your comment (which is true in that cat.) says "more generally". My q to you and the reason for introducing $\bf Top$ is, how is your statement more general? — alancalvitti, Jan 30 '13 at 00:06
@alan, we're going around in circles here. My statement is more general, as I have already pointed out, because my statement is about sets, and every vector space is a set. Now, it happens to be the case that if a linear transformation is invertible when viewed simply as a function on a set then its inverse is also a linear transformation, but that's a theorem about invertible linear maps, not a part of the definition of invertible (in my view). — Gerry Myerson, Jan 30 '13 at 01:07
@Gerry, $\bf Vect(k)$ (for each field $\bf k$) are not $\bf Set$. Your statement is not more general. If it were, there would be no need to distinguish injections and surjections vs. monos and epis. Category theory does not reduce to set theory. — alancalvitti, Jan 30 '13 at 01:15
@Alan, had the original question been, "I need to show that a linear transformation is invertible and its inverse is linear if and only if it is bijective," then my remark would not have been more general, since it would not have addressed the italicized phrase. But that phrase wasn't there, so I took "invertible" to mean "invertible as a function". That makes my comment more general, as it applies to functions whether linear or not, and to sets whether vector spaces or not. — Gerry Myerson, Jan 30 '13 at 03:22
@Gerry, It makes no difference whether the OP explicitly recognizes and mentions that the inverse also be linear, since that condition is determined by the context of the category of vector spaces. I understand that you took invertible to mean iso in sets, but that is not more general than the statement in general categories. So it is the opposite of your claim. Sets and functions is just one cat. among many. — alancalvitti, Jan 30 '13 at 03:30
@alan, from the level of the question, I suspect OP wouldn't know the category of vector spaces if it painted itself purple and danced naked on top of a harpsicord singing "the category of vector spaces is here again". At the level at which the question was asked, my comment was more general. I might not have made the comment had it been Bill Lawvere asking the question, but, then again, it's not a question Bill Lawvere is likely to have to ask. — Gerry Myerson, Jan 30 '13 at 04:11
@GerryMyerson: Oh I love your discussion, and when I google Bill Lawvere I Iaugh. — linear_combinatori_probabi, Apr 10 '18 at 10:26

score 8 · Answer 1 · answered Jan 29 '13 at 04:37

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 $L(W,W)$.

invertible if and only if bijective

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