I'm reading linear algebra, and trying to figure out the relation between bijective(one-to-one and onto) and invertible. I've searched about similar questions and answers of them but I found something differ:

In the first comment of https://math.stackexchange.com/q/530446/390226, I read:

$$\textrm{["]invertible $\iff$ bijective[."].}$$

But I found another answer, https://math.stackexchange.com/a/2415538/390226, I read:

$$\textrm{["]invertible $\Longleftarrow$ bijective[."].}$$

I think they're both right since the first one is about linear transformation and the second one is about a general function. But I don't know the reason/details the two differ.


Bijective implies that there is an inverse function. Define $f^{-1}(x)$ to be the only solution of $f(y)=x$. Such solution exists for each $x$, due to the $f$ being onto, and the solution is unique due to $f$ being one-to-one.

In the context of linear algebra one wants to restrict attention to functions that are linear. A linear and bijective function would have a linear-algebra inverse, by definition, if it has an inverse function that is also linear.

Theorem: A linear function that has an inverse function, has a linear inverse function.

Proof: We just need to show that the inverse function $f^{-1}$ is linear. For $x,y$ vectors and $a$ scalar we have that there are some $X,Y$ such that $f(X)=x$ and $f(Y)=y$. Then $$f^{-1}(ax+y)=f^{-1}(af(X)+f(Y))=f^{-1}(f(aX+Y))=aX+Y=af^{-1}(x)+f^{-1}(y)$$

Therefore, $f^{-1}$ is linear.

Linear algebra is special in this sense. In some other contexts the additional property is not necessarily shared by the inverse function. For example, in topology one cares about continuous functions. A continuous function can have an inverse, but that inverse not be continuous, like $f:[0,2\pi)\to \{(x,y)\in\mathbb{R}^2:\ x^2+y^2=1\}$ defined by $f(x)=(\cos(x),\sin(x))$.

  • 1
    $\begingroup$ @Niing The distinction between $f:X\to Y$ and $f:X\to f(X)$ is useful, specially when functions that preserve some structure are used to study the spaces that they map from and to. Like using linear functions to study vector spaces, or continuous functions to study topological spaces. On the other hand, when the subject of study is the function itself, sometimes that distinction is dropped. Mathematical concepts are not as strict as they might seem. It is human language. What is important is that in each case you and your interlocutor are using the same language. $\endgroup$ – user550929 Apr 10 '18 at 13:33
  • $\begingroup$ Thank you so much... And I'm sorry I deleted my comment because I thought I didn't entirely know what I'm asking... $\endgroup$ – Postal Model Apr 10 '18 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.