10

Let $X$ and $U$ be vector spaces over a field $F$, and let $T : X \to U$.

(a) If there exists an operator $S : U \to X$ such that $S(T(x)) =x$ for all $x \in X$, then $S$ is called a left inverse of $T$.

(b) If there exists an operator $S : U \to X$ such that $T(S(u)) =u$ for all $u \in U$, then $S$ is called a right inverse of $T$.

I'm trying to prove the following theorem.

Let $X$ and $U$ be vector spaces over a field $F$, and let $T: X\to U$ be linear.

(a) There exists a left inverse $S$ of $T$ iff $T$ is injective.

(b) There exists a right inverse $S$ of $T$ iff $T$ is surjective.

So far my study on linear algebra has been largely restricted to finite dimensional vector spaces but this problem, I think applies to general vector spaces. How can I solve this without resorting to basis?

Christoph
  • 25,552
nomadicmathematician
  • 6,309

3 Answers3

6

Here is the proof for (a).

$(\Rightarrow)$ Suppose there exists a left inverse $S$ of $T$. To see that $T$ is injective, let $x,y\in X$ such that $T(x)=T(y)$. Then $$x=S(T(x))=S(T(y))=y$$ Hence $T$ is injective. (Note that this proof requires no linear algebra).

$(\Leftarrow)$ Suppose that $T$ is injective. Let $\{x_i:i\in I\}$ be a basis for $X$. Then $\gamma=\{T(x_i):i\in I\}$ is a linearly independent subset of $U$ (check this!) so there exists a basis $\beta$ of $U$ such that $\gamma\subset\beta$. Now, let $S:U\to X$ be the linear map defined on $\beta$ by $$ S(u)= \begin{cases} v & \text{if } u\in\gamma\text{ with }u=T(v) \\ 0 & \text{if } u\notin\gamma \end{cases} $$ Then for $x=\sum\lambda_i x_i\in X$ we have $$ ST(x)=ST\left(\sum\lambda_i x_i\right)=\sum\lambda_iST(x_i)=\sum\lambda_i x_i=x $$ so that $S$ is a left inverse of $T$. $\Box$

The key result here is that if $\gamma$ is a linearly independent subset of a vector space $V$, then there exists a basis $\beta$ of $V$ such that $\gamma\subseteq\beta$. This is a standard result covered in most linear algebra courses.

The proof for (b) is very similar.

Brian Fitzpatrick
  • 26,933
  • Thanks for you answer! Another question: given a Banach space and a injective linear operator onto another Banach space, is there a left inverse? I'm thinking this proof won't work because a Banach space is not sure to have a basis. –  Oct 26 '17 at 23:31
  • @Hackghost Every vector space has a basis: https://proofwiki.org/wiki/Vector_Space_has_Basis – Brian Fitzpatrick Oct 27 '17 at 02:47
  • What's the point in finding a basis for $U$? The first case (when $u = T(v)$ makes sense; we don't need to even find a basis for this case, right?) but how does finding a basis for $U$ help us with the second case? – beginner Feb 09 '21 at 20:22
  • I've been struggling to understand this problem for a couple hours now and I think why I'm so confused stems from the fact that I don't understand the point of finding a basis for $U$. Why can't you just set $S(u)=0$ w/o finding a basis? Moreover, do you need to prove that this is a linear mapping by showing it fulfills homogeneity/additivity properties? – beginner Feb 09 '21 at 20:29
  • @beginner choosing the basis just makes it easier to define the function. Note that you can't simply set S(u) = 0 for every u not in the image of T or you won't get a linear map. Without choosing a basis what we are doing is setting S to be the inverse of T on the image of T (T is injective so it is invertible on its image) and choosing a complementary subspace and setting S to be some other linear function on this space (in the example we choose the 0 function and we define this complement as the span of some of the basis vectors). We then define S fully by extending linearly. – Callum Feb 19 '21 at 14:49
3

Hint: If $\exists$ left inverse: $$x_1\ne x_2\implies S(T(x_1))\ne S(T(x_2))\implies T(x_1)\ne T(x_2)$$ (Why each step?)

Martín-Blas Pérez Pinilla
  • 42,722
  • 4
  • 51
  • 94
0

Proving that if $T$ has a left inverse, then $f$ is injective actually has very little to do with linearity. It is an immediate set-theoretic consequence. Namely, if $T(x)=T(y)$, what happens when you apply $S$ on the left?

Similarly, if $T$ has a right inverse $S$, then $T$ is surjective. Indeed, if $y$ is an arbitrary element in the codomain of $T$, what can you say about $S(x)$?

The implications in the other directions are also valid set-theoretically and are not hard to prove. That is, for any function $f:A\to B$ if $f$ is injective, then there exists a left inverse $g:B\to A$. You will then need to prove that if there is also a linear structure present on the domain and codomain, and $f$ preserves it, then so does $g$. Similarly for surjectivity.

Ittay Weiss
  • 81,796