Must every right-inverse of a linear transformation be a linear transformation?

Let T be a linear transformation $$\mathbb{R}^2 \rightarrow \mathbb{R}^3$$. Let S be the right-inverse of T. Does S have to be linear transformation?

• I think that since every matrix is a linear transformation and right inverse of a matrix is a matrix itself, hence it can be thought of as a linear transformation. – mathpadawan May 19 at 17:54
• Every matrix is not a linear transformation. A linear transformation needs to have the zero-vector in the range. – James Smith May 19 at 17:56
• ok but multiply the matrix be the zero vector and you get the zero vector in the range? – mathpadawan May 19 at 17:57
• That does not make sense to me. – James Smith May 19 at 18:00
• Any linear map $T$ can be uniquely represented by a matrix $M$ such that $T(v) =M\cdot v$, once bases are fixed in the domain and codomain, and the zero vector is always in the range (which is the column space of the matrix, using the given basis). – Berci May 19 at 18:00

Actually, if the mapping $$S:\Bbb R^3\to\Bbb R^2$$ satisfies $$T(S(v))=v$$ for all $$v\in\Bbb R^3$$, that would imply $$T$$ is surjective, which is impossible if $$T$$ is linear, by considering the dimensions.

However, if $$T$$ is injective, it has left inverses, and it can also have nonlinear left inverses, e.g. if $$T(a, b) =(a, b, 0)$$ and $$S(a,b,c):=(a+c^2, b+c^2)$$

• So you say that T does not have a right-inverse? If so, I would disagree. – James Smith May 19 at 18:09
• $T$ cannot be surjective. – Berci May 19 at 18:11
• It has just come to my attention that T should have been $\mathbb{R}^3 \rightarrow \mathbb{R}^2$. I made a typo. I will not edit the question, because your answer is quite good and may help others. – James Smith May 19 at 18:15
• If $T:\Bbb R^3\to\Bbb R^2$, surjective, e.g. $T(a, b, c) =(a, b)$, then similarly as above, for example $S(a,b)=(a, b, a^2)$ is a nonlinear right inverse of $T$. – Berci May 19 at 18:23

No

(Actually a linear map $$\mathbb R^2 \to \mathbb R^3$$ cannot have a right inverse, since having a right inverse is equivalent to being surjective, and linear maps have the dimension of the range at most the dimension of the domain, so there are no surjective linear maps $$\mathbb R^2 \to \mathbb R^3$$)

BUT ANYWAY. . . .

Consider the projection map $$P: \mathbb R^2 \to \mathbb R$$ given by $$P(x,y) = x$$. Let $$f : \mathbb R \to \mathbb R$$ be your favourite nonlinear function and define the right-inverse $$Q: \mathbb R \to \mathbb R^2$$ by $$Q(x) = (x,f(x))$$.

Then we have $$P\circ Q(x) = P(Q(x))=P(x,f(x))= x$$ so this is indeed a right inverse.

To see $$Q$$ is nonlinear observe the image is the graph of the function $$f$$ which is not a linear subspace of $$\mathbb R^2$$. That means $$Q$$ is nonlinear.

By definition, if $$S$$ is a right-inverse of $$T,$$ then $$S:\Bbb R^3\to\Bbb R^2$$ has the property that for all $$\vec v\in\Bbb R^2,$$ we have $$(T\circ S)(\vec v)=\vec v.$$ However, by Rank-Nullity, there is a vector $$\vec u\in\Bbb R^3$$ with $$\vec u\neq\vec 0_3,$$ such that $$S(\vec u)=\vec 0_2,$$ so that since $$T$$ is a linear transformation, we would have $$\vec u=(T\circ S)(\vec u)=T\bigl(S(\vec u)\bigr)=T(\vec 0_2)=\vec 0_3\neq\vec u.$$ Thus, $$T$$ has no right-inverse.

On the other hand, $$T$$ will have left-inverses (infinitely-many of them, in fact) so long as its null space contains only $$\vec{0}_2,$$ and exactly one of the left-inverses (the one sending all elements outside the range of $$T$$ to $$\vec{0}_2$$) is a linear transformation.

• It has just come to my attention that T should have been $\mathbb{R}^3 \rightarrow \mathbb{R}^2$. I made a typo. I will not edit the question, because your answer is quite good and may help others. – James Smith May 19 at 18:15

If $$T:X\to Y$$ is injective but not surjective, consider a left-inverse $$S$$, i.e. a map $$Y\to X$$ such that $$S(T(x)) = x\text{ for all }x\in X.$$ Note that this equation only says something about how $$S$$ acts on the range (image) of $$T$$. For the points of $$Y$$ that are outside the range of $$T$$, we are free to prescribe absolutely any behavior for $$S$$.

Similarly, if $$T:X\to Y$$ is instead surjective but not injective, we consider a right-inverse $$S$$, $$T(S(y)) = y\text{ for all }y\in Y.$$ This means that for a $$y\in Y$$, the map $$S$$ just picks one of the inverse images of $$y$$ under $$T$$. Every time we have a $$y$$ that is "hit" by more than one $$x$$ under $$T$$, we have the freedom to choose any of them as our value $$S(y)$$.

The above is true in general (in the category of arbitrary sets and maps). If we consider the case where $$X$$ and $$Y$$ are vector spaces (or more concretely $$X=\mathbb{R}^n$$ and $$Y=\mathbb{R}^m$$), and where $$T$$ is a linear map, it is clear (I think; left to the reader) that our freedom to mingle with $$S$$ concerns full subspaces (or complements of subspaces) and we can pick a completely "wild" behavior of $$S$$. In particular, we can ensure that $$S$$ is not a linear map.