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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?

Thanks in Advance.

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    $\begingroup$ 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. $\endgroup$ – mathpadawan May 19 at 17:54
  • $\begingroup$ Every matrix is not a linear transformation. A linear transformation needs to have the zero-vector in the range. $\endgroup$ – James Smith May 19 at 17:56
  • $\begingroup$ ok but multiply the matrix be the zero vector and you get the zero vector in the range? $\endgroup$ – mathpadawan May 19 at 17:57
  • $\begingroup$ That does not make sense to me. $\endgroup$ – James Smith May 19 at 18:00
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    $\begingroup$ 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). $\endgroup$ – Berci May 19 at 18:00
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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 $$S(a,b,c):=(a+c^2, b+c^2)$$

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  • $\begingroup$ So you say that T does not have a right-inverse? If so, I would disagree. $\endgroup$ – James Smith May 19 at 18:09
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    $\begingroup$ $T$ cannot be surjective. $\endgroup$ – Berci May 19 at 18:11
  • $\begingroup$ 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. $\endgroup$ – James Smith May 19 at 18:15
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    $\begingroup$ 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 $T$. $\endgroup$ – Berci May 19 at 18:23
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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – James Smith May 19 at 18:15
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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.

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