A function has a left inverse is equivalent to it is injective，how to use this fact to prove a matrix has a left inverse is equivalent to it is injective，if it can‘t be proven using this fact，how to prove it. Btw, I think the main problem is that you can’t prove the left inverse is linear ,i.e. the left inverse isn’t necessarily a matrix Perhaps the description of this question may cause misunderstandings.As a left inverse of a matrix T ,it has to be a matrix,not all left inverse of funtion T is the left inverse is matrix T.
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This follows from the definitions with very little thought. No mention of matrices or linearity is necessary. – John Douma Apr 06 '23 at 16:16
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@JohnDouma I don't think it's obvious，this theorem can apply to matrixes，but this theorem doesn’t guarantee that the left inverse is linear. – zx Z Apr 09 '23 at 02:51
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@azif00 Thank you very much！This proof is very good（ – zx Z Apr 09 '23 at 03:42
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Definitions: A matrix $A \in F^{m \times n}$ is said to be injective if $Ax=Ay \implies x=y \;\; \forall x,y, \in F^n$. The matrix $A$ is said to have a left-inverse $B \in F^{n \times m}$ if $BA=I$.
Assume $A \in F^{m \times n}$ is injective. We'd like to show it has a left-inverse $B \in F^{n \times m}$. The zero element always maps to zero, and since $A$ is injective, the zero element is the only element that can map to zero. In other words, $Ax=0 \implies x=0\in F^n$ and so $A$ has trivial null space and is of full rank (specifically rank $n$ by rank-nullity). Add $r=m-n$ independent columns to $A$ to make it a square matrix $A'$. $A'$ is a square matrix with full rank, so is invertible with some inverse $B'$. In particular $B'A'_{,i}=I_{,i} \in F^m$. By the rules of matrix multiplication, it is clear the first $m$ rows of $B'$ form a left-inverse of $A$, so we have shown a left-inverse.
Assume $A$ has a left-inverse $B$. We'd like to show it's injective. If $Ax=Ay$, then $[BAx=BAy]\implies [Ix=Iy] \implies [x=y]$ and we're done.
EDS
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of course（so you have to prove the function is linear（matrices are equivalent to linear transformations – zx Z Apr 09 '23 at 02:48
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@zx Z I still don't understand. If a matrix has a left inverse, then that left inverse is a matrix. This left inverse (a matrix) is linear because matrices are equivalent to linear transformations. There is nothing to prove. But more to the point, where is linearity even coming into the picture? It is not necessary to prove the equivalence – EDS Apr 09 '23 at 02:53
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Because you didn’t prove the funtion is the left inverse,the definition of left inverse requires it to be a matrix,it is the left inverse of the funtion T,but it isn’t necessarily the left inverse of the matrix T.(PS:it is very easy to construct a nonlinear left inverse for a matrix. – zx Z Apr 09 '23 at 03:03
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T:R2——R3，it has 2 independent columns ，assume we have already known it has a left inverse （a matrix）A：R3——R2，since the image of T （imT） is a 2 dimension subspace in R3，if you change the value of A on R3（not in the image of T），AT=I is still true ，take v in R3，which isn't in imT，let A‘（v）=（1，0），A'（2v）=（0，0），A'T=I，but A' is no longer linear！ – zx Z Apr 09 '23 at 03:25
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