# Left invertible

## Main Question or Discussion Point

I found on wikipedia that the following statements are equivalant:
1. Matrix A is left invertible
2. Ax=0 => x=0

I couldn't find the proof so I try to do it myself.

From 1. to 2. is easy. Assume A is left invertible. If Ax=0, then x=Ix=A-1Ax=A-10 = 0 .

I can't figure out how to do 2=>1. Any help is appreciated.

Things that might prove useful:
- A is injective
- dim(ker(A))=0
- rank(A)=n (if A is an m x n matrix)
- $$n\leq m$$

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If the solutions to $$Ax=0$$ are only the trivial one, then its RREF is the identity matrix, thus it is invertible.

And another:
$$Ax=0$$ are only the trivial solution, then $$\textrm{det}A\ne 0$$ thus it is invertible.

If the solutions to $$Ax=0$$ are only the trivial one, then its RREF is the identity matrix, thus it is invertible.

And another:
$$Ax=0$$ are only the trivial solution, then $$\textrm{det}A\ne 0$$ thus it is invertible.
Why and why?

Note that A is not (necessarily) a square matrix

Only square matrices can be inverted, thus the statements only make sense for them.

All of the statements above I used can be shown to be equivalent, but I need to know what level you are at so I can know which equivalent statements to use.

$$Ax=0 \Rightarrow \textrm{RREF}(A)=I$$ follows from the rank-nullity theorem, have you seen this theorem?

$$Ax=0 \Rightarrow \textrm{det}A\ne0$$ follows from properties of the determinant and also properties of elementary matrices, are you familiar with these?

Only square matrices can be inverted, thus the statements only make sense for them.

All of the statements above I used can be shown to be equivalent, but I need to know what level you are at so I can know which equivalent statements to use.

$$Ax=0 \Rightarrow \textrm{RREF}(A)=I$$ follows from the rank-nullity theorem, have you seen this theorem?

$$Ax=0 \Rightarrow \textrm{det}A\ne0$$ follows from properties of the determinant and also properties of elementary matrices, are you familiar with these?
This is simply not true. I have a better source ("Linear algebra done wrong"), but I'll use wikipedia to quote:
Non-square matrices (m-by-n matrices for which m ≠ n) do not have an inverse. However, in some cases such a matrix may have a left inverse or right inverse. If A is m-by-n and the rank of A is equal to n, then A has a left inverse: an n-by-m matrix B such that BA = I. If A has rank m, then it has a right inverse: an n-by-m matrix B such that AB = I.
I underlined what I'm trying to proof.

I know the theorems but I don't see how they are applicable to a non-square matrix.

This is simply not true. I have a better source (Linear algebra done wrong), but I'll use wikipedia to quote:

I underlined what I'm trying to proof.
Ah I see, just a left-inverse, in that case then you must avoid mentioning determinants although my first statement still holds.

Ah I see, just a left-inverse, in that case then you must avoid mentioning determinants although my first statement still holds.
That statement does hold indeed, and helped me to get:
rank(A) + dim(ker(A)) = rank(A) = n
rank(At) + dim(ker(At))= rank(A) + dim(ker(At)) = n + dim(ker(At)) = m
Hence: $$m\geq n$$

But it doesn't get me any further.

That statement does hold indeed, and helped me to get:
rank(A) + dim(ker(A)) = rank(A) = n
rank(At) + dim(ker(At))= rank(A) + dim(ker(At)) = n + dim(ker(At)) = m
Hence: $$m\geq n$$

But it doesn't get me any further.
But $$Ax=0$$ having only the trivial solution $$\Rightarrow \textrm{ker}A={ \emptyset }$$ which again seems to imply that its square... I don't know, I'll have to have a think about it.

But $$Ax=0$$ having only the trivial solution $$\Rightarrow \textrm{ker}A={ \emptyset }$$ which again seems to imply that its square... I don't know, I'll have to have a think about it.
It's not true that Ker(A)= empty set, because $$0 \in \textrm{ker}(A)$$ . And why do you think it implies that it's square?

Ax = 0 => x = 0
==> rank(A) = n = rank(ATA)
==> ATA is invertible.

Using this you should get the ball rolling... Or I may be wrong.

(ATA)-1ATA = I

That's it, thanks :)