# Left invertible

1. Feb 28, 2010

### ImAnEngineer

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)
- $$n\leq m$$

2. Mar 1, 2010

### Zorba

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.

3. Mar 1, 2010

### ImAnEngineer

Why and why?

Note that A is not (necessarily) a square matrix

4. Mar 1, 2010

### Zorba

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?

5. Mar 1, 2010

### ImAnEngineer

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.

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

6. Mar 1, 2010

### Zorba

Ah I see, just a left-inverse, in that case then you must avoid mentioning determinants although my first statement still holds.

7. Mar 1, 2010

### ImAnEngineer

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.

8. Mar 2, 2010

### Zorba

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.

9. Mar 2, 2010

### ImAnEngineer

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?

10. Mar 2, 2010

### Lord Crc

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.

11. Mar 3, 2010

### ImAnEngineer

(ATA)-1ATA = I

That's it, thanks :)