Proving 1. $\leftrightarrow$ 2.: A Matrix Invertibility Challenge

[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

 

[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.

 

[ImAnEngineer]

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

 

[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?

 

[ImAnEngineer]

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.

 

[Zorba]

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.

 

[ImAnEngineer]

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.

 

[Zorba]

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.

 

[ImAnEngineer]

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?

 

[Lord Crc]

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

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

 

[ImAnEngineer]

(A^TA)^-1A^TA = I

That's it, thanks :)