Left invertible

  • #1
209
1

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)
- [tex]n\leq m[/tex]
 

Answers and Replies

  • #2
77
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If the solutions to [tex]Ax=0[/tex] are only the trivial one, then its RREF is the identity matrix, thus it is invertible.

And another:
[tex]Ax=0[/tex] are only the trivial solution, then [tex]\textrm{det}A\ne 0[/tex] thus it is invertible.
 
  • #3
209
1
If the solutions to [tex]Ax=0[/tex] are only the trivial one, then its RREF is the identity matrix, thus it is invertible.

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

Note that A is not (necessarily) a square matrix
 
  • #4
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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.

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

[tex]Ax=0 \Rightarrow \textrm{det}A\ne0[/tex] follows from properties of the determinant and also properties of elementary matrices, are you familiar with these?
 
  • #5
209
1
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.

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

[tex]Ax=0 \Rightarrow \textrm{det}A\ne0[/tex] 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.
 
  • #6
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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.
 
  • #7
209
1
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: [tex]m\geq n[/tex]

But it doesn't get me any further.
 
  • #8
77
0
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: [tex]m\geq n[/tex]

But it doesn't get me any further.
But [tex]Ax=0[/tex] having only the trivial solution [tex]\Rightarrow \textrm{ker}A={ \emptyset }[/tex] which again seems to imply that its square... I don't know, I'll have to have a think about it.
 
  • #9
209
1
But [tex]Ax=0[/tex] having only the trivial solution [tex]\Rightarrow \textrm{ker}A={ \emptyset }[/tex] 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 [tex]0 \in \textrm{ker}(A)[/tex] . And why do you think it implies that it's square?
 
  • #10
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29
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
209
1
(ATA)-1ATA = I

That's it, thanks :)
 

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