Symmetrical Matrices and Invertibility: Is A Always Invertible If Ax ≠ Ay?

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    Inverse Matrices
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Discussion Overview

The discussion revolves around the conditions under which a symmetrical matrix A is invertible, particularly in relation to the property that for every pair of distinct vectors x and y, the equation Ax ≠ Ay holds. Participants explore whether this property implies that A is necessarily invertible, considering both singular and non-singular cases.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if Ax ≠ Ay for all distinct vectors x and y, then A is not necessarily invertible, especially if A is singular.
  • Others argue that if A is an n x n matrix and Ax ≠ Ay for all distinct vectors, then A must be invertible.
  • A specific example of a singular matrix is presented, with claims that it satisfies the condition Ax ≠ Ay for distinct vectors, challenging the notion that this condition guarantees invertibility.
  • Some participants clarify that the existence of a nonzero vector x such that Ax = 0 indicates that A is singular, which contradicts the condition of Ax ≠ Ay for all distinct vectors.
  • There is a discussion about whether the condition holds for every pair of distinct vectors if A is non-singular, with some asserting that it does.
  • One participant presents a counterexample involving x ≠ 0 and y = 0, suggesting that if A is singular, it leads to contradictions regarding the condition Ax ≠ Ay.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the condition Ax ≠ Ay implies that A is invertible. There are competing views regarding the implications of singularity and the validity of the condition across different cases.

Contextual Notes

The discussion highlights limitations in the assumptions regarding the nature of the matrix A, particularly its singularity and the implications of the condition Ax ≠ Ay. There are unresolved mathematical steps regarding the proof of invertibility based on the given conditions.

peripatein
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Hello,

Would it be correct to say that if for every two different vectors x and y, A*x ≠ A*y (where A is a symmetrical matrix), then A is NOT necessarily invertible? In other words, albeit for any two different vectors x and y symmetrical matrix A times one of the vectors is not equal to A times the other, A is not necessarily invertible?
 
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Correction, in case A is a square matrix (of order nXn)
 
If A is an n x n - matrix such that Ax ≠ Ay for all pairs of distinct n-vectors x and y, then A is invertible.
 
Let us examine the following singular matrix:
2 0 4
1 -1 3
0 -1 1

For any two different vectors I claim that that matrix multiplied by the first vector will never be equal to the multiplication of that same matrix by the second vector.
Hence, the matrix does not necessarily have to be singular for the proposition to be valid and hold.
Wouldn't you agree?
 
peripatein said:
Let us examine the following singular matrix:
2 0 4
1 -1 3
0 -1 1

For any two different vectors I claim that that matrix multiplied by the first vector will never be equal to the multiplication of that same matrix by the second vector.
Hence, the matrix does not necessarily have to be singular for the proposition to be valid and hold.
Wouldn't you agree?
Certainly not. As you said, the matrix is singular. This means that there is a nonzero vector x such that Ax=0 (such an x can easily be found if we solve the system Ax=0). On the other hand, A0=0, so Ax=A0, despite that x≠0.
 
Okay, so the proposition does not hold in case A is singular, but does that per se guarantee that it holds, for EVERY two different vectors, if A were not singular, i.e. invertible?
 
peripatein said:
Okay, so the proposition does not hold in case A is singular, but does that per se guarantee that it holds, for EVERY two different vectors, if A were not singular, i.e. invertible?
Yes, for if A is invertible and Ax=Ay, then x=Ix=(A-1A)x=A-1(Ax)=A-1(Ay)=(A-1A)y=Iy=y, that is, x=y.

The system Ax=b has a unique solution, x=A-1b, if x is invertible. Otherwise, it has either no solution or infinitely many solutions.
 
Thank you very much! :-)
 
The easy way to see this is false is consider the special case of x ≠ 0 and y = 0.

Ay = 0, so for every x ≠ 0, Ax ≠ 0.

If A is singular, there is a vector x ≠ 0 such that Ax = 0, which is a contradiction. Therefore A is non-singular.
 

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