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

[peripatein]

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?

 

[peripatein]

Correction, in case A is a square matrix (of order nXn)

 

[Erland]

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.

 

[peripatein]

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?

 

[Erland]

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.

 

[peripatein]

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?

 

[Erland]

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.

 

[peripatein]

Thank you very much! :-)

 

[AlephZero]

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.