# Guarantee Ax=Bx implies A=B?

1. Oct 2, 2005

### eddo

Given nxn matrices A and B, and an n-vector x, are there any conditions that can guarantee Ax=Bx implies A=B? I started thinking about this well working on an assignment. It is clearly not always true, since you can easily think up 2x2 examples where it is not. You can also think up examples where the determinants are non-zero, and not equal to each other, where it still doesnt hold. What conditions if any would make this implication true? Thank you.

2. Oct 2, 2005

### hypermorphism

Try multiplying A by the inverse of B.

3. Oct 3, 2005

### DrGreg

Rearrange the equation as (A-B)x = 0. Does that help?

Is this supposed to be true for all values of x or just one specific value?

4. Oct 4, 2005

### HallsofIvy

Staff Emeritus
I presume you mean: are there any conditions on x such that if Ax= Bx for that particular x, then Ax= Bx and the answer is no. If x is any single vector (other than 0. If x= 0 then Ax= Bx= 0 for all A, B.) there exist a basis including x.
Define A by Ax= x, Ay= 0 for all other basis vectors y. Let z be a basis vector other than x, define B by Ax= x, Bz= z, By= 0 for all other basis vectors. Then Ax= Bx but A is not equal to B.

Okay, slight error! In that proof there must be some "other" basis vector- if the vector space is one dimensional, and x is not 0, then Ax=Bx implies A= B!

If Ax= Bx for ALL x, then see what happens when you apply A and B to each of (1, 0, 0,...), (0, 1, 0...)...

Last edited: Oct 4, 2005