Introductory Linear Alebra proof question

[IDumb]

Homework Statement

Let A be an n x n matrix and let x and y be vectors in R^n. Show that if Ax = Ay and x $$\neq$$ y, then the matrix A must be singular.

Homework Equations

So far we have learned the definition of a matrix that has an inverse to be one where: if there exists a matrix B and AB = BA = I. The matrix B is said to be the multiplicative inverse of A.

The Attempt at a Solution

I have done earlier problems that involved proving things have an inverse with the above definition, however I can not think of how to apply it to this problem.

So what I did was use my knowledge from earlier classes (also this concept was touched upon a few problems earlier, but has not yet been defined), that if the determinant = 0 then the matrix does not have an inverse. But I have not found anything that helps me.

A = |a11 a12| x = x1 y = y1
|a21 a22| x2 y2

Ax = Ay

(this is supposed to read as Ax = Ay expanded, sorry.
|a11*x1 + a12*x2| |a11*x1 + a12*x2|
| | = | |
|a21*x1 + a22*x2| |a21*x1 + a22*x2|

I've been getting it into equations like: a11(x1-y1) + a12(x2-y2) = 0, and a few similar things, but I have not yet found something that will fit into the formula for the determinant.

So am I on the right track? If i am please give me some hints on how to proceed, if not then let me know what to do please. Thanks a lot, and sorry if anything is unclear.

 

[Dick]

If Av=0 for some nonzero vector v, then A is not invertible. Can you see why that's true from your definition of invertible? Can you see why that might be true from what you are given?

 

[hunt_mat]

If A has a zero eigenvalue then A is singular (definition) This has shown to be the case in previous posts.

 

[Raskolnikov]

Homework Statement

Let A be an n x n matrix and let x and y be vectors in R^n. Show that if Ax = Ay and x $$\neq$$ y, then the matrix A must be singular.

Homework Equations

So far we have learned the definition of a matrix that has an inverse to be one where: if there exists a matrix B and AB = BA = I. The matrix B is said to be the multiplicative inverse of A.

If you're still stuck, try a proof by contradiction. Assume A is nonsingular, i.e. A^-1 exists. Since this is a proof by contradiction, you should expect to either contradict one of your premises (e.g. x $$\neq$$ y) or to arrive at something that clearly doesn't make sense (e.g. 1 = 0).

Spoiler

You can easily arrive at the former contradiction if you recall that A^-1A = AA^-1 = I.

We often prefer straightforward proofs over proofs by contradiction, but either method seems fine here since no real insight in the theory is lost in this case by the latter approach. I hope this helped a bit.