Can Singular Matrices Satisfy A^3 = A^2 but A^2 ≠ A?

[angrywar]

Homework Statement

the question is :
Get an example of metrics M2X2 that:
A^2 doest equal A but A^3 equals A^2

The Attempt at a Solution

I tried to put matrices A equals :
a b
c d

and tried to solve it , but I get to 4 equations with 4 parameters and I don't know how to solve it , is there any easier way ?

 

[HallsofIvy]

"Metrics"? Do you mean matrix?

If $A^3= A^2$ then $A^3- A^2= A(A^2- A)= 0$.
The requirement that $A^2\ne A$ means that $A^2- A\ne 0$. Now, it does NOT follow that A= 0 ($0^2= 0$ so A= 0 is not a solution)- the ring of matrices has "0 divisors" But that does mean that A cannot be invertible.

 

[angrywar]

yes i ment matrix

i didnt get what u mean by :
"the ring of matrices has "0 divisors""

 

[kai_sikorski]

First prove that both requirements can't be satisfied by a non-singular matrix.

Once you know that A is singular, for a 2x2 this means that one of the columns is a scalar multiple of the other. So A will be of the form

$$A = \left(<br /> \begin{array}{cc}<br /> a & \kappa b \\<br /> a & \kappa b \\<br /> \end{array}<br /> \right)$$

EDIT: not sure why the LaTeX isn't working. The matrix should be
(a) (κ a)
(b) (κ b)Using this assumption calculate A^3-A^2. If this quantity is 0, then in particular the (1,1) component must be 0. This gives you a polynomial which you can solve for a in terms of b (treat $\kappa$ as a parameter). You'll have up to 3 solutions since the polynomial can be at most cubic. Just plug them in and see if any satisfy all the requirements. If so you have your example (in fact you have a family of examples since κ can probably be almost anything), if not you've basically proven that an example doesn't exist for 2x2 matrices.