# Idempotent property

1. Nov 29, 2013

### Maths2468

How do you show that a linear transformation is idempotent?

T:R^3 to R^3 T (x y z)^T = (0.5 (x-z) , y, 0.5 (z-x))

I have no idea where to begin. I know a few facts about idempotent properties e.g such as their eigenvalues are either 0 or 1. How would I show that the above transformation has these eigenvalues. I know how to find them but the above form has thrown me off. Would I have to just prove a couple of idempotent to show it is idempotent?

2. Nov 29, 2013

### LCKurtz

Why don't you just directly show that $T^2=T$, either from the above definition or its matrix representation?

3. Nov 29, 2013

### Maths2468

ok so how would I square this matrix? if I write it in matrix form would it be 0.5x-0.5x
y
-0.5z+0.5z

or do I not expand the bracket because the x variable and z variable cancel out?

4. Nov 29, 2013

### LCKurtz

You have to write it in matrix form if you want to get a matrix to square.$$T(x,y,z) = \begin{bmatrix} a&b&c\\d&e&f\\g&h&i\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}\frac 12(x-z)\\y\\\frac 1 2(z-x)\end{bmatrix}$$Once you figure out that matrix $A$ you can square it to see if $A^2=A$.

Also, as I suggested before, you have the formula for $T(x,y,z)$. You could just apply it twice.

5. Nov 30, 2013

### Maths2468

so if I multiply them I get ax+by+cz 0.5(x-z)
dx+ey+fz = y
gx+hy+iz 0.5(z-x)

Is that what you meant? I am not sure where I am going with this.

6. Nov 30, 2013

### Erland

But it is obvious. If T(x,y,z)=(0.5(x-z), y, 0.5(z-x)) for all x,y,z, then what is T(0.5(x-z), y, 0.5(z-x))?

7. Nov 30, 2013

### Maths2468

I honestly can not see the answer. Could you give me an example if you do not mind? It does not have to be this specific problem. maybe I am looking at the problem from a different angle if it is meant to be that obvious.
is it
0.5 0 -0.5
0 1 0
-0.5 0 0.5

Last edited: Nov 30, 2013
8. Nov 30, 2013

### Erland

You just have to calculate T(0.5(x-z), y, 0.5(z-x)) and see what the answer is.

9. Nov 30, 2013

### Maths2468

ahh ok so the matrix representation is what I said above and then you just check the properties of the idempotent apply

10. Nov 30, 2013

### LCKurtz

Yes, that is the matrix representation. What happens if you square that matrix, which represents $T^2$?

I am curious what course you are taking. Have you talked about matrix representations of linear transformations?

11. Nov 30, 2013

### LCKurtz

You are making this way too difficult. You have the formula for $T$. Could you calculate $T(1,2,3)$? $T(a,b,c)$? What about $T(0.5(x-z), y, 0.5(z-x))$? Just use the formula.

12. Nov 30, 2013

### Maths2468

I am doing maths.I find the course very well except for linear stuff. I can not picture things. Yeah if you square it you get itself again. we have only just touched on it. I am kind of going ahead of the course.