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.