# Projections and diagonal form (Algebra)

1. Jan 17, 2009

### math8

If V has finite dimension n, show that two projections have the same diagonal form if and only if their kernels have the same dimension. (A projection is defined to be a linear transformation P:V-->V for which P^2=P; V is a vector space).

For the forward direction, I was thinking that if T and P are projections that have the same diagonal form, their matrices of transformation are similar.
I think I need to use the formula
dimV= dim ker(T)+ dim Im(T) Knowing that dim Im(T)= rank of the matrix of transformation of T and dim ker(T)= n-dim Im(T)
So the result follows.

But I am not sure about the backward direction.

2. Jan 17, 2009

### Dick

Form a basis of V from vectors in ker(T) and im(T). You know the two sets span V. That's where your dimension formula comes from. Think about what the matrix of T looks like in this basis.

3. Jan 17, 2009

### math8

Since the dim of the kernels are the same, then the dim of the Images are the same as well. We also know that for a projection V=ker T '+' Im T (where '+' denotes the direct sum)
So if {e1,e2,...,ek} is a basis for ker T, we may complete this list by vectors of the basis for Im T to get a basis for V ({e1,e2,...,ek,f1,...,fm}.

Now the matrix of the transformation T with respect to this new basis is a k+m X k+m matrix. I don't know what else to say.

4. Jan 17, 2009

### Dick

T(ei)=0, T(fi)=fi. It looks to me like that means you have a matrix with 0's and 1's on the diagonal. How many zeros and how many ones?

5. Jan 17, 2009

### math8

there should be k 0's and m 1's.

6. Jan 17, 2009

### math8

So the same thing for both the projections T and P, hence they have the same diagonal form?

7. Jan 17, 2009

### Dick

Ok, so k is dim(ker(T)) and m is dim(V)-dim(ker(T)). If you have two diagonal matrices with the same number of 1's and 0's on the diagonal, are they similar?

8. Jan 17, 2009

### Dick

Sure. They have the same matrix in different bases. They are similar.

9. Jan 17, 2009

### math8

Thanks, it makes much more sense now.