1. ## Idempotent linear map

f:V-->V is an idempotent (i.e. f(f(v))=f(v)) linear map of an n-dimensional F-vector space V. I need to prove that there is a basis {v1,v2,...,vn} and a natural number r: 0<=r<=n so that f(vi)=vi for 0<=i<=r and f(vi)=0 for r+1<=i<=n.

I think it has something to do with an intersection of N(f) and V having to have something other than 0, but I'm still struggling with the actual proof. I think it has something to do with the idempotency of f.

An easy to follow formalized proof would be greatly appreciated.

2. Originally Posted by realpart1/2
f:V-->V is an idempotent (i.e. f(f(v))=f(v)) linear map of an n-dimensional F-vector space V. I need to prove that there is a basis {v1,v2,...,vn} and a natural number r: 0<=r<=n so that f(vi)=vi for 0<=i<=r and f(vi)=0 for r+1<=i<=n.

I think it has something to do with an intersection of N(f) and V having to have something other than 0, but I'm still struggling with the actual proof. I think it has something to do with the idempotency of f.

An easy to follow formalized proof would be greatly appreciated.
Can you show $\displaystyle V = \text{Null} (I-T) \oplus \text{Null} (T)$?
(One of the definitions of a direct sum makes it very easy to prove it)

3. No not yet, I'm still in my first semester.

4. Originally Posted by realpart1/2
No not yet, I'm still in my first semester.
I am sorry I was not clear enough. I used T for linear transformation and Null(T) for the kernel of T.

Do you know what a direct sum of vector space is? If yes, it is easy. Or else you have to write a long proof...

Anyway... observe that if f is a linear transformation, then I-f is also a linear transformation, where I is the identity lin.transf. Consider the kernel(I-f). Its a finite dimensional vector space. Let its dimension be r. You can find r basis vectors for this kernel. Observe that all the elements v in that kernel satisfy f(v) = v. Now extend the basis and show that the space spanned by extended vectors is the kernel(f).

5. The proof seems simple enough if the $\displaystyle V=\text{Null} (I-T) \oplus \text{Null} (T)$ part is right (I've never dealt with direct sums before but I'm guessing it's some kind of union between two vector spaces). But how do I prove that part?

6. Originally Posted by realpart1/2
The proof seems simple enough if the $\displaystyle V=\text{Null} (I-T) \oplus \text{Null} (T)$ part is right (I've never dealt with direct sums before but I'm guessing it's some kind of union between two vector spaces). But how do I prove that part?
Definition of direct sum of U and W: If U and W are subspaces of V such that $\displaystyle U \cap W = \{0\}$ and $\displaystyle U + W = V$,then we say $\displaystyle V = U \oplus W$.

[Definition of U+W: $\displaystyle U + W := \{ u + w : u \in U, w \in W\}$ (check that this is a vec. space!)]

Now we claim $\displaystyle V = \text{Null(I - f)} \oplus \text{Null(f)}$.

Applying the definition, we have to prove two things:

(1) $\displaystyle \boxed{\text{Null(I - f)} \cap \text{Null(f)} = \{ 0 \}}$

Proof: This is obvious since $\displaystyle v \in \text{Null(I - f)}\cap \text{Null(f)} \implies (I - f)(v) = 0, f(v) = 0$ $\displaystyle \implies v = f(v), f(v) = 0 \implies v = 0$

(2) $\displaystyle \boxed{\text{Null(I - f)} + \text{Null(f)} = V}$

Proof: $\displaystyle U+W$ is a subspace of V(Why?). Let $\displaystyle v \in V$, then $\displaystyle v = f(v) + (I - f)(v)$. But $\displaystyle f(v) \in \text{Null(I - f)}$ and $\displaystyle (I - f)(v) \in \text{Null(f)}$. Thus every vector can be written as a sum of two vectors, one from $\displaystyle \text{Null(I - f)}$ and the other from $\displaystyle \text{Null(f)}$

From (1) and (2), $\displaystyle V = \text{Null(I - f)} \oplus \text{Null(f)} \, \,\,\,\,\, \blacksquare$