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Math Help - Idempotent linear map

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by realpart1/2 View Post
    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 V = \text{Null} (I-T) \oplus \text{Null} (T)?
    (One of the definitions of a direct sum makes it very easy to prove it)
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    No not yet, I'm still in my first semester.
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    Quote Originally Posted by realpart1/2 View Post
    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).
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  5. #5
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    The proof seems simple enough if the  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?
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  6. #6
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    Quote Originally Posted by realpart1/2 View Post
    The proof seems simple enough if the  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 U \cap W = \{0\} and U + W = V,then we say V = U \oplus W.

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

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

    Applying the definition, we have to prove two things:

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

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

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

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

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