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Let $f \in End_{F}(V)$ be an Endomorphism of a vector spaces $V$ over the field $F$ with the property $f \circ f=id_{V}$. Show that if $char(F) \neq 2$, the following equotation is true: $$V = Ker(f-id_{V}) \oplus Ker(f+id_{V})$$

Notes: $\oplus$ is the direct sum, $id_{V}$ is the identity map, Ker is the kernel and $\circ$ is the function composition.

My thoughts: I think you have to show that every Element in V can be expressed with the direct sum. It is obvious that the direct sum is always an element in V because the Kernel will return an element in V and the sum has to be in V again. But I have no clue where you need that $char(F) \neq 2$

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If $\mathrm{char}\;k=2$, then $\ker(f-\mathrm{id})=\ker(f+\mathrm{id})$, so the sum is rarely direct... – Mariano Suárez-Alvarez Nov 4 '10 at 20:32
up vote 3 down vote accepted

You want to show that two things happen:

  1. Every element of $V$ can be written as a sum $u+w$, where $u\in\mathrm{ker}(f-\mathrm{id}_V)$ and $v\in\mathrm{ker}(f+\ker{id}_v)$; and
  2. The intersection of $\mathrm{ker}(f-\mathrm{id}_V)$ and $\mathrm{ker}(f+\mathrm{id}_V)$ is trivial (contains only the zero vector).

One reason why you need $\mathrm{char}(F)\neq 2$ is that if $\mathrm{char}(F)=2$, then $1=-1$, so $f-\mathrm{id}_V = f+\mathrm{id}_V)$. And the sum of a subspace with itself cannot be a direct sum unless the subspace is trivial (which in this case would require $V$ to be trivial).

The second part is fairly easy: suppose $x$ is in the intersection. Then $(f+\mathrm{id}_V)(x) = \mathbf{0}$ and $(f-\mathrm{id}_V)(x) = \mathbf{0}$; the first equation tells you that $f(x)+x=\mathbf{0}$, the second that $f(x)-x=\mathbf{0}$. Now, since $2\neq 0$, you can conclude that $x=\mathbf{0}$.

So now you "just" need to show that every element of $V$ can be expressed as a sum of vectors in the appropriate place. Now, notice that so far we have not used the fact that $f\circ f=\mathrm{id}_V$, so presumably we are going to need to use it somehow. For one thing, what happens if you take an arbitrary vector $x$, and you look at $x-f(x)$? Then $f(x-f(x)) = f(x)-f(f(x)) = f(x)-x = -(x-f(x))$. So, does $x-f(x)$ lie in one of the two subspaces? What about $x+f(x)$? And, can you express $x$ as a linear combination of $x+f(x)$ and $x-f(x)$ (again, the fact that $2\neq 0$ is going to rear its ugly head...)

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Now that you say it your proof makes perfect sense. Thank you =) – Listing Nov 4 '10 at 20:39

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