Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $char(F) \neq 2$

share|improve this question
    
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
add comment

1 Answer

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...)

share|improve this answer
    
Now that you say it your proof makes perfect sense. Thank you =) –  Listing Nov 4 '10 at 20:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.