1
$\begingroup$

Let V be a $\mathbb{R}-$Vectorspace of arbitrary dimension and $f\in End(V)$ (Set of all endomorphisms) show that if $f \circ f = f$ then V = E(f,1) + E(f,0).

So that means we want to prove that V = fixpoints of f + ker(f)? And f is its on identity but I can't seem to make a connection here.

Thanks.

  • 0
    What are $E(f,1)$ and $E(f,0)$? – 2017-02-02
  • 0
    The set of eigenvalues. So $E(f,1) = \{v \in V :f(v) = 1 * v\}$ and $E(f,0) = \{v \in V : f(v) = 0 *v\}$ – 2017-02-02

1 Answers 1

1

$f\circ f=f$ implies for $x\in V$ that $f(f(x))=f(x)$, so $Im f$ consists of fixpoints. And for $x=f(x)+(x-f(x))$ You have $f(x-f(x))=f(x)-f(f(x))=f(x)-f(x)=0$ and thus $x-f(x)\in Kern(f)$. Of course a fixpoint in the kernel must be $0$. Thus $V=Im(f)\oplus ker(f)=E(f,1)\oplus E(f,0)$ .