Cfrgtkky

Let $V$ be a $mathbb{R}$ vector space, let $fin operatorname{End}(V)$. We define subsets of $V$ as follows:

$V^+={vin V:f(v)=v}$ and $V^-={vin V:f(v)=-v}$

We know that $V^+$ and $V^-$ are vector subspaces of V, and that their intersection is the zero vector.

Prove:

$V^+oplus V^-=V Longleftrightarrow f^2=1_V$

Proving $Longrightarrow$

Hypothesis: $V^+oplus V^-=V$

Then: $forall vin V:v=v^++v^- $ where $ v^+in V^+ ,v^-in V^-$

$f^2(v)=vRightarrow f(f(v))=f(f(v^++v^-))=f(f(v^+))+f(f(v^-))$ and because of the definition of the subsets (subspaces) $V^+$ and $V^-$ we get that:

$f(f(v^+))+f(f(v^-))=f(v^+)+f(-v^-)=v^++v^-=v$

$f^2(v)=v$ We have our proof.

Proving $Longleftarrow$

Hypothesis: $f^2=1_V$

This means that $f^2(v)=f(f(v))=v$ ($f^2$ is bijective) therefore $f$ is bijective.

Well I don't know how to keep going, we are given a hint but I don't know how to apply it.

Hint we are given:

To prove $Longleftarrow$, we must see if $forall vin V$ can be written as $v^++v^-$ where $ v^+in V^+$ and $v^-in V^-$. To get $v^+$ and $v^-$ we suppose that we have $v=v^++v^-$, then apply $f$ to the equality and we would get the second quation for $v^+$ and $v^-$

Not sure what to make out of this.

Note that the polynomial $X^2-1$ kills $f$, and this factors as $(X-1)(X+1)$. Now we can write $frac 1 2 ((X+1)-(X-1)) = 1$, and we observe that for any $v$, this means we can then write
$v= v^+ + v^-$ where $v^+ = frac 12 (f(v)+v)$ and $v^- = frac 12 ((f(v)-v)$. Then check that this gives the desired decomposition.

Assume you have found the wanted $v^+$ and $v^-$; then
$$
f(v)=f(v^+)+f(v^-)=v^+-v^-
$$
Therefore
$$
v+f(v)=2v^+,qquad v-f(v)=2v^-
$$
This gives necessary conditions: if $v^+$ and $v^-$ exist, then
$$
v^+=frac{1}{2}(v+f(v)),qquad v^-=frac{1}{2}(v-f(v))
$$
Do these vectors satisfy the requirements under the assumption that $f^2$ is the identity?