idempotent vector space endomorphism for a finite dimensional vector space gives decomposition

Proposition

Let \(K\) be a field and \(V\) a finite dimension vector space and

\begin{align*} T: V \rightarrow V \end{align*}

an idempotent vectorspace-homomorphism

Then

\begin{align*} T = \langle \mathrm{im}(T), \mathrm{ker}(T)\rangle \end{align*}

with

\begin{align*} \mathrm{im}(T) \cap \mathrm{ker}(T) = (0) \end{align*}

Proof

a) \(\mathrm{im}(T) \cap \mathrm{ker}(T) = (0)\)

Let \(v \in \mathrm{im}(T) \cap \mathrm{ker}(T)\).
Then

\begin{align*} v = T(v') \end{align*}

for some \(v' \in V\) as \(v \in \mathrm{im}(T)\).

Then since \(T\) is idempotent,

\begin{align*} T(v) =& T(T(v')) \\ =& T(v') \\ =& v \end{align*}

But furthermore \(v \in \mathrm{ker}(T)\) hence

\begin{align*} T(v) = 0 \end{align*}

\begin{align*} v = 0 \end{align*}

since

\begin{align*} T(v) = v \end{align*}

b) \(T = \mathrm{im}(T) + \mathrm{ker}(T)\)

It suffices to show that

\begin{align*} \mathrm{dim}(V) = \mathrm{dim}(\mathrm{im}(T) + \mathrm{ker}(T)) \end{align*}

since \(\mathrm{dim}(V) < \infty\) (cf. ?).

Rank Theorem tells us

\begin{align*} \mathrm{dim}(V) = \mathrm{dim}(\mathrm{im}(T)) + \mathrm{dim}(\mathrm{ker}(T)) \end{align*}

Then by dimension der summe von zwei Vektorräumen we get

\begin{align*} \mathrm{dim}(\mathrm{im}(T) + \mathrm{ker}(T)) + \mathrm{dim}(\mathrm{im}(T) \cap \mathrm{ker}(T)) =& \mathrm{dim}(\mathrm{im}(T)) + \mathrm{dim}(\mathrm{ker}(T)) \\ \mathrm{dim}(\mathrm{im}(T) + \mathrm{ker}(T)) + \mathrm{dim}(0) =& \mathrm{dim}(\mathrm{im}(T)) + \mathrm{dim}(\mathrm{ker}(T)) \\ \mathrm{dim}(\mathrm{im}(T) + \mathrm{ker}(T)) =& \mathrm{dim}(\mathrm{im}(T)) + \mathrm{dim}(\mathrm{ker}(T)) \\ \end{align*}

\begin{align*} \mathrm{dim}(V) =& \mathrm{dim}(\mathrm{im}(T)) + \mathrm{dim}(\mathrm{ker}(T)) \\ =& \mathrm{dim}(\mathrm{im}(T) + \mathrm{ker}(T)) \end{align*}