Let $T:V \rightarrow V$ be a linear transformation such that $T^2 =$ $T$. Discuss whether or not there exists a basis of $V$ consisting of eigenvectors of $T$.
The answer in the back of the book says:
The minimal polynomial of $T$ divides $x^2 - x$, and therefore has distinct linear factors in $C[x]$. There does exist a basis of $V$ consisting of eigenvectors.
Please explain to me the process of arriving at this conclusion. Thanks!
We have a linear operator $T \,:\, V \to V$ with $T^2 = T$.
Let $U$ be the range of $T$. Obviously, $T|_U = \textrm{id}_U$, i.e. $T$ acts as the identity on $U$, otherwise we wouldn't have $T^2 = T$.
Let $W$ be the kernel of $T$, i.e. $W = \textrm{ker } T = \{v \in V \,:\, Tv = 0\}$. We have $U \cap W = \{0\}$, because if $x \in U$ then $Tx = x$ (by the above), so $Tx = 0$ only if $x = 0$, meaning $x \notin W$ unless $x = 0$.
For linear operators from a vector space to itself we always have that $$ \dim \underbrace{\textrm{ker } T}_{=W} + \dim \underbrace{\textrm{rng }T}_{=U} = \dim V \text{.} $$ Since $U \cap W = \{0\}$, it follows that $$ V = U \oplus W \text{.} $$
But now we're done. We simply pick a basis $b_1,\ldots,b_k$ of $U$, append a basis $b_{k+1},\ldots,b_n$ of $W$, then $$ T b_i = \begin{cases} b_i &\text{if $b_i \in U$, i.e. if $1 \leq i \leq k$} \\ 0 &\text{if $b_i \in W$, i.e. if $k < i \leq n$.} \end{cases} $$ Obviously, all the $b_i$ are thus eigenvectors of $T$, the first $k$ having eigenvalue $1$ and the others eigenvalue $0$.