I'm trying to understand the proof for this:
A linear transformation $P$ on a vector space $V$ into itself is a projection if and only if $P^2 = P$. The following conditions are equivalent:
- $T:V\rightarrow V$ is a projection
- There exist subspaces $X$ and $Y$ such that $V = X\oplus Y$, and $T(\textbf{v}) = x$, where $\textbf{ v = x+y, x}\in X,\textbf{y}\in Y$ is the unique decomposition of $\bf v$.
I understand why (2) implies (1) - if $\bf v = x+y$, $\bf x = x+0$ are unique decompositions, then $T^2(\textbf{v}) = T(T(\textbf{v})) = T(\textbf{x}) = \textbf{x} = T(\textbf{v})$, so $T^2 = T$.
I'm having difficulty with the proof that (1) implies (2). Here is what my book has:
Define $X = \mathcal{R}(T), y = \mathcal{N}(T).$ From the decomposition $$\textbf{v} = T(\textbf{v}) + \textbf{v} - T(\textbf{v})$$ And the fact that $T(\textbf{v}-T(\textbf{v})) = T(\textbf{v}) - T^2(\textbf{v}) = 0$ follows that $V = X+Y$.
Suppose now that $\textbf{v}\in\mathcal{R}(T)\cap\mathcal{N}(T).$ This implies that there exists $\textbf{w}\in V$ such that $T(\textbf{w})\in\mathcal{N}(T)$, i.e., $T(T(\textbf{w})) = 0$. But $T(T(\textbf{w}))= T^2(\textbf{w}) = T(\textbf{w}) = \textbf{v}$, so $\textbf{v} = 0$ which proves the assertion.
I'm confused starting with the decomposition. I see that $T(\textbf{v})$ is in the range of T, and $\textbf{v}-T(\textbf{v})$ is in the null space. But where does that decomposition come from? Or is it just supposed to be an identity that is used to show that there are components of $\textbf{v}$ in $X$ and $Y$?
And then the last part is used to show that the intersection of $X$ and $Y$ is zero and so the $+$ becomes an $\oplus$?
So I think writing this out helped me understand a bit better, but verification of my thought process would be nice. Thanks!