# Abstract Nonsense

## Projections (Pt. II)

Point of post: This is a continuation of this post.

Combinations of Projections

A clear question one may ask about projections is when the sum of two projections is itself a projection. This and more is answered by the next theorem:

Theorem: Let $\mathscr{V}$ be a finite dimensional $F$-space with $\text{char}\left(F\right)>2$. Suppose then that $\mathscr{U}_i,\mathscr{W}_i\leqslant \mathscr{V},\;\;i=1,2$  are such that

$\mathscr{V}=\mathscr{U}_1+\mathscr{W}_1=\mathscr{U}_2\oplus\mathscr{V}_2$

and $\pi_{\mathscr{U}_i}$ is the projection on $\mathscr{U}_i$ along $\mathscr{W}_i$ for $i=1,2$. Then,

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}\textit{ is a projection if and only if }\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\bold{0}\\ &\quad\quad ,\textit{ if it is a projection then it is the projection on }\mathscr{U}_1+\mathscr{U}_2\textit{ along }\mathscr{W}_1\cap\mathscr{W}_2\\ &\mathbf{(2)}\quad \pi_{\mathscr{U}_1}-\pi_{\mathscr{U}_2}\textit{ is a projection if and only if}\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\pi_{\mathscr{U}_2}\\ &\quad\quad,\textit{ if it is a projection then it is the projection on }\mathscr{U}_1\cap\mathscr{W}_2\textit{ along }\mathscr{W}_1\oplus\mathscr{U}_2\\ &\mathbf{(3)}\quad \textit{If }\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=P\textit{ then }P\textit{ is the projection on }\mathscr{U}_1\cap\mathscr{U}_2\textit{ along }\mathscr{W}_1+\mathscr{W}_2\end{aligned}

Proof:

$\mathbf{(1)}$: Suppose first that $\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}$ was a projection. Note then that

\begin{aligned}\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2} &= \left(\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}\right)^2\\ &= \pi_{\mathscr{U}_1}^2+\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}+\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}^2\end{aligned}

and so upon subtraction we find that $\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}+\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\bold{0}$. But, multiplying this on the left by $\pi_{\mathscr{U}_1}$ gives that

$\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}+\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}+\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\bold{0}$

and multiplying the same initial equation on the right gives

$\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_1}=\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\bold{0}$

and equating these gives us that

$\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}$

but we know that

$\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=-\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}$

from where it follows that $\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\bold{0}$.

Conversely, t if $\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\bold{0}$ then the above argument shows that $\pi_{\mathscr{U}_1}+\pi_{\mathscr{U}_2}\overset{\text{def.}}{=}T$ is idempotent and thus a projection. Thus, it suffices to prove that it’s a projection on $\mathscr{U}_1+\mathscr{U}_2$ along $\mathscr{W}_1\cap\mathscr{W}_2$. It’s clear from our previous theorem that $\mathscr{V}=\mathscr{U}\oplus\mathscr{W}$ where $\mathscr{U}=\left\{v\in\mathscr{V}:T(v)=v\right\}$ and $\mathscr{W}=\ker T$. Thus, it suffices to find $\mathscr{U}$ and $\mathscr{W}$. We note firstly that if $v\in\mathscr{U}$ with $v=v_1+w_1=v_2+w_2$ where $v_i\in\mathscr{U}_i\;\;i=1,2$ and $w_i\in\mathscr{W}_i\;\; i=1,2$. We note then that

$z=T(z)=\pi_{\mathscr{U}_1}(u_2+w_2)+\pi_{\mathscr{U}_2}(u_1+w_1)=\pi_{\mathscr{U}_1}(w_2)+\pi_{\mathscr{U}_2}(w_1)$

where we’ve used the fact that $\pi_{\mathscr{U}_1}(x_2)=\pi_{\mathscr{U}_1}\left(\pi_{\mathscr{U}_2}\left(z\right)\right)=\bold{0}$ and similarly for $\pi_{\mathscr{U}_2}(x_1)$. We note though that $\pi_{\mathscr{U}_1}\left(\pi_{\mathscr{U}_1}\left(w_2\right)\right)=\pi_{\mathscr{U}_1}\left(w_2\right)$ and $\pi_{\mathscr{U}_2}\left(\pi_{\mathscr{U}_2}\left(w_1\right)\right)=\pi_{\mathscr{U}_2}\left(w_1\right)$ from where it follows (from the previous theorem) that $z\in\mathscr{U}_1+\mathscr{U}_2$. The reverse inclusion is clear, and thus it follows that $\mathscr{U}=\mathscr{U}_1+\mathscr{U}_2$.

Now, it suffices to find $\mathscr{W}$. It’s evident that $\mathscr{W}_1\cap\mathscr{W}_2\subseteq\mathscr{W}$. Conversely, if $v\in\mathscr{W}$ then we know that

$\pi_{\mathscr{U}_1}(z)+\pi_{\mathscr{U}_2}(z)=\bold{0}$

multiplying this equation by $\pi_{\mathscr{U}_1}$ on the right (remembering our assumptions) gives that

$\pi_{\mathscr{U}_1}\left(\pi_{\mathscr{U}_1}\left(z\right)\right)+\pi_{\mathscr{U}_1}\left(\pi_{\mathscr{U}_2}\left(z\right)\right)=\pi_{\mathscr{U}_1}(z)=\bold{0}$

and multiplying on the right by $\pi_{\mathscr{U}_2}$ gives

$\pi_{\mathscr{U}_1}\left(\pi_{\mathscr{U}_2}\left(z\right)\right)+\pi_{\mathscr{U}_2}\left(\pi_{\mathscr{U}_2}\left(z\right)\right)=\pi_{\mathscr{U}_2}\left(z\right)=\bold{0}$

from where it follows that $z\in\mathscr{W}_1\cap\mathscr{W}_2$ and thus it follows that $\mathscr{W}=\mathscr{W}_1\cap\mathscr{W}_2$. The conclusion follows.

$\mathbf{(2)}$: We merely note that by our previous theorem $\pi_{\mathscr{U}_1}-\pi_{\mathscr{U}_2}$ is a projection if and only if

$\displaystyle \mathbf{1}-\left(\pi_{\mathscr{U}_1}-\pi_{\mathscr{U}_2}\right)=\left(\mathbf{1}-\pi_{\mathscr{U}_1}\right)+\pi_{\mathscr{U}_2}$

is a projection. But, by $\mathbf{(1)}$ of this theorem we know that this is true if and only if

$\left(\mathbf{1}-\pi_{\mathscr{U}_1}\right)\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\left(\mathbf{1}-\pi_{\mathscr{U}_1}\right)=\bold{0}$

but clearly upon expansion this is true if and only if $\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2} =\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=\pi_{\mathscr{U}_2}$. Moreover, if

$\pi_{\mathscr{U}_1}-\pi_{\mathscr{U}_2}=\left(\mathbf{1}-\pi_{\mathscr{U}_1}\right)+\pi_{\mathscr{U}_2}=\pi_{\mathscr{W}_1}+\pi_{\mathscr{U}_2}$

we have from the previous problem that it is the projection on $\mathscr{W}_1+\mathscr{U}_2$ along $\mathscr{W}_2\cap\mathscr{U}_1$.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print