Abstract Nonsense

Crushing one theorem at a time

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

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January 13, 2011 - Posted by | Algebra, Halmos, Linear Algebra | ,

1 Comment »

  1. […] Point of post: This is a continuation of this post. […]

    Pingback by Projections (Pt. III) « Abstract Nonsense | January 13, 2011 | Reply


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