Abstract Nonsense

Crushing one theorem at a time

Projections (Pt. III)


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

\mathbf{(3)}: It’s clear that if \pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}=\pi_{\mathscr{U}_2}\pi_{\mathscr{U}_1}=P then P is a projection since

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

 

We use the same methodology to find the \mathscr{U} and \mathscr{W} with which P is on and along respectively. We claim that \mathscr{U}=\mathscr{U}_1\cap\mathscr{U}_2. Indeed, if z\in\mathscr{U} then

\pi_{\mathscr{U}_1}(z)=\pi_{\mathscr{U}_1}P(z)=\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}(z)=\pi_{\mathscr{U}_1}\pi_{\mathscr{U}_2}(z)=P(z)=z

and so z\in\mathscr{U}_1\cap\mathscr{U}_2. The reverse inclusion is clear from where it follows that \mathscr{U}=\mathscr{U}_1\cap\mathscr{U}_2.

Now we next claim that \mathscr{W}=\mathscr{W}_1+\mathscr{W}_2. Indeed, if v\in\mathscr{W} then we have that

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

and thus \pi_{\mathscr{U}_1}(z)\in\mathscr{W}_2 and \pi_{\mathscr{U}_2}(z)\in\mathscr{W}_1. We note though that

v=\pi_{\mathscr{U}_2}(z)+\left(\mathbf{1}-\pi_{\mathscr{U}_2}\right)

and since \left(\mathbf{1}-\pi_{\mathscr{U}_2}\right)\in\mathscr{W}_2 it follows that v\in\mathscr{W}_1+\mathscr{W}_2. The revere inclusion is evident. The conclusion follows. \blacksquare

Projections and Invariances

Lastly we discuss how projections tie into the invariance of subspaces. Namely, we have the following theorem:

Theorem: If \mathscr{V} is a finite dimensional vector space and \mathscr{U} is invariant under T\in\text{End}\left(\mathscr{V}\right) then \pi_{\mathscr{U}}T\pi_{\mathscr{U}}=T\pi_{\mathscr{U}} for any projection on \mathscr{U}. Conversely, if \pi_{\mathscr{U}}T\pi_{\mathscr{U}}=T\pi_{\mathscr{U}} for some projection \pi_{\mathscr{U}} on \mathscr{U} along \mathscr{W} then T is invariant under \mathscr{U}.

Proof: Suppose first that T is invariant under \mathscr{U} and \pi_{\mathscr{U}} is a projection on \mathscr{U} on any complement \mathscr{W} of \mathscr{U}. We see then that (remembering that T\left(\mathscr{U}\right)\subseteq\mathscr{U})

\pi_{\mathscr{U}}T\pi_{\mathscr{U}}\left(u+w\right)=\pi_{\mathscr{U}}T(u)=T(u)=T\pi_{\mathscr{U}}(u+w)

and since u+w\in\mathscr{V} was arbitrary it follows that \pi_{\mathscr{U}}T\pi_{\mathscr{U}}=T\pi_{\mathscr{U}}.

Conversely, if \pi_{\mathscr{U}}T\pi_{\mathscr{U}}=T\pi_{\mathscr{U}} then for any u\in\mathscr{U} we have that

T(u)=T\pi_{\mathscr{U}}=\pi_{\mathscr{U}}T\pi_{\mathscr{U}}=\pi_{\mathscr{U}}T(u)

but this is true if and only if T(u)\in\mathscr{U}. It follows from the arbitrariness of u that \mathscr{U} is invariant under T. \blacksquare

Our next theorem gives us a similar formulation for reducibility of a linear transformation in terms of projections. Namely:

Theorem: Let \mathscr{V} be a finite dimensional F-space and \mathscr{V}=\mathscr{U}\oplus\mathscr{W} with \mathscr{U},\mathscr{W}\leqslant \mathscr{V}. Then, \mathscr{U} and \mathscr{W} reduce T if and only if T\pi_{\mathscr{U}}=\pi_{\mathscr{U}}T where \pi_{\mathscr{U}} is the projection on \mathscr{U} and along \mathscr{W}.

Proof: Suppose first that \mathscr{U} and \mathscr{W} reduce T then we have that u+w\in\mathscr{V} that

\begin{aligned}\pi_{\mathscr{U}}T(u+w) &=\pi_{\mathscr{U}}\left(T(u)+T(w)\right)\\ &=\pi_{\mathscr{U}}T(u)+\pi_{\mathscr{U}}T(v)\\ &=T(u)\\ &=T\pi_{\mathscr{U}}(u+w)\end{aligned}

Conversely, suppose that \pi_{\mathscr{U}}T=T\pi_{\mathscr{U}} it’s evident from the last theorem that \mathscr{U} is invariant under T since

\pi_{\mathscr{U}}T\pi_{\mathscr{U}}=T\pi_{\mathscr{U}}\pi_{\mathscr{U}}=T\pi_{\mathscr{U}}

 

But, \mathscr{W} is also invariant under T. Indeed, let w\in\mathscr{W}. Then,

\begin{aligned}\pi_{\mathscr{W}}(T(w)) &= \left(\mathbf{1}-\pi_{\mathscr{U}}(T(w))\right)\\ &= T(w)-\pi_{\mathscr{U}}T(w)\\ &= T(w)-T(\pi_{\mathscr{U}}(w))\\ &= T(w)-T(\bold{0})\\ &= T(w)\end{aligned}

from where it follows that T(w)\in\mathscr{W}. Since w was arbitrary the conclusion follows. \blacksquare

The proof of \mathbf{(3)} is continued here.

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 | ,

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