# Abstract Nonsense

## 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 -

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