Projection Operators

# Projection Operators

We have already seen that if $V$ is a finite-dimensional nonzero vector space over the complex numbers, then $V$ has at least one eigenvalue, however, if $V$ is a finite-dimensional nonzero vector space over the real numbers, then any linear operator $T \in \mathcal L (V)$ need not have an eigenvalue, however, we will subsequently see that this only applies when the dimension of $V$ is even - that is, every nonzero real vector space with odd dimension has an eigenvalue.

Before we look at the proof, we will need to find look at a type of operator on $V$ known as a projection operator. Let $U$ and $W$ be subspaces of $V$ such that $V$ is the direct sum of $U$ and $W$, that is $V = U \oplus W$. Then if $v \in V$, we have that $v$ can be written uniquely as the sum of an element $u \in U$ and an element $w \in W$:

(1)
\begin{align} \quad v = \underbrace{u}_{\in U} + \underbrace{w}_{\in W} \end{align}
 Definition: Let $V$ be a vector space and let $U$ and $W$ be subspaces of $V$ such that $V = U \oplus W$. Then $v$ can be written uniquely as $v = u + w$ where $u \in U$ and $w \in W$. The Projection Operator Onto $U$ is the linear operator $P_{U,W} \in \mathcal L(V)$ defined by $P_{U, W}(v) = u$ for all $v \in V$.

The following proposition verifies that $P_{U, W}$ is indeed a linear operator on $V$.

 Proposition 1: Let $V$ be a vector space and let $U$ and $W$ be subspaces of $V$ such that $V = U \oplus W$ then for all $v \in V$, $v = u + w$ where $u \in U$ and $w \in W$ and thus the transformation $P_{U, W}$ defined by $P_{U, W}(v) = u$ is a linear operator.
• Proof: We first that clearly $P_{U,W} : V \to V$ since for every vector $v \in V$ we have that $P_{U,W} (v) \in V$, so $P_{U,W}$ maps elements in $V$ back to elements in $V$. To show that $P_{U,W}$ is a linear operator, we must show that both the additivity and homogeneity properties hold.
• To show that the additivity property holds, let $v_1, v_2 \in V$. Then both of these vectors can be written uniquely as $v_1 = u_1 + w_1$ and $v_2 = u_2 + w_2$ for $u_1, u_2 \in U$ and $w_1, w_2 \in W$. Now:
(2)
\begin{align} \quad P_{U, W} (v_1 + v_2) = P_{U, W} ((u_1 + u_2) + (w_1 + w_2)) = u_1 + u_2 = P_{U,W}(v_1) + P_{U,W}(v_2) \end{align}
• Therefore the additivity property holds. Now to show that the homogeneity property holds, let $a \in \mathbb{F}$. Thus we have that $a v_1 = au_1 + aw_1$ and so:
(3)
\begin{align} \quad P_{U,W} (av_1) = P_{U,W} (au_1 + aw_1) = au_1 = aP_{U,W}(v_1) \end{align}
• Therefore the homogeneity property holds. Therefore $P_{U, W} \in \mathcal L (V)$. $\blacksquare$

The following simple proposition will tell us that the range of $P_{U,W}$ is precisely $U$ and that the null space of $P_{U,W}$ is precisely $W$.

 Proposition 2: Let $V$ be a vector space and let $U$ and $W$ be subspaces of $V$ such that $V = U \oplus W$. Then $\mathrm{range} ( P_{U,W} ) = U$ and $\mathrm{null} ( P_{U,W}) = W$.
• Proof: We will first show that $\mathrm{range} ( P_{U,W} ) = U$. Let $u \in \mathrm{range} ( P_{U,W} )$. Then there exists a $v \in V$ such that $P_{U,W}(v) = u$. So for $v \in V$ we have that $v = \underbrace{u}_{\in U} + \underbrace{0}_{\in W}$. Thus $u \in U$. Now let $u \in U$. Thus since $U$ is a subspace of $V$ we have that $u \in V$. Since $V = U \oplus W$ we have that $u$ can be uniquely written as $u = \underbrace{u}_{\in U} + 0$. Thus $P_{U,W}(u) = u$ so $u \in \mathrm{range} ( P_{U,W} )$. Thus $\mathrm{range} ( P_{U,W} ) = U$.
• We will now show that $\mathrm{null} ( P_{U,W}) = W$. Let $w \in \mathrm{null} (P_{U, W})$. Then $P_{U,W}(w) = 0$. Thus we have that $w = \underbrace{0}_{\in U} + \underbrace{w}_{\in W}$ so $w \in W$. Now let $w \in W$. We have that $w \in V$ as well since $W$ is a subspace of $V$. Thus since $V = U \oplus W$ we have that $w \in V$ can be uniquely written as $w = \underbrace{0}{\in U} + \underbrace{w}_{\in W}$. So $P_{U,W} (w) = 0$ and hence $w \in \mathrm{null} ( P_{U,W})$. $\blacksquare$
 Proposition 3: Let $V$ be a vector space and let $U$ and $W$ be subspaces of $V$ such that $V = U \oplus W$. Then $P_{U,W} (v) = v$ if and only if $v \in U$.
• Proof: $\Rightarrow$ Suppose that $P_{U,W}(v) = v$. Since $v \in V$ and $V = U \oplus W$, then we have that $v = u + w$ where $u \in U$ and $w \in W$. By how $P_{U,W}$ is defined, we have that $P_{U,W} = u$. Thus $v = u \in U$.
• $\Leftarrow$ Suppose that $v \in U$. Since $U$ is a subspace of $V$, we have that $v \in V$. Since $V = U \oplus W$, then for $v \in V$ we have that $v$ can be written uniquely as $v = u + w$ where $u \in U$ and $w \in W$. Since $v \in U$, we have that $v = u = P_{U,W} (v)$ is the unique way to write $v$. (Note that if $u \neq v$ then $w \neq 0$ and we would have more than one way to express $v$ which would contradict the fact that $V = U \oplus W$). $\blacksquare$