Characterization of Projections

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be an idempotent operator.
Then the following are equivalent:
:$(1): \quad A$ is a projection in the Hilbert space sense
:$(2): \quad A$ is the orthogonal projection onto $\Img A$
:$(3): \quad \norm A = 1$, where $\norm {\, \cdot \,}$ is the norm on bounded linear operators.
:$(4): \quad A$ is Hermitian
:$(5): \quad A$ is normal
:$(6): \quad \forall h \in \HH: \innerprod {A h} h_\HH \ge 0$
:$(7): \quad \forall h \in \HH: \innerprod {A h} h_\HH = \norm {A h}^2$
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $K = \Img A$.
From Range of Idempotent is Kernel of Complementary Idempotent: Corollary 2, $K$ is a closed linear subspace of $\HH$. 
Hence we can define the orthogonal projection $P_K$. 
We just need to show that:
:$A = P_K$
From Unique Point of Minimal Distance to Closed Linear Subspace of Hilbert Space iff Orthogonal, for each $h \in \HH$:
:$P_K h$ is the unique element of $K$ such that $h - P_K h \perp K$
In other words:
:$P_K h$ is the unique element of $K$ such that $h - P_K h \in K^\bot$, where $K^\bot$ is the orthocomplement of $K$. 
Since $A$ is a projection, we have:
:$K^\bot = \paren {\Img A}^\perp = \map \ker A$ 
Now note that we have:
:$\map A {h - A h} = A h - A^2 h = A h - A h = {\mathbf 0}_\HH$
Since $P_K h$ is unique, we have $A h = P_K h$. 
Hence $A = P_K$.
=== $(2)$ implies $(3)$ ===
This is shown in Orthogonal Projection is Bounded.
=== $(3)$ \implies $(1)$ ===
Suppose that $\norm A = 1$. 
We aim to show that:
:$\paren {\map \ker A}^\perp = \Img A$ 
Let $h \in \paren {\map \ker A}^\perp$. 
From Range of Idempotent is Kernel of Complementary Idempotent: Corollary 1, we have $\Img {I - A} = \map \ker A$. 
Hence $h - A h \in \map \ker A$. 
Hence $h \perp h - A h$. 
It follows that: 
:$0 = \innerprod {h - A h} h = \norm h^2 - \innerprod {A h} h$
by the definition of the inner product norm.
Hence, we have:
:$\innerprod {A h} h = \norm h^2$ 
Now, by Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces, we have:
:$\innerprod {A h} h \le \norm {A h} \norm h$ 
From Fundamental Property of Norm on Bounded Linear Transformation, we have:
:$\norm {A h} \norm h \le \norm h^2$ 
Hence we have:
:$\norm h^2 = \innerprod {A h} h \le \norm {A h} \norm h \le \norm h^2$ 
We can therefore conclude that:
:$\norm {A h} \norm h = \norm h^2$ 
so that:
:$\norm {A h} = \norm h$ for all $h \in \paren {\map \ker A}^\perp$
We now have, for $h \in \paren {\map \ker A}^\perp$:
:$\norm {h - A h}^2 = \norm h^2 - 2 \map \Re {\innerprod {A h} h} + \norm {A h}^2$ 
from Square of Inner Product Norm of Sum. 
Now, since: 
:$\innerprod {A h} h = \norm h^2$ 
we have:
:$\norm {h - A h}^2 = 0$
Hence from , we have $\paren {I - A} h = 0$ for all $h \in \paren {\map \ker A}^\perp$. 
Hence we have $\paren {\map \ker A}^\perp \subseteq \ker {I - A}$. 
Using Range of Idempotent is Kernel of Complementary Idempotent: Corollary 1 again we have $\paren {\map \ker A}^\perp \subseteq \Img A$. 
Now let $g \in \Img A = \ker {I - A}$. 
From Direct Sum of Subspace and Orthocomplement, there exists $g_1 \in \map \ker A$ and $g_2 \in \paren {\map \ker A}^\perp$ such that $g = g_1 + g_2$. 
Since $g \in \ker {I - A}$, we have $g = A g$. 
From linearity, we have $A g = A g_1 + A g_2 = A g_2$, since $g_1 \in \map \ker A$. 
We have already shown that $\paren {\map \ker A}^\perp \subseteq \ker {I - A}$, so $E g_2 = g_2$. 
So we have $\Img A \subseteq \paren {\map \ker A}^\perp$. 
Hence we have $\Img A = \paren {\map \ker A}^\perp$.
We conclude that $A$ is a projection.
=== $(2)$ implies $(7)$ ===
Let $h \in \HH$. 
Again from Direct Sum of Subspace and Orthocomplement, we can pick $h_1 \in \Img A$ and $h_2 \in \paren {\Img A}^\perp$ such that:
:$h = h_1 + h_2$
So that, from linearity:
:$A h = A h_1 + A h_2 = A h_1$ 
Since $A$ is a projection, we have:
$\paren {\Img A}^\perp = \map \ker A$
so that $A h_2 = {\mathbf 0}_\HH$. 
We therefore have:
:$\innerprod {A h} h = \innerprod {\map A {h_1 + h_2} } {h_1 + h_2} = \innerprod {A h_1} {h_1 + h_2}$ 
Since the inner product is linear in its first argument we have:
:$\innerprod {A h_1} {h_1 + h_2} = \innerprod {A h_1} {h_1} + \innerprod {A h_1} {h_2}$
Since $h_2 \in \paren {\Img A}^\bot$, we have $\innerprod {A h_1} {h_2} = 0$ and hence:
:$\innerprod {A h} h = \innerprod {A h_1} {h_1} = \norm {h_1}^2$ 
From Fixed Points of Idempotent Operator, since $h_1 \in \Img A$, we have:
:$h_1 = A h_1 = A h$
So we have:
:$\innerprod {A h} h = \norm {A h}^2$ 
from the definition of the inner product norm. 
=== $(7)$ implies $(6)$ ===
Immediate since $\norm {A h}^2 \ge 0$ for each $h \in \HH$. 
=== $(6)$ implies $(1)$ ===
We again wish to show:
:$\paren {\map \ker A}^\perp = \Img A$ 
We first show that:
:$\map \ker A \perp \Img A$ 
Let $h_1 \in \map \ker A$ and $h_2 \in \Img A$. 
From $(6)$, we have:
:$0 \le \innerprod {\map A {h_1 + h_2} } {h_1 + h_2}$
while:
Hence we have:
:$-\norm {h_1}^2 \le \innerprod {h_1} {h_2}$ 
Suppose that:
:$\innerprod {h_1} {h_2} = \alpha \ne 0$ for some $h_1 \in \Img A$ and $h_2 \in \map \ker A$. 
From Kernel of Linear Transformation is Linear Subspace, we have:
:$-2 \alpha^{-1} \norm {h_1}^2 h_2 \in \map \ker A$ 
Hence by swapping $h_2$ for $-2 \alpha^{-1} \norm {h_1}^2 h_2$, we have:
:$-\norm {h_1}^2 \le -2 \alpha^{-1} \norm {h_1}^2 \innerprod {h_1} {h_2}$ 
Hence:
:$-\norm {h_1}^2 \le -2 \norm {h_1}^2$ 
Then we have $\norm {h_1}^2 = 0$ and hence $h_1 = {\mathbf 0}_\HH$ from . 
This contradicts that $\innerprod {h_1} {h_2} \ne 0$. 
So we have $\innerprod {h_1} {h_2} = 0$ for all $h_1 \in \Img A$ and $h_2 \in \map \ker A$. 
Now for general $x \in \HH$, we can again use Direct Sum of Subspace and Orthocomplement to decompose:
:$x = x_1 + x_2$ 
where $x_1 \in \Img A$ and $x_2 \in \map \ker A$. 
Then we have $\innerprod x {h_2} = \innerprod {x_1} {h_2} + \innerprod {x_2} {h_2} = \innerprod {x_2} {h_2}$. 
So if $\innerprod x {h_2} = 0$ for all $h_2 \in \map \ker A$, we have $\innerprod {x_2} {h_2}$ for all $x_2 \in \map \ker A$. 
Then $x_2 \in \map \ker A \cap \paren {\map \ker A}^\perp = \set 0$. 
So $x_2 = {\mathbf 0}_\HH$ and $x \in \Img A$. 
Hence $\paren {\map \ker A}^\perp = \Img A$.
=== $(1)$ implies $(4)$ ===
Let $h, g \in \HH$. 
By Direct Sum of Subspace and Orthocomplement, decompose:
:$h = h_1 + h_2$
and:
:$g = g_1 + g_2$
for $h_1, g_1 \in \Img A$ and $h_2, g_2 \in \map \ker A$. 
Since $A$ is a projection, we have:
:$\map \ker A = \paren {\Img A}^\perp$
Hence we have: 
Similarly, we have:
Hence we have:
:$\innerprod {A^\ast h} g = \innerprod {A h} g$ 
so:
:$\innerprod {A^\ast h - A h} g = 0$ 
Setting $g = A^\ast h - A h$ and applying we have:
:$A^\ast h = A h$ for each $h \in \HH$. 
Hence $A$ is Hermitian.
=== $(4)$ implies $(5)$ ===
Suppose that $A$ is Hermitian. 
We have:
:$A^\ast A = A^2 = A A^\ast$ 
So $A$ is normal.
=== $(5)$ implies $(1)$ ===
From Characterization of Normal Operators, we have:
:$\norm {A h} = \norm {A^\ast h}$ for each $h \in \HH$. 
Hence we have $h \in \map \ker A$ $\norm {A h} = 0$ $\norm {A^\ast h} = 0$. 
That is, from , we have $\map \ker A = \ker {A^\ast}$. 
From Kernel of Linear Transformation is Orthocomplement of Image of Adjoint and Adjoint is Involutive, we have:
:$\ker {A^\ast} = \paren {\Img A}^\perp$ 
Hence:
:$\map \ker A = \paren {\Img A}^\perp$
So $A$ is a projection.
== Sources ==
* $II.3.3$
Category:Linear Transformations on Hilbert Spaces