Orthogonal Projection

linear-algebra

Definition

Orthogonal Projection

Let $P:\mathcal{H}\to \mathcal{H}$ be a linear operator on an inner product space $\mathcal{H}$. The operator is an orthogonal projection if it is Hermitian and idempotent, that is,

$$P=P^{†}\text{and}{P}^2=P$$

Equivalently, $P$ projects each vector onto a subspace along its orthogonal complement.

A projection keeps one subspace and removes the orthogonal complement.

The Kept Subspace

The subspace that is kept is the image of the projection:

$$U=\mathrm{im}(P)=\{Px\mid x\in \mathcal{H}\}.$$

This means $U$ is the set of all possible outputs of $P$. If $u\in U$, then $u=Px$ for some $x\in \mathcal{H}$. Since $P$ is idempotent,

$$Pu=P(Px)=P^{2}x=Px=u.$$

So every vector in $U$ is fixed by $P$. This is the sense in which $P$ keeps $U$.

For an orthogonal projection, the removed directions are exactly perpendicular to $U$. Those directions form the orthogonal complement $U^{\perp }$.

Image and Kernel

Let $U=\mathrm{im}(P)$. An orthogonal projection splits the space into two perpendicular parts:

$$\mathcal{H}=U\oplus U^{\perp }.$$

Thus every $x\in \mathcal{H}$ decomposes uniquely as

$$x=u+w$$

with $u\in U$ and $w\in U^{\perp }$. The projection acts by

$$P(u+w)=u.$$

So the sentence means:

$$Pu=u\text{and}Pw=0.$$

The part in $U$ is fixed. The part in $U^{\perp }$ is sent to the zero vector.

Equivalently, every $x\in \mathcal{H}$ decomposes as

$$\begin{array}{rl}x& =Px+x-Px\\ & =Px+(I-P)x\end{array}$$

where $Px\in \mathrm{im}(P)$ and $(I-P)x\in \ker (P)$. Since $P$ is Hermitian, the kernel is the orthogonal complement of the image:

$$\ker (P)=\mathrm{im}(P{)}^{\perp }.$$

Thus $P$ keeps the component in its image (eigenvalue $1$) and removes the component in the orthogonal complement (eigenvalue $0$).

Eigenvalues

Eigenvalues of a Projection

Projections have eigenvalues $0$ and $1$.

Eigenvalues of a Projection

Let $|v⟩$ be a non-zero vector and $P$ be a projection

$$P^2=P.$$

Suppose

$$P|v⟩=\lambda |v⟩.$$

Apply $P$ again:

$$P^2|v⟩=P(\lambda |v⟩)=\lambda P|v⟩={\lambda }^2|v⟩.$$

Since $P^2=P$,

$$P^2|v⟩=P|v⟩=\lambda |v⟩.$$

Therefore:

$${\lambda }^2|v⟩=|v⟩.$$

Since $|v⟩\ne 0$,

$${\lambda }^2=\lambda .$$

So:

$$\lambda (\lambda -1)=0$$

Therefore:

$$\lambda =0\text{or}\lambda =1.$$

Example

Projection onto one basis state

Let $|e_i⟩$ be a normalised vector in an inner product space, so

$$⟨e_i\mid e_i⟩=1.$$

The operator

$$P_i=|e_i⟩⟨e_i|$$

is an orthogonal projection onto the one-dimensional subspace $\mathrm{span}(|e_i⟩)$.

Acting on a vector $|\psi ⟩$, it gives

$$P_i|\psi ⟩=|e_i⟩⟨e_i\mid \psi ⟩=⟨e_i\mid \psi ⟩|e_i⟩.$$

Thus $P_i$ keeps the component of $|\psi ⟩$ in the $|e_i⟩$ direction and removes the orthogonal component.

It is idempotent:

$$P_i^2=(|e_i⟩⟨e_i|)(|e_i⟩⟨e_i|)=|e_i⟩⟨e_i\mid e_i⟩⟨e_i|=|e_i⟩⟨e_i|=P_i.$$

It is also Hermitian:

$$P_i^{†}=(|e_i⟩⟨e_i|{)}^{†}=|e_i⟩⟨e_i|=P_i.$$

Therefore $P_i$ is an orthogonal projection.

Example

The operator $T:{\mathbb{C}}^2\to {\mathbb{C}}^2$, given by the matrix:

$$A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$$

is an orthogonal projection.

In particular, $T$ corresponds to the projection to the first component:

$$T\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}{x}_1\\ 0\end{array}\right)$$