Projectors

A projector is a linear operator $P$ that is idempotent:

$$P^2=P.$$

Applying it once separates a vector into a retained component and a discarded component; applying it again changes nothing. This simple algebraic condition organizes subspace decompositions, constrained approximation, spectral resolutions, and ideal quantum alternatives.

This page develops the linear algebra. The interpretation of projectors as yes–no quantum propositions belongs to Projectors in Core Formalism, and probabilities and conditional states belong to Projective Measurement.

The Range–Kernel Decomposition

Let $P:V\to V$ satisfy $P^2=P$. Its two distinguished subspaces are

$$\begin{aligned} \operatorname{Ran}P &=\{P v:v\in V\},\\ \ker P &=\{v\in V:Pv=0\}. \end{aligned}$$

The operator is the identity on its range. Indeed, if $u=Pv$, then $Pu=P^2v=Pv=u$. Every $v\in V$ has the decomposition

$$\begin{aligned} v &=Pv+(I-P)v,\\ Pv &\in\operatorname{Ran}P,\\ (I-P)v &\in\ker P. \end{aligned}$$

The intersection of these subspaces is trivial: if $w=Pv$ and $Pw=0$, then $w=Pw=0$. Consequently,

$$V=\operatorname{Ran}P\oplus\ker P.$$

Conversely, a direct-sum decomposition $V=M\oplus N$ defines a unique projector onto $M$ along $N$:

$$P(m+n)=m, \qquad m\in M,\quad n\in N.$$

Thus a projector remembers two subspaces, not merely its range. Different choices of the complementary subspace $N$ give different projectors with the same range. See Direct Sums for the underlying decomposition.

Spectrum and Canonical Form

If $Pv=\lambda v$ for a nonzero vector $v$, idempotence gives

$$\lambda^2v=P^2v=Pv=\lambda v.$$

Therefore every eigenvalue is either $0$ or $1$. More strongly, the minimal polynomial of $P$ divides $x(x-1)$. Because this polynomial has distinct roots, every projector in finite dimensions is diagonalizable, even when it is not Hermitian. In a basis adapted to the range–kernel decomposition,

$$[P]= \begin{pmatrix} I_r & 0\\ 0 & 0 \end{pmatrix}, \qquad r=\dim\operatorname{Ran}P.$$

It follows that

$$\operatorname{tr}P =\operatorname{rank}P =r.$$

The equality between trace and rank is basis independent. It is also a useful consistency check for an exact finite-dimensional projector, although floating-point traces need not be exact integers.

Orthogonal Projectors

On an inner-product space, an orthogonal projector satisfies

$$P^2=P, \qquad P^\dagger=P.$$

Self-adjointness forces the discarded and retained subspaces to be orthogonal. For $u=Px$ and $w\in\ker P$,

$$\langle w,u\rangle =\langle w,Px\rangle =\langle Pw,x\rangle =0.$$

Hence

$$\begin{aligned} \ker P &=(\operatorname{Ran}P)^\perp,\\ V &=\operatorname{Ran}P \mathbin{\oplus_\perp}\ker P. \end{aligned}$$

This is an equivalence: an idempotent is orthogonal precisely when its range is perpendicular to its kernel. There is therefore exactly one orthogonal projector onto a specified subspace $M$.

The complementary operator

$$P^\perp=I-P$$

is the orthogonal projector onto $M^\perp$. For every $v$,

$$\begin{aligned} \langle v,Pv\rangle &=\lVert Pv\rVert^2\ge 0,\\ \langle v,(I-P)v\rangle &=\lVert(I-P)v\rVert^2\ge 0. \end{aligned}$$

In operator order this is written $0\le P\le I$. Orthogonal projection cannot increase a norm:

$$\lVert Pv\rVert\le\lVert v\rVert.$$

If $P\ne0$, its operator norm is exactly $\lVert P\rVert_{\mathrm{op}}=1$, because equality holds for every nonzero vector in its range.

Constructing Orthogonal Projectors

Let $\{u_1,\ldots,u_r\}$ be an orthonormal basis for a subspace $M$. The orthogonal projector onto $M$ is

$$P_M =\sum_{a=1}^{r} \lvert u_a\rangle\langle u_a\rvert.$$

This formula is independent of the orthonormal basis chosen inside $M$. For a one-dimensional subspace spanned by a nonzero, not necessarily normalized vector $u$,

$$P_u =\frac{\lvert u\rangle\langle u\rvert} {\langle u\vert u\rangle}.$$

Omitting the denominator is a common source of a non-idempotent outer product.

There is also a useful matrix formula. Put linearly independent spanning vectors into the columns of an $n\times r$ matrix $B$. The Gram matrix $B^\dagger B$ is invertible, and

$$P_M =B(B^\dagger B)^{-1}B^\dagger.$$

For any $x$, the residual is orthogonal to every column of $B$:

$$B^\dagger(x-P_Mx)=0.$$

The formula is exact mathematics, but explicitly forming $(B^\dagger B)^{-1}$ is usually a poor numerical algorithm. A stable QR factorization $B=QR$ gives

$$P_M=QQ^\dagger,$$

where the columns of $Q$ are orthonormal. A singular-value decomposition is preferable when the spanning vectors may be nearly linearly dependent.

Best Approximation

Orthogonal projection gives the nearest vector in a subspace. Let $M$ be a subspace and let $P_M$ be its orthogonal projector. For any $m\in M$,

$$v-m =\bigl(v-P_Mv\bigr) +\bigl(P_Mv-m\bigr).$$

The two terms on the right are orthogonal: the first lies in $M^\perp$ and the second lies in $M$. The Pythagorean theorem therefore gives

$$\begin{aligned} \lVert v-m\rVert^2 &=\lVert v-P_Mv\rVert^2\\ &\quad+\lVert P_Mv-m\rVert^2. \end{aligned}$$

The first term is independent of $m$, and the second vanishes only for $m=P_Mv$. Thus

$$P_Mv =\underset{m\in M}{\operatorname{arg\,min}}\, \lVert v-m\rVert.$$

This variational characterization is often more useful than the equation $P^2=P$. It underlies least-squares methods, basis truncation, and many variational approximations in quantum mechanics.

Worked Example: A Complex Line

Consider the line in $\mathbb C^3$ spanned by

$$u= \begin{pmatrix} 1\\ 1\\ i \end{pmatrix}, \qquad u^\dagger u=3.$$

The orthogonal projector onto this line is

$$P_u =\frac13 \begin{pmatrix} 1 & 1 & -i\\ 1 & 1 & -i\\ i & i & 1 \end{pmatrix}.$$

The conjugation in the row $u^\dagger$ is essential. The matrix is Hermitian, has trace one, and satisfies $P_u^2=P_u$. Acting on the first coordinate vector gives

$$P_u \begin{pmatrix} 1\\0\\0 \end{pmatrix} =\frac13 \begin{pmatrix} 1\\1\\i \end{pmatrix}.$$

The residual is orthogonal to $u$, as the best-approximation theorem requires.

Oblique Projectors

Idempotence alone does not imply orthogonality. For any $a\in\mathbb C$,

$$P_a= \begin{pmatrix} 1 & a\\ 0 & 0 \end{pmatrix}$$

satisfies $P_a^2=P_a$. Its range and kernel are

$$\begin{aligned} \operatorname{Ran}P_a &=\operatorname{span}\{(1,0)^T\},\\ \ker P_a &=\operatorname{span}\{(-a,1)^T\}. \end{aligned}$$

Unless $a=0$, these subspaces are not orthogonal and $P_a^\dagger\ne P_a$. The operator projects onto the first coordinate axis along a tilted direction. Its norm is

$$\lVert P_a\rVert_{\mathrm{op}} =\sqrt{1+\lvert a\rvert^2},$$

so an oblique projection can amplify vectors substantially. This is one reason nearly parallel complementary subspaces lead to ill-conditioned decompositions.

Resolutions of the Identity

A finite family of mutually orthogonal projectors satisfies

$$P_aP_b=\delta_{ab}P_a.$$

If the ranges span the whole space, the family is a resolution of the identity:

$$\sum_a P_a=I.$$

Every vector then decomposes into orthogonal components,

$$\lvert\psi\rangle =\sum_a P_a\lvert\psi\rangle, \qquad \lVert\psi\rVert^2 =\sum_a\lVert P_a\psi\rVert^2.$$

Rank-one projectors from an orthonormal basis are the simplest example, but the $P_a$ may have ranks greater than one. Higher-rank projectors encode degenerate subspaces without selecting a preferred basis inside them. Their role in decomposing Hermitian operators is developed in Spectral Decomposition.

Commuting Projectors

Let $P$ and $Q$ be orthogonal projectors. If they commute, then

$$R=PQ=QP$$

is an orthogonal projector. Its range is the intersection

$$\operatorname{Ran}R =\operatorname{Ran}P\cap\operatorname{Ran}Q.$$

Commutation is essential: $(PQ)^\dagger=QP$, so the product can be orthogonal only when $PQ=QP$. For commuting projectors,

$$P+Q-PQ$$

is the orthogonal projector onto $\operatorname{Ran}P+\operatorname{Ran}Q$. These formulas are the subspace analogs of intersection and union for compatible yes–no alternatives.

Bridge to Quantum Measurements

For a self-adjoint matrix with distinct eigenvalues,

$$A=\sum_a aP_a, \qquad P_aP_b=\delta_{ab}P_a, \qquad \sum_aP_a=I.$$

Functional calculus then gives $f(A)=\sum_a f(a)P_a$. When the distinct eigenvalues are known, one can recover a projector algebraically:

$$P_a=\prod_{b\ne a}\frac{A-bI}{a-b}.$$

For a state $\rho$, a yes–no projector has probability $p=\operatorname{Tr}(\rho P)$ and variance

$$(\Delta_\rho P)^2=p(1-p).$$

These Born and state-update rules are physical postulates layered on top of the projector algebra.

If two orthogonal projectors do not commute, $PQ$ is generally neither Hermitian nor idempotent. The sandwich $PQP$ is positive and obeys

$$0\le PQP\le P,$$

but is not generally a projector. Sequential quantum measurements therefore cannot be treated as ordinary set intersection unless the relevant projectors commute.

Unitary conjugation preserves orthogonal projection:

$$P\longmapsto UPU^\dagger, \qquad \operatorname{Ran}(UPU^\dagger)=U(\operatorname{Ran}P).$$

A projective measurement is specified by an orthogonal resolution of the identity $\{P_a\}$. The projector algebra ensures that the outcome subspaces are exclusive and complete. The physical postulates then assign outcome probabilities and conditional states to those subspaces.

Those postulates are not consequences of idempotence. See Projective Measurement for the Born probabilities, repeatability, degeneracy, and Lüders state update. Generalized measurement effects are positive operators and need not be projectors.

Projectors also have a distinct approximation-theory role. Complementary subspaces can separate retained and eliminated sectors of a Hamiltonian; see Projection Methods.

Infinite-Dimensional Qualification

In a Hilbert space, every closed subspace $M$ has a unique bounded orthogonal projector $P_M$. Closedness matters: the nearest-point limit must remain in $M$. A nonclosed proper subspace has no bounded orthogonal projector whose range is exactly that subspace.

For any bounded idempotent $P$, both $\ker P$ and $\operatorname{Ran}P$ are closed because

$$\begin{aligned} \ker P &=P^{-1}(\{0\}),\\ \operatorname{Ran}P &=\ker(I-P). \end{aligned}$$

Infinite resolutions of the identity require a convergence statement. For a countable orthogonal family, the partial sums converge strongly when

$$\sum_a P_a\psi=\psi$$

for every vector $\psi$; norm convergence of the operators is generally too strong. The spectral theorem extends this idea from sums to projection-valued measures.

Numerical Diagnostics

For a computed matrix $\widetilde P$, inspect both

$$\epsilon_{\mathrm{id}} =\lVert\widetilde P^2-\widetilde P\rVert, \qquad \epsilon_{\mathrm{H}} =\lVert\widetilde P^\dagger-\widetilde P\rVert.$$

The first tests idempotence; the second distinguishes an approximate orthogonal projector from an approximate oblique one. Eigenvalues should cluster near $0$ and $1$, and the trace should be close to the intended rank. These checks must be interpreted relative to matrix size, norm convention, and floating-point precision.

When constructing a projector from spanning vectors:

• orthonormalize with QR when the numerical rank is clear;
• use an SVD and an explicit singular-value tolerance when rank is uncertain;
• avoid explicit matrix inversion;
• check the residual orthogonality $B^\dagger(x-\widetilde Px)$;
• do not repair a poor basis by merely rounding eigenvalues.

See Matrix Diagonalization for related conditioning and residual checks.

Common Mistakes

• Assuming $P^2=P$ automatically implies $P^\dagger=P$.
• Saying “the projector onto $M$” without specifying the complementary direction unless orthogonal projection is intended.
• Using $\lvert u\rangle\langle u\rvert$ for an unnormalized vector.
• Forgetting complex conjugation when forming an outer product.
• Applying the best-approximation theorem to an oblique projector.
• Multiplying orthogonal projectors and assuming the product is a projector without checking commutation.
• Treating every positive measurement effect as a projector.
• Expecting an infinite resolution of identity to converge in operator norm.

Exercises

1. Let

   $$P= \begin{pmatrix} 1 & 2\\ 0 & 0 \end{pmatrix}.$$

   Verify idempotence, find the range and kernel, and show that $P$ is not an orthogonal projector. Compute its operator norm.

Solution

Direct multiplication gives

$$P^2= \begin{pmatrix} 1 & 2\\ 0 & 0 \end{pmatrix} =P.$$

The output $P(x,y)^T=(x+2y,0)^T$ lies on the first coordinate axis, so

$$\operatorname{Ran}P =\operatorname{span}\{(1,0)^T\}.$$

The kernel condition $x+2y=0$ gives

$$\ker P =\operatorname{span}\{(-2,1)^T\}.$$

These lines are not orthogonal, and

$$P^\dagger= \begin{pmatrix} 1 & 0\\ 2 & 0 \end{pmatrix} \ne P.$$

Finally,

$$PP^\dagger= \begin{pmatrix} 5 & 0\\ 0 & 0 \end{pmatrix},$$

so the largest singular value is $\sqrt5$ and $\lVert P\rVert_{\mathrm{op}}=\sqrt5$.

2. Let $M\subset\mathbb R^3$ be spanned by $b_1=(1,0,1)^T$ and $b_2=(0,1,1)^T$. Find the orthogonal projector onto $M$ without first orthonormalizing $b_1,b_2$.

Solution

A normal vector to both spanning vectors is $n=(1,1,-1)^T$. Therefore the projector onto $M=n^\perp$ is the complement of the rank-one projector onto $n$:

$$\begin{aligned} P_M &=I-\frac{nn^T}{n^Tn}\\ &=\frac13 \begin{pmatrix} 2 & -1 & 1\\ -1 & 2 & 1\\ 1 & 1 & 2 \end{pmatrix}. \end{aligned}$$

The matrix is symmetric, has trace two, fixes both $b_1$ and $b_2$, and annihilates $n$. These facts also verify that its range is $M$.

3. Prove the best-approximation theorem directly from the normal equation $P_M^\dagger=P_M$: show that $P_Mv$ is the unique minimizer of $\lVert v-m\rVert$ over $m\in M$.

Solution

Because $P_M$ fixes $M$, for any $m\in M$,

$$P_M(P_Mv-m)=P_Mv-m.$$

The residual obeys

$$P_M(v-P_Mv)=0,$$

so it belongs to $\ker P_M=M^\perp$. Hence $v-P_Mv$ is orthogonal to $P_Mv-m$, and

$$\begin{aligned} \lVert v-m\rVert^2 &=\lVert v-P_Mv\rVert^2\\ &\quad+\lVert P_Mv-m\rVert^2. \end{aligned}$$

The right side is minimized exactly when the second term vanishes, which occurs only for $m=P_Mv$.

4. Let $P$ and $Q$ be commuting orthogonal projectors. Prove that $PQ$ is the orthogonal projector onto $\operatorname{Ran}P\cap\operatorname{Ran}Q$. Then show that $P+Q-PQ$ is an orthogonal projector.

Solution

Commutation and idempotence imply

$$\begin{aligned} (PQ)^2 &=P^2Q^2=PQ,\\ (PQ)^\dagger &=QP=PQ. \end{aligned}$$

Thus $PQ$ is an orthogonal projector. If $x=PQy$, then $Px=x$ and $Qx=x$, so its range lies in the intersection. Conversely, if $Px=Qx=x$, then $PQx=x$, so every vector in the intersection belongs to the range.

Set $S=P+Q-PQ$. Since all factors commute and are Hermitian, $S^\dagger=S$. Expanding and using $P^2=P$, $Q^2=Q$, and $(PQ)^2=PQ$ gives

$$S^2=P+Q-PQ=S.$$

Therefore $S$ is an orthogonal projector. It fixes both $\operatorname{Ran}P$ and $\operatorname{Ran}Q$, and its image lies in their sum, so its range is $\operatorname{Ran}P+\operatorname{Ran}Q$.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• G. H. Golub and C. F. Van Loan, Matrix Computations, 4th ed., Johns Hopkins University Press, 2013.
• P. R. Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, 2nd ed., Chelsea, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.