Orthogonal projection matrix

A matrix represents a linear projection when applying it twice has the same effect as applying it once. For an orthogonal projection with the standard Hermitian inner product on ${\mathbb{C}}^m$, the additional condition is that the conjugate transpose equals the original matrix:Thus $P$ is both idempotent and a Hermitian matrix. Over the real numbers this says $P^2=P=P^T$. Its image of a linear map is the retained linear subspace, while its kernel of a linear map is the orthogonal complement: vectors in the image are unchanged, and vectors in the kernel become zero.

For a matrix $A$ with full column rank, projecting onto its column space givesThe invertibility of $A^{†}A$ follows from linear independence of the columns. If those columns are an orthonormal set, $A^{†}A=I$ and $P=A{A}^{†}$. These formulas and the conditions $P^2=P=P^T$ in the real case are developed in MIT 18.06SC, Projections onto Subspaces, pp. 1–3.

For the line spanned by $e=(1,2{)}^T$,In particular, $Pe=e$ and $P(2,-1{)}^T=0$. The eigenvalues are $1$ on the retained direction and $0$ on the removed direction. This exposes the meaning of an orthogonal projection as retaining selected components, rather than stretching them.

Symmetry refers to the specified inner product. In the real weighted setting, the formula becomes $P_M=A(A^{T}MA{)}^{-1}{A}^{T}M$ and satisfies $P_M^2=P_M$ and $P_M^{T}M=M{P}_M$. For the preceding weighted line example,It is not symmetric under the ordinary transpose, but it is an orthogonal projection for the weighted Euclidean inner product. The matrix criterion $P=P^{†}$ should therefore be used with the standard Hermitian inner product, not silently transferred to a different geometry.

Under orthogonal projection from one plane onto another, a circle of radius $R$ becomes an ellipse with semiaxes $R$ and $R|\cos \alpha |$, where $\alpha$ is the angle between the planes' unit normal vectors. To prove this, choose one in-plane unit vector parallel to the planes' intersection and the other perpendicular to it. The first vector is unchanged by the projection; the second has projected length $|\cos \alpha |$ and remains perpendicular to the first. Parametrizing the circle in that orthonormal basis therefore gives projected coordinates $(R\cos t,R\cos \alpha \sin t)$. Parallel planes have a circular image, and perpendicular planes give the degenerate case of a line segment. An arbitrary center simply projects to the center of the image.

For a nonnull vector $k$ in a space with a nondegenerate bilinear form, $P_{\mu \nu }={\eta }_{\mu \nu }-k_{\mu }{k}_{\nu }/k^2$ projects onto the subspace transverse to $k$. It obeys $P_{\mu \nu }{k}^{\nu }=0$ and $P^2=P$.