Abstract Linear Algebra 16 | Gramian Matrix [dark version]

Orthogonal projection onto a finite-dimensional subspace can be computed by solving a linear system built from inner products—using a special matrix called the Gramian (often written as G(B,B)). The core idea is to express the projection p of a vector X onto a K-dimensional subspace U as a linear combination of a chosen basis B = {B1, …, BK}. The coefficients in that combination aren’t guessed; they’re determined by enforcing orthogonality between the residual X − p and every basis direction in U.

Start with the standard decomposition X = p + n, where p lies in U and n lies in the orthogonal complement U⊥. Because n is orthogonal to all vectors in U, the inner product ⟨Bj, X⟩ equals ⟨Bj, p⟩ for each basis vector Bj. Writing p = Σ(i=1..K) Λi Bi and using linearity of the inner product in the second argument turns each condition ⟨Bj, X⟩ = ⟨Bj, p⟩ into a linear equation in the unknown scalars Λ1, …, ΛK.

Collecting these K equations produces a matrix-vector system. The Gramian matrix G(B,B) has entries Gij = ⟨Bi, Bj⟩, so each row corresponds to one basis vector Bj tested against the linear combination defining p. The unknown coefficients Λ form a column vector, and the right-hand side is the vector of inner products ⟨Bj, X⟩. Solving G(B,B)Λ = (⟨B1, X⟩, …, ⟨BK, X⟩)ᵀ yields the coefficients Λi, and therefore the orthogonal projection p.

A key question is whether this system has a unique solution. Uniqueness comes down to whether the Gramian matrix is invertible. For square G(B,B), invertibility is equivalent to having a trivial kernel. The transcript shows this by assuming a coefficient vector β in the kernel and forming the linear combination Y = Σ(i=1..K) βi Bi. Kernel membership implies ⟨Bj, Y⟩ = 0 for every basis vector Bj, which forces Y to be orthogonal to all of U. But Y also lies in U by construction, so Y must lie in U ∩ U⊥. That intersection contains only the zero vector, so Y = 0, and because the basis vectors are linearly independent, all βi must be zero. Hence the kernel is trivial and G(B,B) is invertible, guaranteeing a unique orthogonal projection for any K-dimensional subspace in an inner-product space.

An example in R³ illustrates the method. Take U as the yz-plane, spanned by two vectors (the transcript uses basis vectors whose Gramian entries become 2, 1, and 1). The Gramian is a 2×2 matrix built from inner products of the basis with itself, while the right-hand side uses inner products with X. Solving the resulting 2×2 linear system produces coefficients that reconstruct p as a combination of the basis vectors; the resulting projection removes the x-component, matching the geometric expectation that projecting onto the yz-plane kills the x direction.