GraphMath

Projection

From one direction to a subspace: orthogonal and general projection

How do we split a vector into a projected part and a residual?

Projection decomposes a vector into two parts: one part lying in the chosen subspace and the other lying in a complementary residual subspace. This chapter starts with orthogonal projection onto one vector, extends it to the columns of a matrix, compares orthogonal and general projection, develops the main properties of projection matrices and ends with the serial-transformation view of P = U(UᵀU)⁻¹Uᵀ.

Key ideas

Projection separates a vector into a part that we keep and a part that we discard. In orthogonal projection, the discarded part is perpendicular to the target subspace.

  • For one direction u, the vector v decomposes as v = v + v, where v is a scalar multiple of u and v is orthogonal to u
  • For a matrix U with independent columns, the orthogonal projection of v onto col(U) is U(UᵀU)⁻¹Uᵀv
  • General projection also projects onto col(U), but its residual lies in null(Sᵀ), where SᵀU is invertible
  • Orthogonal projection is the special case S = U, so the residual lies in null(Uᵀ), the orthogonal complement of col(U)
  • Every projection matrix is idempotent: P² = P; orthogonal projection matrices are also symmetric
  • The only eigenvalues of a projection matrix are 0 and 1, corresponding to discarded and preserved directions
  • The complementary projection I − P keeps the residual component instead of the projected component
  • P = U(UᵀU)⁻¹Uᵀ can be read as a serial transformation from ℝ³ to ℝ² and back to ℝ³

The chapter moves from vector form to matrix form, compares orthogonal and general projection side by side, develops the algebraic properties of P and ends with the transformation view of the projection matrix.

What makes orthogonal projection special?

Orthogonal projection chooses the projected vector so that the residual is perpendicular to the target subspace. That perpendicularity makes the projected vector unique and gives the standard formula P = U (Uᵀ U)⁻¹ Uᵀ.

Related chapters

Chapter contents

Why are the only projection eigenvalues 0 and 1?

Because P² = P. If Pv = λv, then applying P again gives λ²v = λv, so λ² = λ. Therefore λ is either 0, for directions eliminated by the projection, or 1, for directions preserved by it.

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