Oblique Projections

Choose both a target line and a projection direction to see why an idempotent map need not be orthogonal.

Oblique Projections: Not All Shadows Are Straight

A projection matrix P satisfies P² = P (idempotence). Geometrically, this means applying the transformation a second time does not change the result, because the output is already in the target subspace. However, a projection does not have to be an orthogonal projection (where the matrix is symmetric and the residual x - Px is perpendicular to the range of P).

Defining an Oblique Projection

Instead of just projecting "straight down" onto a subspace, we can project onto a target range spanned by u along a complementary direction spanned by v. For 2D vectors, as long as u and v are linearly independent, they form a basis, meaning det[u v] ≠ 0.

We can calculate the projection matrix P onto span(u) along span(v) directly: P = u * row₁( [u v]^-1 ). This maps the u-basis vector to itself (Pu = u) and the v-basis vector to zero (Pv = 0), matching the geometric intent of projecting along v.

Worked example: Projecting onto y = x/2

Let the target line be spanned by u = (2, 1) and project along the direction v = (1, -1). Project the vector x = (-1, 2).

The matrix [u v] is [[2, 1], [1, -1]]. Its determinant is (2)(-1) - (1)(1) = -3.
The inverse of [u v] is (-1/3) * [[-1, -1], [-1, 2]] = [[1/3, 1/3], [1/3, -2/3]].
The first row of the inverse is [1/3, 1/3].
The projection matrix P = u * row₁ = [[2], [1]] * [1/3, 1/3] = [[2/3, 2/3], [1/3, 1/3]].
The projected vector is Px = [[2/3, 2/3], [1/3, 1/3]] * [-1, 2] = [2/3, 1/3].
The residual is x - Px = [-1, 2] - [2/3, 1/3] = [-5/3, 5/3]. Notice this residual is a scalar multiple (-5/3) of the projection direction v = (1, -1).

Explore oblique vs orthogonal projections

Open Oblique Projections. Choose the target line and direction to see how the residual and the projection matrix behave.

Set Range u1 = 1, Range u2 = 0, Direction v1 = 1, Direction v2 = 1. This projects onto the x-axis along the line y = x.
Set Vector x1 = 2, Vector x2 = 3. The projected vector is (-1, 0). The residual is (3, 3). Notice the residual is parallel to v, but not perpendicular to u. The "Orthogonal projection" status is false.
Change Direction v1 = 1, Direction v2 = -1, and set Range u1 = 1, Range u2 = 1. Observe that u and v are perpendicular.
With Vector x1 = 2, x2 = 0, the product Px is (1, 1). The "Orthogonal projection" status is true.
If you set both u1 = 1, u2 = 1 and v1 = 2, v2 = 2, the status reads "Range and direction do not form a basis" because they point in the same direction.

Assumptions and Limitations

This model assumes the target subspace and direction subspace are both 1-dimensional lines in 2D space. In higher dimensions, an oblique projection involves projecting onto an m-dimensional subspace along an (n-m)-dimensional complementary subspace.

The determinant domain test uses exact bounded integers. Displayed entries use twelve significant digits from unrounded calculations. The displayed P²−P is the actual numerical difference, so floating-point roundoff can appear as a tiny residual; it is not silently replaced by zero. An oblique projection may increase norm: with u=(1,0), v=(1,1) and x=(-1,1), Px=(-2,0) has norm2 greater than √2 and is not the nearest point on the x-axis.

References

Numerical Linear Algebra (Eric Darve): Projections discusses basis-adapted projection, oblique complement, symmetry, and closest-point distinction. It verifies that oblique projection can increase norm, unlike orthogonal projection. The exact coordinate idempotence and residual-direction witnesses independently verify our calculated case.