Orthogonal Projection Calculator

Math Linear Algebra • Applications and Advanced Linear Algebra (capstone)

Written by STEM Calculators Team

Project a vector \(v\) onto a line or a subspace. For a line spanned by \(u\), \(\mathrm{proj}(v)=\dfrac{v\cdot u}{\lVert u\rVert^2}u\). For a subspace with basis columns \(U\), \(\mathrm{proj}(v)=U(U^{\mathsf T}U)^{-1}U^{\mathsf T}v\). The orthogonal component is \(r=v-\mathrm{proj}(v)\).

Dimension

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Vector \(v\)

The vector you want to project.

Line direction \(u\)

Projection onto the line through the origin spanned by \(u\).

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Results

Projection

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Orthogonal component

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Lengths

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Orthogonality check

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Key matrices / scalars

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Graph

2D only. Shows the “shadow” projection: \(v\), \(\mathrm{proj}(v)\), and \(r=v-\mathrm{proj}(v)\). Zoom with wheel/trackpad, drag to pan, double-click to reset. Play animates the drop to the subspace.

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Step-by-step

Enter inputs and click “Calculate”.

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Theory

Projection onto a line

The orthogonal projection of a vector \(v\) onto a line through the origin spanned by a nonzero vector \(u\) is the “shadow” of \(v\) on that line. The formula comes from scaling \(u\) so that the residual is perpendicular to the line: \[ \mathrm{proj}_{\mathrm{span}\{u\}}(v)=\frac{v\cdot u}{u\cdot u}\,u. \] The coefficient \(c=\dfrac{v\cdot u}{\lVert u\rVert^2}\) is chosen so that \(r=v-cu\) satisfies \(r\cdot u=0\).

The orthogonal component \(r=v-\mathrm{proj}(v)\) measures the distance from \(v\) to the line: \[ \text{dist}(v,\mathrm{span}\{u\})=\lVert r\rVert. \]

Projection onto a subspace spanned by a basis

Suppose a subspace \(S\subset\mathbb{R}^n\) is spanned by basis vectors \(u_1,\dots,u_k\). Put them into a matrix \(U=[u_1\;u_2\;\cdots\;u_k]\in\mathbb{R}^{n\times k}\). The projection of \(v\) onto \(S\) is the vector \(p\in S\) that minimizes \(\lVert v-p\rVert\). If we write \(p=U\alpha\), then \(\alpha\) solves the normal equations: \[ (U^{\mathsf T}U)\alpha = U^{\mathsf T}v. \] Once \(\alpha\) is found, the projection is \(p=U\alpha\) and the orthogonal component is \(r=v-p\).

If the columns of \(U\) are orthonormal, then \(U^{\mathsf T}U=I\) and the projection simplifies to \[ \mathrm{proj}_S(v)=UU^{\mathsf T}v. \] For a general (non-orthonormal) basis, the full matrix form is \[ \mathrm{proj}_S(v)=U(U^{\mathsf T}U)^{-1}U^{\mathsf T}v. \]

Least squares connection

The subspace projection is a least squares problem: among all vectors \(U\alpha\) in the span of \(U\), the projection minimizes \(\lVert v-U\alpha\rVert^2\). This is exactly the geometric idea behind least squares regression and many numerical methods in linear algebra. The condition \(U^{\mathsf T}(v-U\alpha)=0\) means the residual is orthogonal to the entire subspace.

Frequently Asked Questions

What does orthogonal projection mean?

It is the closest point in the target line/subspace to the vector v, measured by Euclidean distance. The difference r=v-proj(v) is perpendicular to the target.

Why does subspace projection use (U^T U)α=U^T v?

Writing proj(v)=Uα and minimizing ||v-Uα||^2 yields the normal equations. The residual must be orthogonal to every column of U.

What if U^T U is singular?

That indicates the basis vectors are linearly dependent or nearly dependent. The calculator may use a tiny stabilization to compute a least-squares style projection.

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Frequently Asked Questions

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