Linear Algebra · Advanced · 5 min

Gram–Schmidt

Remove the part of a vector that points along another. What remains is perpendicular.

WHAT HAPPENS IF…Does subtracting half the projection make the remainder perpendicular?

w=v-\frac{v\cdot u}{u\cdot u}u,\qquad w\cdot u=0

Current dot product with u3.48

Final perpendicular w(-0.208, 0.416)

‖w‖0.465

Green is u; gray is the original v; gold is its projection onto u; blue is the remaining vector. If w = 0 the inputs are dependent and cannot produce two orthonormal basis vectors.

Preparing your experiment…

Connect the picture to the idea

\mathbf w=\mathbf v-\frac{\mathbf v\cdot\mathbf u}{\mathbf u\cdot\mathbf u}\mathbf u

For u≠0, dot the expression for w with u. It gives v·u−(v·u)/(u·u)(u·u)=0. Thus w is orthogonal to u. If w≠0, divide u and w by their respective lengths to obtain an orthonormal pair with the same span as the original vectors.

A detail that matters

If the original vectors are dependent, w=0 and cannot be normalized. Zero is orthogonal to every vector but cannot be a basis vector.

A bridge to another ideaFirst, find the part pointing along u ↗Orthogonalization subtracts precisely that projection.

Keep the connection going

One idea leads to another.

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Gram–Schmidt

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