Carry out Gram-Schmidt orthogonalization on a basis in $\mathbb{R}^3$

Primary Tool

Math Workspace (CAS)

CAS is useful here because every projection coefficient has to be exact. The embed shows the vector subtraction and then checks the dot products, so the method is not reduced to memorizing a formula.

Problem

Apply Gram-Schmidt to $v_{1} = 110$ , $v_{2} = 101$ , and $v_{3} = 011$ .

The goal is not to change the subspace. These vectors already form a basis of $R^{3}$ . The goal is to produce a new basis for the same space whose vectors meet at right angles.

Projection idea

Set $u_{1} = v_{1}$ . To build $u_{2}$ , remove from $v_{2}$ the component that points along $u_{1}$ : $u_{2} = v_{2} - \frac{v _{2} \cdot u _{1}}{u _{1} \cdot u _{1}} u_{1}$ .

For $u_{3}$ , remove the components pointing along both $u_{1}$ and $u_{2}$ . That is the only new ingredient in three dimensions.

Step-by-step walkthrough

The fractions are not the point. The point is that each new vector is cleaned of every direction already chosen.

1Start with $u_{1} = v_{1} = 110$ .
2Compute $u_{2} = v_{2} - \frac{v _{2} \cdot u _{1}}{u _{1} \cdot u _{1}} u_{1} = 1/2 - 1/2 1$ .
3Check $u_{1} \cdot u_{2} = 0$ , so the first two directions are orthogonal.
4Compute $u_{3} = v_{3} - \frac{v _{3} \cdot u _{1}}{u _{1} \cdot u _{1}} u_{1} - \frac{v _{3} \cdot u _{2}}{u _{2} \cdot u _{2}} u_{2} = - 2/3 2/3 2/3$ .
5Check $u_{1} \cdot u_{3} = 0$ and $u_{2} \cdot u_{3} = 0$ .
6The result is an orthogonal basis. Divide each $u_{i}$ by its length if you need an orthonormal basis.

Common Pitfall

Gram-Schmidt does not merely normalize vectors. Normalizing changes lengths; Gram-Schmidt first changes directions by removing projections so the vectors become orthogonal.

Try a Variation

Normalize the three $u_{i}$ vectors from the example. What are their lengths, and what orthonormal basis do you get?

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