How to use the Gram–Schmidt process to find an orthonormal basis

Gram–Schmidt keeps the span of a basis while turning its vectors into perpendicular unit vectors. At each step, we subtract the parts that point along the directions we already have. Then we divide by the remaining vector’s length.

Normalize the first vector

q_{1} = \frac{v_{1}}{‖ v_{1} ‖}

Remove the first direction from the second vector

\begin{array}{l} w_{2} = v_{2} - (v_{2} \cdot q_{1}) q_{1} \\ q_{2} = \frac{w_{2}}{‖ w_{2} ‖} \end{array}

For the next vector, subtract its projections onto both earlier unit vectors before normalizing. Continue this way through an independent basis.

Apply the process to a basis for a plane

Let $v_{1} = (1, 2, 0)$ and $v_{2} = (- 2, 1, - 5)$ . The first length is $\sqrt{5}$ , so $q_{1} = \frac{1}{\sqrt{5}} (1, 2, 0)$ . Here $v_{2} \cdot q_{1} = 0$ , so $w_{2} = v_{2}$ . Its length is $\sqrt{30}$ , giving

q_{2} = \frac{1}{\sqrt{30}} (- 2, 1, - 5)

These unit vectors are perpendicular and span the same plane as the original pair.

See a nonzero projection being removed

For $v_{1} = (1, 0)$ and $v_{2} = (1, 1)$ , the first unit vector is $q_{1} = (1, 0)$ . The dot product $v_{2} \cdot q_{1}$ is 1, so $w_{2} = (1, 1) - (1, 0) = (0, 1)$ . It already has length 1, giving $q_{2} = (0, 1)$ .

Learn mathKrista KingAugust 11, 2021math, learn online, online course, online math

How to use the Gram–Schmidt process to find an orthonormal basis

Normalize the first vector

Remove the first direction from the second vector

Apply the process to a basis for a plane

See a nonzero projection being removed

Courses