Worked examples · Sheet 3

Orthonormal bases with the Gram-Schmidt process

Theory used: Gram-Schmidt process Gram-Schmidt in computations Orthonormal basis Norm and distance

Exercise 1

Find an orthonormal basis of the subspace $W = ⟨ v_{1}, v_{2} ⟩ \subseteq R^{3}$ , with $v_{1} = (1, 1, 0)$ and $v_{2} = (2, 0, 1)$ .

u_{1} = v_{1} = (1, 1, 0), ⟨ v_{2}, u_{1} ⟩ = 2, ⟨ u_{1}, u_{1} ⟩ = 2

u_{2} = v_{2} - \frac{2}{2} u_{1} = (2, 0, 1) - (1, 1, 0) = (1, - 1, 1)

∥ u_{1} ∥ = 2, ∥ u_{2} ∥ = 3

e_{1} = \frac{1}{2} (1, 1, 0), e_{2} = \frac{1}{3} (1, - 1, 1)

The vectors $v_{1}$ and $v_{2}$ are not proportional, therefore they are linearly independent and the Gram-Schmidt process applies. The first vector stays unchanged. From the second one subtracts its projection onto $u_{1}$ , which is $u_{1}$ itself because the coefficient $\frac{⟨ v _{2} , u _{1} ⟩}{⟨ u _{1} , u _{1} ⟩}$ equals $1$ . The vector $u_{2} = (1, - 1, 1)$ is orthogonal to $u_{1}$ , as confirmed by the scalar product $1 - 1 + 0 = 0$ . Dividing each vector by its norm one obtains the orthonormal basis ${e_{1}, e_{2}}$ of $W$ .

Exercise 2

Apply the Gram-Schmidt process to the basis $v_{1} = (1, 1, 1)$ , $v_{2} = (1, 1, 0)$ , $v_{3} = (1, 0, 0)$ of $R^{3}$ and obtain an orthonormal basis.

u_{1} = (1, 1, 1), ⟨ u_{1}, u_{1} ⟩ = 3

v_{2} - \frac{⟨ v _{2} , u _{1} ⟩}{⟨ u _{1} , u _{1} ⟩} u_{1} = (1, 1, 0) - \frac{2}{3} (1, 1, 1) = (\frac{1}{3}, \frac{1}{3}, - \frac{2}{3}) ⟹ u_{2} = (1, 1, - 2)

⟨ v_{3}, u_{1} ⟩ = 1, ⟨ v_{3}, u_{2} ⟩ = 1, ⟨ u_{2}, u_{2} ⟩ = 6

v_{3} - \frac{1}{3} u_{1} - \frac{1}{6} u_{2} = (1, 0, 0) - (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) - (\frac{1}{6}, \frac{1}{6}, - \frac{1}{3}) = (\frac{1}{2}, - \frac{1}{2}, 0) ⟹ u_{3} = (1, - 1, 0)

e_{1} = \frac{1}{3} (1, 1, 1), e_{2} = \frac{1}{6} (1, 1, - 2), e_{3} = \frac{1}{2} (1, - 1, 0)

At each step one subtracts from the next vector its projections onto the vectors already built. The vector obtained at the second step has fractional components, therefore it is replaced by the multiple $u_{2} = (1, 1, - 2)$ ; the replacement is legitimate because the projection onto a vector does not change if the vector is multiplied by a nonzero number. At the third step one subtracts from $v_{3}$ its projections onto $u_{1}$ and onto $u_{2}$ , and here too one chooses the integer multiple $u_{3} = (1, - 1, 0)$ . The three vectors are pairwise orthogonal, as checked with the scalar products $⟨ u_{1}, u_{2} ⟩ = 0$ , $⟨ u_{1}, u_{3} ⟩ = 0$ , $⟨ u_{2}, u_{3} ⟩ = 0$ . The final normalization gives the orthonormal basis ${e_{1}, e_{2}, e_{3}}$ of $R^{3}$ .

Exercise 3

Find an orthonormal basis of the subspace

W = {(x_{1}, x_{2}, x_{3}, x_{4}) \in R^{4} : x_{1} + x_{2} + x_{3} + x_{4} = 0} .

v_{1} = (1, - 1, 0, 0), v_{2} = (0, 1, - 1, 0), v_{3} = (0, 0, 1, - 1)

u_{1} = v_{1}, v_{2} - \frac{⟨ v _{2} , u _{1} ⟩}{⟨ u _{1} , u _{1} ⟩} u_{1} = (0, 1, - 1, 0) + \frac{1}{2} (1, - 1, 0, 0) = (\frac{1}{2}, \frac{1}{2}, - 1, 0) ⟹ u_{2} = (1, 1, - 2, 0)

⟨ v_{3}, u_{1} ⟩ = 0, ⟨ v_{3}, u_{2} ⟩ = - 2, ⟨ u_{2}, u_{2} ⟩ = 6

v_{3} - \frac{- 2}{6} u_{2} = (0, 0, 1, - 1) + \frac{1}{3} (1, 1, - 2, 0) = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, - 1) ⟹ u_{3} = (1, 1, 1, - 3)

e_{1} = \frac{1}{2} (1, - 1, 0, 0), e_{2} = \frac{1}{6} (1, 1, - 2, 0), e_{3} = \frac{1}{2 3} (1, 1, 1, - 3)

The subspace is described by one homogeneous equation in four unknowns, therefore it has dimension $4 - 1 = 3$ . The vectors $v_{1}$ , $v_{2}$ , $v_{3}$ satisfy the equation and are linearly independent, because the matrix that has them as rows is in echelon form with three pivots; so they form a basis of $W$ . The Gram-Schmidt process applied to this basis gives the orthogonal vectors $u_{1}$ , $u_{2}$ , $u_{3}$ , which still belong to $W$ , since the sum of the components of each is zero. The norms are $2$ , $6$ and $12 = 23$ , and normalization gives the orthonormal basis ${e_{1}, e_{2}, e_{3}}$ of $W$ .