Worked examples · Sheet 2

Orthogonal projections onto a vector and onto a subspace

Theory used: Orthogonal projection onto a vector Orthogonal projection onto a subspace Distance from a subspace Dimension of the orthogonal complement Complement of a subspace in Cartesian form

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute the orthogonal projection of $v = (4, 3)$ onto the vector $u = (1, 2)$ and write $v$ as the sum of a vector parallel to $u$ and a vector orthogonal to $u$ .

⟨ v, u ⟩ = 4 + 6 = 10, ⟨ u, u ⟩ = 1 + 4 = 5

pr_{u} (v) = \frac{10}{5} (1, 2) = (2, 4), v - pr_{u} (v) = (4, 3) - (2, 4) = (2, - 1)

(4, 3) = (2, 4) + (2, - 1), ⟨(2, - 1), (1, 2)⟩ = 2 - 2 = 0

The projection

p = (2, 4)

v = (4, 3)

onto the line spanned by

u = (1, 2)

and the orthogonal component

v - p = (2, - 1)

The orthogonal projection of $v$ onto $u$ is the multiple $\frac{⟨ v , u ⟩}{⟨ u , u ⟩} u$ , and the coefficient equals $2$ . Subtracting the projection from $v$ one obtains the orthogonal component $(2, - 1)$ ; its scalar product with $u$ is zero, therefore the decomposition $v = (2, 4) + (2, - 1)$ has the first summand parallel to $u$ and the second orthogonal to $u$ , as required.

Exercise 2

Decompose $v = (2, 3, 1)$ into the sum of a vector parallel to $u = (1, 1, 1)$ and a vector orthogonal to $u$ .

⟨ v, u ⟩ = 2 + 3 + 1 = 6, ⟨ u, u ⟩ = 3

pr_{u} (v) = \frac{6}{3} (1, 1, 1) = (2, 2, 2), v - pr_{u} (v) = (0, 1, - 1)

(2, 3, 1) = (2, 2, 2) + (0, 1, - 1), ⟨(0, 1, - 1), (1, 1, 1)⟩ = 0

The part parallel to $u$ is the orthogonal projection of $v$ onto $u$ , that is, $2 u = (2, 2, 2)$ . The orthogonal part is the difference $v - pr_{u} (v) = (0, 1, - 1)$ , and checking that the scalar product is zero confirms that it is orthogonal to $u$ . The decomposition is unique, because the projection is the only multiple of $u$ that leaves a difference orthogonal to $u$ .

Exercise 3

Let $W = ⟨(1, 1, 1), (1, - 1, 0)⟩ \subseteq R^{3}$ . Compute the orthogonal projection of $v = (4, 0, - 1)$ onto $W$ and the distance of $v$ from $W$ .

w_{1} = (1, 1, 1), w_{2} = (1, - 1, 0), ⟨ w_{1}, w_{2} ⟩ = 1 - 1 + 0 = 0

⟨ v, w_{1} ⟩ = 4 + 0 - 1 = 3, ⟨ w_{1}, w_{1} ⟩ = 3, ⟨ v, w_{2} ⟩ = 4, ⟨ w_{2}, w_{2} ⟩ = 2

P_{W} (v) = \frac{3}{3} (1, 1, 1) + \frac{4}{2} (1, - 1, 0) = (1, 1, 1) + (2, - 2, 0) = (3, - 1, 1)

v - P_{W} (v) = (1, 1, - 2), d (v, W) = ∥ (1, 1, - 2) ∥ = 1 + 1 + 4 = 6

The formula for the projection as a sum of projections onto the generators holds only for an orthogonal basis, therefore the first step checks that $w_{1}$ and $w_{2}$ are orthogonal; their scalar product is zero, so the theorem on orthogonal projection onto a subspace applies directly. One obtains $P_{W} (v) = (3, - 1, 1)$ . As a check, the difference $(1, 1, - 2)$ has zero scalar product both with $w_{1}$ , since $1 + 1 - 2 = 0$ , and with $w_{2}$ , since $1 - 1 + 0 = 0$ . The projection is the point of $W$ closest to $v$ , therefore the distance of $v$ from $W$ is the norm of the difference and equals $6$ .

Exercise 4

Let $W = {(x, y, z) \in R^{3} : x + 2 y + 2 z = 0}$ . Compute the orthogonal projection of $v = (3, 1, 2)$ onto $W$ and the distance of $v$ from $W$ .

W^{⊥} = ⟨ n ⟩, n = (1, 2, 2), ⟨ n, n ⟩ = 9

⟨ v, n ⟩ = 3 + 2 + 4 = 9 ⟹ pr_{n} (v) = \frac{9}{9} (1, 2, 2) = (1, 2, 2)

P_{W} (v) = v - pr_{n} (v) = (3, 1, 2) - (1, 2, 2) = (2, - 1, 0)

d (v, W) = ∥ pr_{n} (v) ∥ = ∥ (1, 2, 2) ∥ = 3

The plane $W$ is given in Cartesian form, therefore its orthogonal complement is spanned by the row of coefficients, that is, by the normal vector $n = (1, 2, 2)$ . Since $R^{3} = W \oplus W^{⊥}$ , the vector $v$ can be written uniquely as the sum of its projection onto $W$ and its projection onto $W^{⊥}$ ; so $P_{W} (v) = v - pr_{n} (v)$ . In this way the computation of an orthogonal basis of the plane is avoided. The result $(2, - 1, 0)$ belongs to $W$ , since $2 - 2 + 0 = 0$ . The distance of $v$ from $W$ is the norm of $v - P_{W} (v) = pr_{n} (v)$ , and equals $3$ .