The all-inclusive span

Ayoub B. Ayoub asked his students to check whether the vector (2, 4, 1, 5) belonged to the span of {(1, 1, 1, 1), (1, 0, 1, −2), (1, −3, 1, 1)} in R4. A student who noticed that the three vectors are pairwise orthogonal, used the dot product to calculate the scalar factors x, y, z in the linear combination

(2, 4, 1, 5) = x(1, 1, 1, 1) + y(1, 0, 1, −2) + z(1, −3, 1, 1).

Taking the dot product of both sides with (1, 1, 1, 1) yields x = 3. Similarly, he got y = −7/6 and z = −1/3. Then the student claimed that the vector belonged to the subspace. □

However, this is not true, as one can see by noting that the first and third components of the three vectors are the same. Equating components gives an inconsistent system of four equations for the three unknowns x, y, z. The method actually delivers the orthogonal projection of (2, 4, 1, 5) on the span of the three given vectors.

Rotating a vector

Consider the set A of three-dimensional real vectors (a, b, c) for which ab + bc + ca = 1. The set A is not empty, since it contains the vector (1, 1, 0). However, one can interpret the set A as consisting of those vectors (a, b, c) whose scalar product with the vector (b, c, a) is equal to 1.