Find an orthonormal basis for the subspace of R4 consisting

Find an orthonormal basis for the subspace of R_4 consisting of all vectors [a b c d] such that a - b - 2c - d - 0

Solution

Solution :

Solving this problem in two steps :

(1) Find a basis (any basis at all) for that subspace.
(2) Apply the Gram-Schmidt process to the basis produced in (1).

To do step (1) it helps to note that a vector (a,b,c,d) satisfies a - b - 2c - d = 0 if and only if the column vector (a,b,c,d) is in the nullspace of the 1x4 matrix [1, -1, -2, -1]. So step (1) is a special case of finding a basis for the nullspace of a given matrix (or equivalently, finding a basis for the set of solutions to Ax = 0, where the matrix A is given).

Using standard algorithms to do that in (1), I find that the vectors v_1 = (1,0,0,1), v_2 = (1,1,0,0), and v_3 = (2,0,1,0) form a basis for your subspace. Applying the Gram-Schmidt process to this basis produces

w_1 = v_1 = (1,0,0,1),
w_2 = v_2 - ((v_2 dot w_1)/(w_1 dot w_1)) w_1
= (1,1,0,0) - (1/2) (1,0,0,1) = (1/2,1,0,-1/2),
w_3 = v_3 - ((v_3 dot w_1)/(w_1 dot w_1)) w_1 - ((v_3 dot w_2)/(w_2 dot w_2)) w_2
= (2,0,1,0) - (2/2) (1,0,0,1) - (1/(3/2)) (1/2,1,0,-1/2)
= (2,0,1,0) - (1,0,0,1) - (1/3,2/3,0,-1/3)
= (2/3,-2/3,1,-2/3)

and then

u_1 = w_1/||w_1|| = ((1/2) sqrt(2), 0, 0, (1/2) sqrt(2))
u_2 = w_2/||w_2|| = ((1/6) sqrt(6), (1/3) sqrt(6), 0, -(1/6) sqrt(6))
u_3 = w_3/||w_3|| = ((2/21) sqrt(21), -(2/21) sqrt(21), (1/7) sqrt(21), -(2/21) sqrt(21)).

So these three vectors are an orthonormal basis for that subspace of R⁴.

It is worth pointing out that, like almost any question that asks for a basis of a vector space, there is more than one possible right answer. (Different algorithms for computing a basis for [1,-1,-2,-1] will produce different bases of its nullspace, and any one of them is as good as any other for computing an orthonormal basis via the Gram-Schmidt process.)