Gram-Schmidt Orthogonalization

Question

How does Gram-Schmidt Orthogonalization work, and why should we do it?

Answer 1

For example, take the basis S = {u_{1 =} (1,-1,-1), u₂ = (0,3,3), u₃ = (3,2,4)} and find an orthogonal basis - we'll call it V = {v₁, v₂, v₃} - from it.

OK, let's start with the first vector in the basis. We will let v₁ = u₁. That part is easy.

Now, we need another vector that is orthogonal to v₁. Remember that the definition of orthogonal is that the inner product of the two vectors equals 0. So, we need our vector to satisfy the formula

v₂ = c₁v₁ + c₂u_•2 where v₁•v₂=0. So,

v₂ = (0,3,3) - [(0,3,3)•(1,-1,-1)] / [(1,-1,-1)•(1,-1,-1)] (1,-1,-1)

v₂ = (0,3,3) - [-6] / [3] (1,-1,-1)

v₂ = (0,3,3) - (-2)(1,-1,-1)

v₂ = (0,3,3) - (-2,2,2)

v₂ = (2,1,1)

We can now check to make sure that v₁•v₂=0

(1,-1,-1)•(2,1,1)= 2 - 1 - 1 = 2 - 2 = 0, so these two vectors are indeed orthogonal.

Now, we need a third vector that is orthogonal to both v₁ and v₂.

v₃ = (3,2,4) - [u₃•v₁/v₁•v₁]v₁ - [u₃•v₂/v₂•v₂]v₂

v₃ = (3,2,4) - [-3/3](1,-1,-1) - [12/6](2,1,1)

v₃ = (3,2,4) - [-1](1,-1,-1) - [2](2,1,1)

v₃ = (3,2,4) - (-1,1,1) - (4,2,2)

v₃ = (4,1,3) - (4,2,2)

v₃ = (0,-1,1)

Again, I would encourage you to check to make sure that v₃•v₂=0 and v₃•v₁=0.

If you need an orthonormal basis, you would then take each of these vectors we just found and divide them by their norms (also called length or magnitude).