Problem of the week - Gram-Schmidt process

As stated in a previous question, orthogonal bases have some practical advantages and are very useful when dealing with projections onto subspaces. These bases are defined in spaces equipped with an inner product also called a dot product, and by definition, a basis is called orthogonal if every pair of basis vectors are orthogonal, that is, their inner product is 0. When the length of each vector is 1 (vectors are normalized), the basis is called an orthonormal basis.

In an inner product space, it is always possible to get an orthonormal basis starting from any basis, by using the Gram-Schmidt algorithm. This process is a method for orthonormalising a set of vectors, most commonly the Euclidean space Rⁿ equipped with the standard inner product, and it can be guaranteed that if the set of vectors is a basis, at the end of the process, the resulting set will continue being a basis.

To solve this problem of the week you will need to prove you master the Gram–Schmidt process. You also need to compute the change of basis matrix, a concept tackled in another question.

Let A = {(0 2 0), (4 -1 -3), (3 2 4)} a basis of R³. ¿Find the change of basis matrix from the basis A to the orthonormal basis obtained when orthogonalize A by using the Gram-Schmidt process?

The first step is to use the Gram-Schmidt process to get an orthogonal basis from the basis A.

Let's call V = {v₁,v₂,v₃} to this basis:

v1 = a1 = (0 2 0)

          <a2,v1>
v2 = a2 - ——————— • v1
           ‖v1‖²  

<a2,v1> = <(4 -1 -3),(0 2 0)> = -2
‖v1‖² = 4

v2 = (4 -1 -3) - (-2)•(0 2 0)/4
v2 = (4 -1 -3) - (0 -1 0)
v2 = (4 0 -3)

          <a3,v1>        <a3,v2> 
v3 = a3 - ——————— • v1 - ——————— • v2
           ‖v1‖²          ‖v2‖²      

<a3,v1> = <(3 2 4),(0 2 0)> = 4          
‖v1‖² = 4

<a3,v2> = <(3 2 4),(4 0 -3)> = 0
‖v2‖² = 25

v3 = (3 2 4) - 4•(0 2 0)/4 - 0•(4 0 -3)/25
v3 = (3 2 4) - (0 2 0)
v3 = (3 0 4)

Then, we need to normalize the orthogonal basis, by dividing each vector by its norm.

‖v1‖ = 2
‖v2‖ = 5
‖v3‖ = 5

Thus, the orthonormal basis B, obtained after normalizing all vectors in the basis V is:

B = {(0 1 0), (4/5 0 -3/5), (3/5 0 4/5)}

The final step is to find the change of basis matrix from base A to B.

The problem can be solved using different methods, here we will use the matrix way. The change of basis matrix C[A->B] from the basis A to the basis B, can be computed by multiplying the inverse of the basis matrix of B by the basis matrix of A.

The basis matrix of B (the change of basis matrix from the basis B to the standard basis N) is:

          ┌              ┐
          │ 0   4/5  3/5 │
C[B->N] = │ 1     0    0 │
          │ 0  -3/5  4/5 │
          └              ┘

Let's compute its inverse:

│ 0   4/5  3/5 │
│ 1     0    0 │ =  -1
│ 0  -3/5  4/5 │

               ┌                ┐
               │  0  -4/5  -3/5 │
Cof(C[B->N]) = │ -1     0     0 │
               │  0   3/5  -4/5 │
               └                ┘
               ┌                ┐
               │    0  -1     0 │
Adj(C[B->N]) = │ -4/5   0   3/5 │
               │ -3/5   0  -4/5 │
               └                ┘
                         ┌              ┐
                         │   0  1     0 │
Inv(C[B->N]) = C[N->B] = │ 4/5  0  -3/5 │
                         │ 3/5  0   4/5 │
                         └              ┘

The basis matrix of A (the change of basis matrix from the basis A to the standard basis N) is:

          ┌          ┐
          │ 0   4  3 │
C[A->N] = │ 2  -1  2 │
          │ 0  -3  4 │
          └          ┘

Finally, let's multiply the matrices C[N->B] and C[A->N].

C[A->B]11 = 0•0 + 1•2 + 0•0 = 2
C[A->B]12 = 0•4 + 1•(-1) + 0•(-3) = -1
C[A->B]13 = 0•3 + 1•2 + 0•4 = 2

C[A->B]21 = 4/5•0 + 0•2 + (-3/5)•0 = 0
C[A->B]22 = 4/5•4 + 0•(-1) + (-3/5)•(-3) = 5
C[A->B]23 = 4/5•3 + 0•2 + (-3/5)•4 = 0

C[A->B]31 = 3/5•0 + 0•2 + 4/5•0 = 0
C[A->B]32 = 3/5•4 + 0•(-1) + 4/5•(-3) = 0
C[A->B]33 = 3/5•3 + 0•2 + 4/5•4 = 5
          ┌          ┐
          │ 2  -1  2 │
C[A->B] = │ 0   5  0 │
          │ 0   0  5 │
          └          ┘

Which is the final answer.