Orthonormal Bases in Rⁿ 

Orthonormal Bases

We all understand what it means to talk about the point (4,2,1) in R³.  Implied in this notation is that the coordinates are with respect to the standard basis (1,0,0), (0,1,0), and (0,0,1).  We learn that to sketch the coordinate axes we draw three perpendicular lines and sketch a tick mark on each exactly one unit from the origin.  What makes the standard basis easy to use is that the vectors are mutually orthogonal and are all unit vectors.  We take these two properties to defined what it means to be a nice basis.

Definition

Let  S  =  {v₁, v₂, ... , v_k} be a set of vectors in Rⁿ, then S is called an orthogonal if 

        v_i . v_j  =  0

for all i not equal to j.  An orthogonal set of vectors is called orthonormal if all vectors in S are unit vectors.

Example

Let 

        S  =  {(1,2,0,0), (0,0,3,4), (4,-2,-4,3)}

is an orthogonal set in R⁴ (Check this) but not an orthonormal set since the vectors are not unit vectors.

Example

    S  =  {(1/2,/2), (-/2, 1/2)}

are an orthonormal set of vectors in R³.  

Theorem

Any orthogonal set of vectors S  =  {v₁, v₂, ... , v_k} are linearly independent.

Proof

Let

        c₁v₁ + ... + c_nv_k  =  0

Since the vectors are orthogonal, we have 

        v_i ^. v_j  =  0  (for i  j)

and when we dot both sides of the equation with v_i all the terms drop out except the i^th term.

        c₁v₁ ^. v_i  + ... + c_nv_k ^. v_i  =  0 ^. v_i 

        =  c_i

And we see that 

        c_i  =  0

Hence S is linearly independent.

Finding the coordinate of a vector with respect to a basis can be computationally difficult, usually including the inverse of a matrix.  However if the basis is orthonormal, then the computation is simple.  

Theorem

Let S  =  {v₁, v₂, ... , v_n} be an orthonormal basis for Rⁿ and v be a vector in Rⁿ then the i^th coordinate of v with respect to the basis S is 

        ([v]_S)_i  =  v . v_i 

Proof

Since S is a basis, it spans Rⁿ hence we can write

        v  =  c₁v₁ + ... + c_nv_k 

now just take the dot product with v_i on both sides and note and use the fact that S is orthonormal to get

        v ^. v_i  =  c_i 

Example

Find the coordinates for the vector (5,10) with respect to the basis 

        S  =  {(3/5, 4/5), (-4/5, 3/5)}

Solution

We find 

        (5,10) ^. (3/5,4/5)  =  11        (5,10) ^. (-4/5, 3/5)  =  2

so that

        [(5,10)]_S  =  (11,2)

The Gram Schmidt Process

If we are given a basis 

        S  =  {u₁, u₂, ... , u_n} 

that is not orthonormal, and want to find a basis for the space that is orthonormal, we follow the following process.  

First, if we can find an orthogonal basis, we can always divide each of the basis vectors by their magnitudes to arrive at an orthonormal basis.  Hence we have reduced the problem to finding an orthogonal basis.  

Here is how to find an orthogonal basis   T  =  {v₁, v₂, ... , v_n} given any basis S.

1.  Let the first basis vector be

        v₁  =  u₁ 

2.  Let the second basis vector be

                            u₂ ^. v₁ 
        v₂  =  u₂ -                 v₁ 
                            v₁ ^. v₁

Notice that 

        v₁ ^. v₂  =  0

3.  Let the third basis vector be

                            u₃ ^. v₁               u₃ ^. v₂    
        v₃  =  u₃ -                 v₁ -                    v₂ 
                            v₁ ^. v₁                 v₂ ^. v₂ 

You can easily check that these three vectors are orthogaonal.

4.  Let the fourth basis vector be

                            u₄ ^. v₁               u₄ ^. v₂                 u₄ ^. v₃    
        v₄  =  u₄ -                 v₁ -                    v₂   -                    v₃                         
                            v₁ ^. v₁                 v₂ ^. v₂                      v₃ ^. v₃ 

Example

Find an orthogonal basis for the vector space spanned by (1,2,4,0), (0,1,1,0), (0,3,1,4).  We can check that these three vectors are linearly independent.  We have

        v₁  =  (1,2,4,0)

                                     (0,1,1,0) ^.  (1,2,4,0)
        v₂  = (0,1,1,0)  -                                       (1,2,4,0)
                                      (1,2,4,0) ^.  (1,2,4,0)

                                 6
        =  (0,1,1,0)  -          (1,2,4,0)  =  (-2/7, 3/7, -1/7, 0)
                                21

We can replace this vector by a multiple of it to get

        v₂  = (-2,3,-1,0)

Now we find the third and last basis vector

                                     (0,3,1,4) ^.  (1,2,4,0)                           (0,3,1,4) ^. (-2,3,-1,0)
        v₃  = (0,3,1,4)  -                                       (1,2,4,0)  -                                        (-2,3,-1,0)
                                      (1,2,4,0) ^.  (1,2,4,0)                         (-2,3,-1,0) ^. (-2,3,-1,0)

                                10                        8
        =  (0,3,1,4)  -          (1,2,4,0) -          (-2,3,-1,0)  =  (14/21, 7/21, -7/21, 4)  =  (2/3, 1/3, -1/3, 4)
                                21                      14

Multiplying by 3 gives

        v₃  = (2,1,-1,12)

Hence an orthogonal basis is given by

        (1,2,4,0, (-2,3,-1,0), (2,1,-1,12)

Note:  If we want to find an orthonormal basis we just divide each of these vectors by their magnitudes.

Back to the Vectors Home Page

Back to the Linear Algebra Home Page

Back to the Math Department Home Page

e-mail Questions and Suggestions