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, ... , vk} be a set of vectors in Rn, then S is called an orthogonal if          vi . vj  =  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, ... , vk} are linearly independent.   Proof Let         c1v1 + ... + cnvk  =  0 Since the vectors are orthogonal, we have          vi . vj  =  0  (for i  j) and when we dot both sides of the equation with vi all the terms drop out except the ith term.         c1v1 . vi  + ... + cnvk . vi  =  0 . vi          =  ci And we see that          ci  =  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, ... , vn} be an orthonormal basis for Rn and v be a vector in Rn then the ith coordinate of v with respect to the basis S is          ([v]S)i  =  v . vi  Proof Since S is a basis, it spans Rn hence we can write         v  =  c1v1 + ... + cnvk  now just take the dot product with vi on both sides and note and use the fact that S is orthonormal to get         v . vi  =  ci    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, ... , un}  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, ... , vn} 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  Continue until you have exhausted all the vectors.   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 Linear Algebra Home Page