Orthogonal Complements Definition of the Orthogonal Complement Geometrically, we can understand that two lines can be perpendicular in R2 and that a line and a plane can be perpendicular to each other in R3.  We now generalize this concept and ask given a vector subspace, what is the set of vectors that are orthogonal to all vectors in the subspace. Definition Let V be a vector space and W be a subspace of V.  Then the orthogonal complement of W in V is the set of vectors u such that u is orthogonal to all vectors in W.   Example Let V  =  R2  and W be the subspace spanned by (1,2).  Then  is the set of vectors (a,b) with         (a,b) . c(1,2)  =  0   or         ac + 2bc  =  0        a + 2b  =  0 This is a 1 dimensional vector space spanned by         (-2,1) In the example above the orthogonal complement was a subspace.  This will always be the case.   Theorem Let W be a subspace of a vector space V.  Then the orthogonal complement of W is also a subspace of V.  Furthermore, the intersection of W and its orthogonal complement is just the zero vector.   Proof Let u1 and u2 be vectors in the orthogonal complement of W and c be a constant.  Then  1.  If w is in W, then         (u1 + u2) . w  =  u1. w + u2 . w  =  0 2.  If w is in W, then         (cu1) . w  =  c(u1 . w)  =  c(0)  =  0 Now we prove that the intersection is zero.  If v is in the intersection then we think of v first as being in W and second as being in the orthogonal complement of W.  Hence          v . v  =  0 This implies that         v  =  0 The next theorem states that if w1, ... ,wr is a basis for W and u1, ... ,uk is a basis for then          {w1, ... ,wr, u1, ... ,uk} is a basis for Rn.  In symbols, we write   Theorem           We leave it up to you to look up the proof of this statement.  What this means is that every vector v in Rn can be uniquely written in the form         v  =  w + u  with w in W and u in . A corollary of this theorem is the following   Corollary         Proof First if a vector is in W then it is orthogonal to every vector in the orthogonal complement of W.  If a vector v is orthogonal to every vector in the orthogonal complement of W, and also by the theorem above we have          v  =  w + u with w in W and u in the orthogonal complement of W.  Since u is in the orthogonal complement of W, we have          0  =  v . u  =  (w + u) . u  =  w . u + u . u  =  u . u Hence u  =  0 and v  =  w. Matrices and Complements If we think of matrix multiplication as a collection of dot products then if         Ax  =  0 then x is orthogonal to each of the rows of A.  Also if          ATy  =  0 then y is orthogonal to each of the columns of A.  More precisely we have    Theorem 1.  The null space of A is the orthogonal complement of the row space of A. 2.  The null space of AT is the orthogonal complement of the column space of A.   Example Find a basis for the orthogonal complement of the space spanned by (1,0,1,0,2), (0,1,1,1,0) and (1,1,1,1,1).   Solution We find the null space of the matrix         We find the rref of A.         We get a basis          {(0,-1,0,1,0), (-1,1,-1,0,1)}   Projections         Given a vector v and a subspace W with orthogonal basis w1, ... , wn, we are often interested in finding in finding the vector in W that is closest to v.  This closest vector is                               v . w1               v . w2                            v . wn             projWv  =                   w1+                    w2   + ... +                    wn                                                       w1 . w1               w2 . w2                                  wn . wn   We will use this formula when we talk about inner product spaces and Fourier coefficients.  Notice that if W is orthonormal then the denominators are all equal to one.   Back to the Linear Algebra Home Page