Lemma if v is a subspace of r n then v v further if p

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Lemma. If V is a subspace of R n , then ( V ) = V . Further, if P V is the orthogonal projection of R n onto V , then I - P V is the orthogonal projection of R n onto V . Proof. If v V , then v · w = 0 for all w V , so v ( V ) . Thus V ( V ) . Let x ( V ) . Write x = v + w , where v V and w V . Since x V , we have 0 = x · w = v · w + w · w = || w || 2 . Hence w = 0 and x V . It follows that ( V ) V , and consequently ( V ) = V . Now x = w + P V ( x ) where w V and P V ( x ) ( V ) . Thus w = P V ( x ). Since x = P V ( x ) + P V ( x ) for all x R n , P V + P V is the identity I , and P V = I - P V . There is another useful characterization of the orthogonal projection, which follows from the Pythagorean theorem. Theorem 10. Let V be a subspace of R n , and let P V be the orthogonal projection of R n onto V . If x R n , then P V ( x ) is the nearest point in V to x , that is, || x - P V ( x ) || ≤ || x - y || for all y V , with equality if and only if y = P V ( x ) . Proof. If y V , then P V ( x ) - y V , so x - P V ( x ) P V ( x ) - y . Since x - y = ( x - P V ( x )) + ( P V ( x ) - y ), we obtain from the Pythagorean theorem that || x - y || 2 = || x - P V ( x ) || 2 + || P V ( x ) - y || 2 . Thus || x - y || 2 ≥ || x - P V ( x ) || 2 , and equality holds if and only if || P V ( x ) - y || 2 = 0, that is, P V ( x ) = y . 4. Orthogonal Matrices Definition: A square matrix A is orthogonal if A t = A - 1 . Theorem 11. The following are equivalent, for an n × n matrix A : (1) A is orthogonal, that is, A t = A - 1 , (2) the columns of A form an orthonormal basis for R n , (3) the rows of A form an orthonormal basis for R n . Proofsketch. The matrix equation AA t = I is equivalent to n 2 scalar equations, which are precisely the equations for orthonormality of the columns of A . The matrix equation A t A = I is equivalent to the equations for the orthonormality of the rows of A . Example: Permutation matrices are orthogonal. Each column of a permutation matrix has one 1 and other entries 0, so each column vector has unit length. Different columns have the 1’s in different places, so the column vectors are orthogonal. Example: Rotation matrices are orthogonal. If a 2 × 2 matrix represents a rotation by θ , the two columns represent the rotates of the unit vectors e 1 and e 1 by θ , and since e 1 and e 1 are orthogonal, their rotates by the same angle are orthogonal. 7
Lemma . Inverses of orthogonal matrices are orthogonal. Proof . If A is orthogonal, then AA t = I , so ( A t ) t = A = ( A t ) - 1 , and A - 1 is orthogonal. Lemma . Products of orthogonal matrices are orthogonal. Proof . If A and B are orthogonal, then ( AB ) - 1 = B - 1 A - 1 = B t A t = ( AB ) t , so AB is orthogonal. Lemma . The determinant of an orthogonal matrix is ± 1 . Proof . From I = AA t and the multiplicativity of the determinant, we obtain 1 = det ( AA t ) = det ( A )det ( A t ) = [det ( A )] 2 . It follows that det ( A ) = ± 1. Example: The determinant of a permutation matrix is the sign of the permutation.

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