Projection orthogonale vecteur

Calculate projb(a) and proja(b) for a = (3; 4) and b = ( 5; 2) Then projb(a) = (35=29; 14=29) and proja(b) = ( 21=25; 28=25) One can also de ne the (orthogonal) projection projW (y) of the vector y onto the vector space W If we think of y as a point then the projection of it onto W is the closest point of W to it

Orthogonal Projection on a line •Orthogonal projection of a vector on a line v: any vector u: any nonzero vector on L w: orthogonal projection of v onto L w = cu z: v −w L u v w z Distance from tip of v to L: − ∙ = − ∙ = ∙ − ∙ = ∙ − 2 = ∙ 2 = = ∙ 2 z ∙ =0 =0 = −

1 Orthogonal Projections We shall study orthogonal projections onto closed subspaces of H In summary we show: • If X is any closed subspace of H then there is a bounded linear operator P : H → H such that P = X and each element x can be written unqiuely as a sum a + b with a ∈ Im(P) and b ∈ ker(P); explicitly a = Px and b = x − Px

Sep 13 2021 · Orthogonal Projection Using the Dot Product ~b ~a p ~a(~b) = (c ~a(~b))~u ‘ ~b p ~a(~b) ~u I Since u~is orthogonal to ~b p ~a(~b) 0 = ~u(~b p ~a(~b)) = ~u~b ~u((c ~a(~b))~u) = ~b~u c ~a(~b) I Therefore c ~a(~b) = ~bu~and p ~a(~b) = (~bu~)~u

Find the orthogonal projection P: R3 ¡! W Solution We notice flrst that ((1;1;1);(1;0;1)) = 2 6= 0 so this is not an orthogonal basis Using Gram-Schmidt we get: v1 = (1;1;1) v2 = (1;0;1)¡ 2 3 (1;1;1) = (1 3;¡2 3); 1 3 = 1 3 (1;¡2;1) To avoid fractions we can use (1;¡2;1) instead of 1 3 (1;¡2;1) Thus the orthogonal projection is: P