Consequently r v vectorx 2 vectorx vectoru 1 vectoru

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Consequently, R V vectorx = 2( ( vectorx, vectoru 1 ) vectoru 1 + ( vectorx, vectoru 2 ) vectoru 2 ) vectorx. 6. [Bretscher, Sec. 5.2 #32] Find an orthonormal basis for the plane x 1 + x 2 + x 3 = 0. Solution: Pick any point in the plane, say vectorv 1 = (1 , 1 , 0). This will be the first vector in our orthogonal basis. We use the Gram-Schmidt process to extend this to an orthogonal basis for the plane. Pick any other point in the plane, say vectorw 1 := (1 , 0 1). Write it as vectorw 1 = avectorv 1 + vector z , where vector z is perpendicular to vectorv 1 . Note that, although unknown, vector z will also be in the plane since it will be a linear combination of vectorv 1 and vectorw , both of which are in the plane. As usual, by taking the inner product of both sides of vectorw 1 = avectorv 1 + vector z with vectorv 1 , we find a = ( vectorw 1 , vectorv 1 ) / bardbl vectorv 1 bardbl 2 = 1 2 . Thus vector z = vectorw 1 1 2 vectorv 1 = 1 / 2 1 / 2 1 is in the plane and orthogonal to vectorv 1 . The vectors vectorv 1 and vector z are an orthogonal basis for this plane. To get an orthonormal basis we just make these into unit vectors vectoru 1 := vectorv 1 bardbl vectorv 1 bardbl = 1 2 1 1 0 and vectoru 2 := vector z bardbl vector z bardbl = 1 radicalbig 3 / 2 1 / 2 1 / 2 1 4
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7. Let V be a linear space. A linear map P : V V is called a projection if P 2 = P (this P is not necessarily an “orthogonal projection”). a) Show that the matrix P = ( 0 1 0 1 ) is a projection. Draw a sketch of R 2 showing the vectors (1 , 2), ( 1 , 0), (3 , 1) and (0 , 3) and their images under the map P . Also indicate both the image, V , and kernel, W , of P . Solution: The image of P is the line x 1 = x 2 (the subspace V ); the kernel is the x 1 axis (the subspace W ). See the figure below. b) Repeat this for the complementary projection Q := I P . Solution: The image of Q is the x 1 axis (the subspace W ); its kernel is the line x 1 = x 2 (the subspace V ). -1 -3 -2 -1 1 2 1 2 3 P: Q: W 3 V c) If the image and kernel of a projection P are orthogonal then P is called an orthogonal projection . [This of course now assumes that V has an inner product.] Let M = ( 0 a 0 c ). For which real value(s) of a and c is this a projection? An orthogonal projection? Solution: M 2 = parenleftbigg 0 ac 0 c 2 parenrightbigg so M 2 = M requires that ac = a and c 2 = c . The first requires that either a = 0 or c = 1. If a = 0 the second equation is satisfied if either c = 0 or c = 1. If a negationslash = 0, then c = 1. Thus, the possibilities are: P 1 := parenleftbigg 0 0 0 0 parenrightbigg , or P 2 := parenleftbigg 0 a 0 1 parenrightbigg (for any a ) . For an orthogonal projection P , its image and kernal must be orthogonal. Since for P 1 its image is just 0, which is orthogonal to everything, it is an orthogonal projection. The kernel of P 2 is the horizontal axis. Its image consists of points of the form t ( a, 1) for any scalar t . This straight line is perpendicular to the horizontal axis if (and only if) a = 0. Thus P 2 is an orthogonal projection if and only if a = 0.
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