Preparation problems for the discussion sections on October 28th and 30th1. Let u1=120, u2=2−12. Let v =1−21. Can you find real numbers c1, c2such thatv = c1u1+ c2u2?Solution: Since u1and u2are orthogonal (i.e. u1· u2= 0), we have that ifv = c1u1+ c2u2for some real number c1, c2, thenv · u1= c1u1· u1+ c2u2· u1= c1u1· u1andv · u2= c1u1· u2+ c2u2· u2= c2u2· u2.Hencec1=v · u1u1· u1=1−21·120120·120= −35andc2=v · u2u2· u2=1−21·2−122−12·2−12=69=23.However, we see that −35u1+23u26= v, so it is not possible to find real numbers c1, c2suchthat v = c1u1+ c2u2.The numbers that we found, however, are “best possible” in the sense that the two sides areas close as possible. In other words, −35u1+23u2is the orthogonal projection of v onto thespace spanned by u1and u2.[Note that you can solve this problem in many other ways. The way above serves to makeus more familiar with notions such as orthogonal projections.]2. Let W = Span{v}, where v =111, be a subspace of R3. Find the projections aW, bW, cWof the vectorsa =123, b =2−1−1, c =222onto the subspace W . Interpret your results geometrically.1Solution: We have,aW=a · vv · vv =123·111111·111111=63111=222,bW=b · vv · vv =2−1−1·111111·111111=03111=000,cW=c · vv · vv =222·111111·111111=63111=222.The fact that bWis zero means that b is orthogonal to W . In this, and the other two cases,we obtain the vector in W which is closest to the vector that we start with.3. Let W = Span{1111,1−100} be a subspace of R4.(i) Find the closest point to1010on the subspace W .(ii) Find the projection matrix, P , corresponding to the projection onto W .(iii) Use the projection matrix, P , to find the projection of1010onto the subspace W .Solution:(i) The closest point is the orthogonal projection:1010·11111111·11111111+1010·1−1001−100·1−1001−100=241111+121−100=1012122(ii) The projections of the four standard basis vectors are1000W=1000·11111111·11111111+1000·1−1001−100·1−1001−100=141111+121−100=34−141414,0100W=0100·11111111·11111111+0100·1−1001−100·1−1001−100=141111+−121−100=−14341414,0010W=0010·11111111·11111111+0010·1−1001−100·1−1001−100=141111+ 01−100=14141414,0001W=0001·11111111·11111111+0001·1−1001−100·1−1001−100=141111+ 01−100=14141414.Hence, the projection matrix is:P =34−141414−143414141414141414141414(iii) Using P , we find that the orthogonal projection is1010W=34−141414−1434141414141414141414141010=10121234. Let W = Span{111,1−10} and V = Span{111,11−2} be subspaces of R3.(i) Find the projection matrices, P and Q, corresponding to the projections onto W andV , respectively.(ii) Check that P Q = QP . Can you interpret P Q as a projection matrix?Solution:(i) The projections onto W of the three standard basis vectors are100W=100·111111·111111+100·1−101−10·1−101−10=13111+121−10=56−1613,010W=010·111111·111111+010·1−101−10·1−101−10=13111−121−10=−165613,001W=001·111111·111111+001·1−101−10·1−101−10=13111+ 01−10=131313.Hence, the projection matrix corresponding to the orthogonal projection onto W is:P =56−1613−165613131313On the other hand, the projections onto V of the three standard basis vectors
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