Comments on midtermSupposeV is a vector space, and you are asked to give a basis.• CORRECT: V has basis110,101• CORRECT: V has basis(110,101)• OK: V = span(110,101)(but you really should point out that the two vectors are independent)• INCORRECT: V =(110,101)• INCORRECT: basis =1 11 00 1Reviewyxˆxx⊥• Orthogonal projection of x onto y:xˆ=x · yy · yy.“Error” x⊥= x − xˆis orthogonal to y.• If y1,, ynis an orthogonal basis of V , and x is in V ,thenx = c1y1++ cnynwith cj=x · yjyj· yj.x decomposes as the sum of its projections onto each vectorin the orthogonal basis.Example 1. Express211xin terms of the basis1−10y1,110y2,001y3.Solution. Note that y1, y2, y3is an orthogonal basis of R3.Armin [email protected]211= c11−10+ c2110+ c3001=211·1−101−10·1−101−10projection of x onto y1+211·110110·110110projection of x onto y2+211·001001·001001projection of x onto y3=121−10+32110+001Orthogonal projection on subspacesTheorem 2. Let W be a su bspace of Rn. Then, each x i n Rncan be uniquely written asx = xˆin W+ x⊥in W⊥.v1v2xˆxx⊥• xˆis the orthogonal projection of x onto W .xˆis the point in W closest to x. For any other y in W , dist(x, xˆ) < dist(x, y).• If v1,, vmis an orthogonal ba sis of W , thenxˆ=x · v1v1· v1v1++x · vmvm· vmvm.Once xˆis determined, x⊥= x − xˆ.(This is also the orthogonal p roje ction ofx onto W⊥.)Example 3. L e t W = span(301,010), and x =0310.• Find the orthogonal projection of x onto W .(or: find the vector inW which is closest to x)• Write x as a vector in W plus a vector orthogonal to W .Armin [email protected] that w1=301and w2=010are an ortho gonal basis for W .[We will soon learn how t o construct orthogonal bases ourselves.]Hence, the orthogonal projection ofx onto W is:xˆ=x · w1w1· w1w1+x · w2w2· w2w2=0310·301301·301301+0310·010010·010010=1010301+ 3010=331xˆis the vector in W which be st approximates x.Orthogonal projection ofx onto the orthogonal complement of W :x⊥=0310−331=−309. Hence, x =0310=331in W+−309in W⊥.Note: Indeed,−309is orthogonal to w1=301and w2=010.Definition 4. Let v1,, vmbe an orthogonal basis of W , a subspace of Rn. Note thatthe proje c tion map πW: Rn→ Rn, given byxxˆ=x · v1v1· v1v1++x · vmvm· vmvmis linear. The matrix P representing πWwith respect to the standard basis i s thecorresponding projection matrix.Example 5. Find the projection matrix P which corresponds to orthogonal projectionontoW = span(301,010)in R3.Solution. Standard basis :100,010,001.The first column ofP encodes the projection of100:100·301301·301301+100·010010·010010=310301. Hence P =91 0∗ ∗0 ∗ ∗31 0∗ ∗.Armin [email protected] second column of P encod es the projection of010:010·301301·301301+010·010010·010010=010. Hence P =91 00 ∗0 1 ∗31 00 ∗.The third column ofP encodes the projection of001:001·301301·301301+001·010010·010010=110301. Hence P =91 0031 00 1 031 0011 0.Example 6. (again)Find the orthogonal projection of x =0310onto W = span(301,010).Solution. xˆ= Px =91 0031 00 1 031 0011 00310=331, as in the previous example.Example 7. Compute P2for the projection matrix we just found. Explain!Solution.91003100 1 0310011091003100 1 03100110=91003100 1 03100110Projecting a s ec o n d time does not change anyth i ng anymore.Practice prob lemsExample 8. Find the closest poin t to x in span{v1, v2}, wherex =240−2, v1=1100, v2=1011.Solution. This is the orthogonal proje c tion of x on t o span{v1, v2}.Armin
View Full Document
Unlocking...