Solutions to Review Problems for Chapter 6 ( ), 7.1

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1 Solutions to Review Problems for Chapter (-, 7 The Final Exam is on Thursday, June,, : AM : AM at NESBITT Final Exam Breakdown Sections % -,7-9,- - % -9,-,7,-,-7 - % -, 7 - % Let u u and v Let x x x x, v x + x x + x Thus, x x Describe the set of all vectors in R that are orthogonal to both be a vector orthogonal to both u and v Then x u and x v Thus,, where x is any real number The vectors in R orthogonal to both u and v are the scalar multiples of x x x x x x free Let S be the two-dimensional subspace of R spanned by u, v Find a basis for S Give a geometric description of S and S This is just question ( We have that S Span A basis for S is S is the plane in R spanned by the vectors u and v, and S is the line through the origin and the vector Let y, u Let L Span{u}

2 (a Find the orthogonal projection of y onto L (b Write y as a sum of a vector in L and a vector orthogonal to L (c What is the distance from y to L? (a ŷ proj L y ( y u u u u (b y ŷ + z, ŷ (c ŷ y z 7, z 7 ( + ( Let x, u, v, and W Span{u, v} Note that u v Find a vector a in W and a vector b that is orthogonal to W, such that x a + b a proj W x ( ( x u u u u + x v v v v, b x a a + +, b x a Given a vector x, and a line L { x x } : x + x in R : (a Find the vector in L that is closest to x; that is, find the orthogonal projection of x onto the line L (b Write x as a vector sum x p + z, where p is in L and z is in L We have that L Span{u} where u (a ˆx proj L x ( x u u u u (b x ˆx + z, ˆx, z Let A, b (a Write the normal equations for the least-squares solution of Ax b (b Find the least-squares solution of Ax b, using the normal equations (c Find the closest point to b in the column space of A

3 (a The normal equations are A T Ax A T b, where A T A and A T b (b Row reduction: 7 The least-squares solution is ˆx 7/ 7/, 7/ (c The least-squares solution ˆx has the property that Aˆx is the closest point in Col A to b So / Aˆx 7/ / 7 Describe all least-squares solutions of Ax b: A, b The normal equations are A T Ax A T b, where A T A and Row reduction: A T b The least-squares solutions are vectors of the form ˆx + x where x is any real number,,

4 Let A A x x x, where x Find an orthonormal basis for the column space of A, x, x Let s use the Gram-Schmidt process to orthogonalize {x, x, x } : v x v x ( x v v v v 9 9 ( ( v x x v v v v x v v v v 9 9 An orthonormal basis for the column space of A is { } v v, v v, v v,, 9 Let A, b (a Find the orthogonal projection of b onto Col A (b Use (a to find the least-squares solution of Ax b (a Let A u u u Then u u, u u and u u, so {u, u, u } is an orthogonal basis for Col A Thus, ˆb proj Col A b ( b u ( b u ( b u u u u + u u u + u u u + +

5 (b From (a, we have ˆb u + u u The least-squares solution ˆx is the solution to Aˆx ˆb, that is, the entries of ˆx are the weights when the projection ˆb is written as a combination of the columns of A Thus, ˆx Find the equation y c +c x of the least-squares line for the data: (,, (,, (,, and (, We need to find c and c that best approximates c c } {{ } A The normal equations are: A T Aˆx A T b, where A T A }{{} b, and A T b Row reduction: The least-squares line that best approximates the data points is y x Let A Then v, v, v are linearly independent eigenvectors of A Orthogonally diagonalize A (That is, find an orthogonal matrix U u u u whose columns are an orthonormal set of eigenvectors of A, and a diagonal matrix D such that A UDU UDU T Check that Av v Av v Av v

6 Thus, v and v are linearly independent eigenvactors corresponding to the eigenvalue, and v is an eigenvector corresponding to the eigenvalue So, we may let D Note that v is orthogonal to both v and v To find the columns of U, first orthogonalize {v, v } by Gram-Schmidt: z v z v ( v z z z z Now, {z, z, v } is an orthogonal set of eigenvectors of A Finally, let u z z z u z z z u z z z z Thus, A UDU T, where D, U Let v be a unit eigenvector for A T A corresponding to the eigenvalue λ What is the length of the vector Av? Make calculations that justify your answer We have that A T Av v Multiply both sides on the left by v T : v T A T Av v T v (Av T (Av v T v (Av (Av v v Av v Since v, then Av Thus, Av

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