Introduction to Linear algebra-Strang-Solutions-Manual_ver13

1 1 0 and 1 0 1 if s is spanned by 2 0 0 and 0

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Unformatted text preview: aces would have dimension 2 which is impossible for 3 by 3. (a) If Ax D b has a solution and AT y D 0, then y is perpendicular to b. bT y D .Ax /T y D x T .AT y / D 0. (b) If AT y D .1; 1; 1/ has a solution, .1; 1; 1/ is in the row space and is orthogonal to every x in the nullspace. Solutions to Exercises 43 6 Multiply the equations by y1 ; y2 ; y3 D 1; 1; 1. Equations add to 0 D 1 so no solution: 7 8 9 10 11 12 13 14 15 16 y D .1; 1; 1/ is in the left nullspace. Ax D b would need 0 D .y T A/x D y T b D 1. Multiply the 3 equations by y D .1; 1; 1/. Then x1 x2 D 1 plus x2 x3 D 1 minus x1 x3 D 1 is 0 D 1. Key point: This y in N .AT / is not orthogonal to b D .1; 1; 1/ so b is not in the column space and Ax D b has no solution. x D x r C x n , where x r is in the row space and x n is in the nullspace. Then Ax n D 0 and Ax D Ax r C Ax n D Ax r . All Ax are in C .A/. Ax is always in the column space of A. If AT Ax D 0 then Ax is also in the nullspace of AT . So Ax is perpendicular to itself. Conclusion: Ax D 0 if AT Ax D 0. (a) With AT D A, the column and row spaces are the same (b) x is in the nullspace and z is in the column space = row space: so these “eigenvectors” have x T z D 0. For A: The nullspace is spanned by . 2 ; 1/, the row space is spanned by .1; 2/. The column space is the line through .1; 3/ and N .AT / is the perpendicular line through .3; 1/. For B: The nullspace of B is spanned by .0; 1/, the row space is spanned by .1; 0/. The column space and left nullspace are the same as for A. x splits into x r C x n D .1; 1/ C .1; 1/ D .2; 0/. Notice N .AT / is a plane .1; 0/ D .1; 1/=2 C .1; 1/=2 D x r C x n . V T W D zero makes each basis vector for V orthogonal to each basis vector for W . Then every v in V is orthogonal to every w in W (combinations of the basis vectors).   x Ax D B b means that Œ A B  x D 0. Three homogeneous equations in four b x unknowns always have a nonzero solution. Here x D .3; 1/ and b D .1; 0/ and x Ax D B b D .5; 6; 5/ is in both column spaces. Two planes in R3 must share a line. x A p-dimensional and a q -dimensional subspace of Rn share at least a line if p C q &gt; n. (The p C q basis vectors of V and W cannot be independent.) AT y D 0 leads to .Ax /T y D x T AT y D 0. Then y ? Ax and N .AT / ? C .A/. 17 If S is the subspace of R3 containing only the zero vector, then S ? is R3 . If S is spanned by .1; 1; 1/, then S ? is the plane spanned by .1; 1; 0/ and .1; 0; 1/. If S is spanned by .2; 0; 0/ and .0; 0; 3/, then S ? is the line spanned by .0; 1; 0/.   151 . Therefore S ? is a subspace even if S is not. 18 S ? is the nullspace of A D 222 19 L? is the 2-dimensional subspace (a plane) in R3 perpendicular to L. Then .L? /? is a 1-dimensional subspace (a line) perpendicular to L? . In fact .L? /? is L. 20 If V is the whole space R4 , then V ? contains only the zero vector. Then .V ? /? D R4 D V .  1223 21 For example . 5; 0; 1; 1/ and .0; 1; 1; 0/ span S D nullspace of A D . 1332   22 .1; 1; 1; 1/ is a basis for P ? . A D 1 1 1 1 has P as its...
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