Introduction to Linear algebra-Strang-Solutions-Manual_ver13

1 1 0 and 3 4 1 nullspace basis 1 0 0 0 0 0

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Unformatted text preview: many independent 4 by 4 permutation matrices? Solutions to Exercises 40 (a) zero when x D 0 (b) one when x D .1; 1; 1; 1/ (c) three when x D .1; 1; 1; 1/ because all rearrangements of this x are perpendicular to .1; 1; 1; 1/ (d) four when the x ’s are not equal and don’t add to zero. No x gives dim S D 2. I owe this nice problem to Mike Artin—the answers are the same in higher dimensions: 0; 1; n 1; n. The problem is to show that the u’s, v’s, w’s together are independent. We know the u’s and v’s together are a basis for V , and the u’s and w’s together are a basis for W . Suppose a combination of u’s, v’s, w’s gives 0. To be proved: All coefficients D zero. Key idea: In that combination giving 0, the part x from the u’s and v’s is in V . So the part from the w’s is x . This part is now in V and also in W . But if x is in V \ W it is a combination of u’s only. Now the combination uses only u’s and v’s (independent in V !) so all coefficients of u’s and v’s must be zero. Then x D 0 and the coefficients of the w’s are also zero. The inputs to an m by n matrix fill Rn . The outputs (column space!) have dimension r . The nullspace has n r special solutions. The formula becomes r C .n r / D n. If the left side of dim.V/ C dim.W/ D dim.V \ W/ C dim.V C W/ is greater than n, then dim.V \ W/ must be greater than zero. So V \ W contains nonzero vectors. If A2 D zero matrix, this says that each column of A is in the nullspace of A. If the column space has dimension r , the nullspace has dimension 10 r , and we must have r  10 r and r  5. 42 The dimension of S spanned by all rearrangements of x is 43 44 45 46 Problem Set 3.6, page 190 1 (a) Row and column space dimensions D 5, nullspace dimension D 4, dim.N .AT // 2 3 4 5 6 D 2 sum D 16 D m C n (b) Column space is R3 ; left nullspace contains only 0. A: Row space basis D row 1 D .1; 2; 4/; nullspace . 2 ; 1; 0/ and . 4 ; 0; 1/; column space basis D column1 D .1; 2/; left nullspace . 2 ; 1/. B : Row space basis D both rows D .1; 2; 4/ and .2; 5; 8/; column space basis D two columns D .1; 2/ and .2; 5/; nullspace . 4 ; 0; 1/; left nullspace basis is empty because the space contains only y D 0. Row space basis D rows of U D .0; 1; 2; 3; 4/ and .0; 0; 0; 1; 2/; column space basis D pivot columns (of A not U ) D .1; 1; 0/ and .3; 4; 1/; nullspace basis .1; 0; 0; 0; 0/, .0; 2; 1; 0; 0/, .0; 2; 0; 2 ; 1/; left nullspace .1; 1; 1/ D last row of E 1 ! # "   10 9 3 (a) 1 0 (b) Impossible: r C.n r / must be 3 (c) Œ 1 1  (d) 3 1 01 (e) Impossible Row space D column space requires m D n. Then m r D n r ; nullspaces have the same dimension. Section 4.1 will prove N .A/ and N .AT / orthogonal to the row and column spaces respectively—here those are the same space.     111 2 1 has the AD has those rows spanning its row space B D 1 210 same rows spanning its nullspace and BAT D 0. A: dim 2; 2; 2; 1: Rows .0; 3; 3; 3/ and .0; 1; 0; 1/; columns .3; 0; 1/ and .3; 0; 0/; nullsp...
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