Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# The same vectors happen to be in the nullspace an

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Unformatted text preview: ace .1; 0; 0; 0/ and .0; 1; 0; 1/; N .AT / .0; 1; 0/. B : dim 1; 1; 0; 2 Row space (1), column space .1; 4; 5/, nullspace: empty basis, N .AT / . 4 ; 1; 0/ and . 5; 0; 1/. Solutions to Exercises 41 7 Invertible 3 by 3 matrix A: row space basis D column space basis D .1; 0; 0/, .0; 1; 0/,   8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 .0; 0; 1/; nullspace basis and left nullspace basis are empty. Matrix B D A A : row space basis .1; 0; 0; 1; 0; 0/, .0; 1; 0; 0; 1; 0/ and .0; 0; 1; 0; 0; 1/; column space basis .1; 0; 0/, .0; 1; 0/, .0; 0; 1/; nullspace basis . 1; 0; 0; 1; 0; 0/ and .0; 1; 0; 0; 1; 0/ and .0; 0; 1; 0; 0; 1/; left nullspace basis is empty.      I 0 and I I I 0 0 and 0 D 3 by 2 have row space dimensions D 3; 3; 0 D column space dimensions; nullspace dimensions 2; 3; 2; left nullspace dimensions 0; 2; 3. (a) Same row space and nullspace. So rank (dimension of row space) is the same (b) Same column space and left nullspace. Same rank (dimension of column space). For rand .3/, almost surely rankD 3, nullspace and left nullspace contain only .0; 0; 0/. For rand .3; 5/ the rank is almost surely 3 and the dimension of the nullspace is 2. (a) No solution means that r < m. Always r  n. Can’t compare m and n here. (b) Since m r > 0, the left nullspace must contain a nonzero vector. " # " #  11 221 101 A neat choice is 0 2 D 2 4 0 ; r C .n r / D n D 3 does 120 10 101 not match 2 C 2 D 4. Only v D 0 is in both N .A/ and C .AT /. (a) False: Usually row space ¤ column space (same dimension!) (b) True: A and A have the same four subspaces (c) False (choose A and B same size and invertible: then they have the same four subspaces) Row space basis can be the nonzero rows of U : .1; 2; 3; 4/, .0; 1; 2; 3/, .0; 0; 1; 2/; nullspace basis .0; 1; 2 ; 1/ as for U ; column space basis .1; 0; 0/, .0; 1; 0/, .0; 0; 1/ (happen to have C.A/ D C.U / D R3 ); left nullspace has empty basis. After a row exchange, the row space and nullspace stay the same; .2; 1; 3; 4/ is in the new left nullspace after the row exchange. If Av D 0 and v is a row of A then v  v D 0. Row space D yz plane; column space D xy plane; nullspace D x axis; left nullspace D z axis. For I C A: Row space D column space D R3 , both nullspaces contain only the zero vector. Row 3 2 row 2 C row 1 D zero row so the vectors c.1; 2 ; 1/ are in the left nullspace. The same vectors happen to be in the nullspace (an accident for this matrix). (a) Elimination on Ax D 0 leads to 0 D b3 b2 b1 so . 1; 1; 1/ is in the left nullspace. (b) 4 by 3: Elimination leads to b3 2 b1 D 0 and b4 C b2 4 b1 D 0, so . 2 ; 0; 1; 0/ and . 4 ; 1; 0; 1/ are in the left nullspace. Why? Those vectors multiply the matrix to give zero rows. Section 4.1 will show another approach: Ax D b is solvable (b is in C .A/) when b is orthogonal to the left nullspace. (a) Special solutions . 1; 2; 0; 0/ and . 1 ; 0; 3; 1/ are perpendicular to the rows of 4 R (and then ER). (b) AT y D 0 has 1 independent solution D last row of E 1 . (E 1 A D R has a zero row, which is just the...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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