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Unformatted text preview: c2 D c3 D 0. Only this combination of v1 ; v2 ; v3 gives 0. 9 (a) The four vectors in R3 are the columns of a 3 by 4 matrix A. There is a nonzero solution to Ax D 0 because there is at least one free variable (b) Two vectors are
dependent if Œ v 1 v2 has rank 0 or 1. (OK to say “they are on the same line” or “one is
a multiple of the other” but not “v2 is a multiple of v1 ” —since v1 might be 0.) (c) A
nontrivial combination of v1 and 0 gives 0: 0v1 C 3.0; 0; 0/ D 0.
10 The plane is the nullspace of A D Œ 1 2 3 1 . Three free variables give three
solutions .x; y; z; t / D .2; 1 0 0/ and .3; 0; 1; 0/ and .1; 0; 0; 1/. Combinations
of those special solutions give more solutions (all solutions). 11 (a) Line in R3 (b) Plane in R3 (c) All of R3 (d) All of R3 . 12 b is in the column space when Ax D b has a solution; c is in the row space when AT y D c has a solution. False. The zero vector is always in the row space. 13 The column space and row space of A and U all have the same dimension = 2. The row spaces of A and U are the same, because the rows of U are combinations of the rows
of A (and vice versa!).
1
.v
2 C w/ C 1 .v w/ and w D 1 .v C w/ 1 .v w/. The two pairs span the
2
2
2
same space. They are a basis when v and w are independent. 14 v D 15 The n independent vectors span a space of dimension n. They are a basis for that space. If they are the columns of A then m is not less than n .m n/. Solutions to Exercises 38 (a) .1; 1; 1; 1/ for the space of all constant vectors
.c; c; c; c/
(b) .1; 1; 0; 0/; .1; 0; 1; 0/; .1; 0; 0; 1/ for the space of vectors with
sum of components = 0
(c) .1; 1; 1; 0/; .1; 1; 0; 1/ for the space perpendicular to .1; 1; 0; 0/ and .1; 0; 1; 1/
(d) The columns of I are a basis for its column
space, the empty set is a basis (by convention) for N .I / D {zero vector}.
10101
17 The column space of U D
is R2 so take any bases for R2 ; (row 1
01010
and row 2) or (row 1 and row 1 C row 2) and (row 1 and row 2) are bases for the row
spaces of U . 16 These bases are not unique! 18 (a) The 6 vectors might not span R4 (b) The 6 vectors are not independent (c) Any four might be a basis.
19 nindependent columns ) rank n. Columns span Rm ) rank m. Columns are basis for Rm ) rank D m D n. The rank counts the number of independent columns. 20 One basis is .2; 1; 0/, . 3; 0; 1/. A basis for the intersection with the xy plane is .2; 1; 0/. The normal vector .1; 2 ; 3/ is a basis for the line perpendicular to the plane.
21 (a) The only solution to Ax D 0 is x D 0 because the columns are independent (b) Ax D b is solvable because the columns span R5 . Key point: A basis gives
exactly one solution for every b. 22 (a) True (b) False because the basis vectors for R6 might not be in S. 23 Columns 1 and 2 are bases for the (different) column spaces of A and U ; rows 1 and 2 are bases for the (equal) row spaces of A and U ; .1; 1; 1/ is a basis for the (equal)
nullspaces.
24 (a) False A D Œ 1 1 has dependent
columns, independent row
(b) False column
01
(c) True: Both dimensions D 2 if A is inspace ¤ row space for A D
00
vertible, dimensions D 0 if A D 0, otherwise dimensions D 1
(d) False, columns
may be dep...
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 Spring '12
 Minki
 Mass

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