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Unformatted text preview: endent, in that case not a basis for C .A/.
cd
has rank 2 except when c D d or
25 A has rank 2 if c D 0 and d D 2; B D
dc
c D d.
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26 (a) 0 0 0 ; 0 1 0 ; 0 0 0
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(b) Add 1 0 0 ; 0 0 0 , 0 0 1
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0 01
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0 0 0; 0 0 1.
(c)
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These are simple bases (among many others) for (a) diagonal matrices (b) symmetric
matrices (c) skewsymmetric matrices. The dimensions are 3; 6; 3. Solutions to Exercises 39 " #"
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27 I , 0 1 0 , 0 2 0 , 0 1 0 , 0 1 0 , 0 1 1 ; echelon matri002
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ces do not form a subspace; they span the upper triangular matrices (not every U is
echelon).
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and
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29 (a) The invertible matrices span the space of all 3 by 3 matrices (b) The rank one
matrices also span the space of all 3 by 3 matrices (c) I by itself spans the space of
all multiples cI .
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31 (a) y.x/ D constant C (b) y.x/ D 3x this is one basis for the 2 by 3 matrices with
.2; 1; 1/ in their nullspace (4dim subspace). (c) y.x/ D 3x C C D yp C yn solves
dy=dx D 3. 32 y.0/ D 0 requires A C B C C D 0. One basis is cos x 33 (a) y.x/ D e 2x cos 2x and cos x cos 3x . 0 is a basis for, all solutions to y D 2y (b) y D x is a basis for all
solutions to dy=dx D y=x (Firstorder linear equation ) 1 basis function in solution
space). 34 y1 .x/; y2 .x/; y3 .x/ can be x; 2x; 3x .dim 1/ or x; 2x; x 2 .dim 2/ or x; x 2 ; x 3 .dim 3/.
35 Basis 1, x , x 2 , x 3 , for cubic polynomials; basis x 1, x 2 1, x 3 1 for the subspace with p.1/ D 0. 36 Basis for S: .1; 0; 1; 0/, .0; 1; 0; 0/, .1; 0; 0; 1/; basis for T: .1; 1; 0; 0/ and .0; 0; 2; 1/; S \ T D multiples of .3; 3; 2; 1/ D nullspace for 3 equation in R4 has dimension 1. 37 The subspace of matrices that have AS D SA has dimension three. 38 (a) No, 2 vectors don’t span R3 (b) No, 4 vectors in R3 are dependent (c) Yes, a
basis (d) No, these three vectors are dependent 39 If the 5 by 5 matrix Œ A b is invertible, b is not a combination of the columns of A. 40 If Œ A b is singular, and the 4 columns of A are independent, b is a combination of
those columns. In this case Ax D b has a solution.
(a) The functions y D sin x , y D cos x , y D e x , y D e
to d 4 y=dx 4 D y.x/. x are a basis for solutions (b) A particular solution to d 4 y=dx 4 D y.x/ C 1 is y.x/ D 1. The complete
solution is y.x/ D 1 C c; sin x C c2 cos x C c3 e x C c4 e x (or use another basis
for the nullspace of the 4th derivative).
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The six P ’s
1C
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41 I D 1
C
.
.
are dependent
1
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Those ﬁve are independent: The 4th has P11 D 1 and cannot be a combination of the
others. Then the 2nd cannot be (from P32 D 1) and also 5th (P32 D 1). Continuing,
a nonzero combination of all ﬁve could not be zero. Further challenge: How...
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 Spring '12
 Minki
 Mass

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