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Unformatted text preview: EXERCISE SET 5.2 [I ti El 0
F]—f[?]—irl]+[—l]=reg1ud63=[gld[ﬂ—h[ﬂ:—b::]=TE3;thlﬁ{Tr]=
I) 0 Li H l —l —1 —
—1CI I'J
[0—1 III].
11' lIii—1 lﬁ. {a} r" 26. The matrix A is a rotation rnatrhr since it is orthogonal and det{A] = l. The axis of rotation is —1 I) .1:
found by solving the system [I  A}: = ti, which is [ ﬂ 1 —1: [a] = Fl. A general solution
1 {I I. s o r
of this system is x = [a] = t [I]; thus the axis of rotation is the line passing through the origin
1
and the point (111.1}. The plane passing through the origin that is perpendicular to this line has
equation : +y + z = ﬂ, and w = {4,1113} is a vector in this plane. Writing w in eolurnn vector
form, we have 1 [I 1 {l —'l l l
w = 1 1 Aw = I} {l l l = IJ , w x Aw = l
[i 1 Li I] [l' —1 1
Thus the rotation angle 5 relative to the orientation u = w x Aw = [11 1,1} is determined by
eosﬁ' = ﬁ‘ﬁ” = JEL, and so 9: “T'ilﬂﬂﬁ. D3. The two columns of this matrix are orthogonal for any values of a and b. Thus1 for the matrix to
he I::rthop_1:onal_.2 all that is required is that the column vectors he of length 1. Thus a and 5 must
eatrefy [a + tr} + (a — M2 = l, or [eqtuvalentlﬂ 2n2 + 2.53 = I. Dr. From the polarization identity. we have x . y = HM}: + y"? — “x — gin”) = the m o = s. EXERCISE SET 6.3 . . 1 r x
E. The kernel of T Is equal to the solution spaoe of the system [I “I [y] = [3]. The augmented
ii'..' . . 1 a i '0
matrix of this system can he reduced to [0 l D E 0]; thus keriT} consists of all vectors of the form —]
$=t[ 0] where—meteor).
1 12. The vector w is in the range of the linear operator T if and on];r if the linear system 3—3; = i
I+y+z= 2
5..“ +2z=—1 is consistent. The augmented matrix of this system can he reduoed to long
i 010:; stung Thusthe.stemhasauniue [' =1g_§ _ T q 50L].th H: agar all! and “'2' have EllL .—s = 19. {a} The linear transformation T; : R: —> RE ha onetoone if and only if the linear system An = II}
has only the trivial solution. The augmented matrix of Ar: = {I is 11:e
s sin
sMe and the redunei rottr echelon form of this matrix is
1 1]
I] l}
I] I} alll: From this we ooneluele that A: = D has onl}r the trivial solutionT and so T1 is onetoone.
{In} The augmented matrix of the system Ax = I} is [ 1 2 s 1 I]
—1 l} 4 i [I
and the reduced row echelon form of this matrix is
1 o 4 : o
o 1 —% i o
"4
From this are eonelude that Ax = [I has the general solution x = t i . In particular, the
1 spatem Ax = D has nontrivial solutions and so the transformation TA : R3 r R2 is not one
to—nne. D1. {a} True. If T is onetoone, then T): = a if and oni},r if x = I]; thus Tin — vi = 0 implies
n — tr 2 ﬂ and u = o.
(h) True. If T . R“ —t R" is onto1 then {from Theorem $3.14} it is oneto—one and so the argu
ment givmi in part {a} applies.
{c} True. See Theorem 6.3.1.5.
(:1) True. If TA is not oneto—one. then the heterogeneous linear system Air = D has infinitelyr
many nontrivial solutions. e True. The standard matrix of a shear operator T is of the form A = l k or A = J u . In
e 1 k 1
either case, we hate det{A} = 1 of El and so ’1' = TA is oneto—one. P1. If B}: = ID. then [213]): = Elﬁn} = At] = I}; thus x is in the nullspaoe of AB. ...
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 Spring '08
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