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Unformatted text preview: MAW (O3 til/J5 Soil/H9145 14.1’es I I / i
In {0,0= E]5 . ..,0, . . .) converges to 0. I Iflilnnqm In = 0 and iil'ﬂnﬁoo Z 0‘ then “lug—109':ng + = Hnlrtwoo In + “mt! mo y‘ri : II If limﬂhm z.“ = U and t". is any constant, then lil1‘l,._.m[las:?.) — Mimi,th inn = 0. '30. TlNeare lookingch the matrices a hisuehthat 2 a ‘5 = o+2e b+2d
c ci‘J .3 6 c d [I E]
[0 0]. It is required that a 2 —2e and b =
—2e —2d;I—20 d0 —2 c d ‘° 1 n " 0 1‘ , —2 0 — .
'Ihus [ l U] , i3 3. basis, and the dimension is 2‘ —2n’. Thus the general element is 4!]. Let V be the space of all n x n matrices A such that At?“ = [3. We look at some possibilities
for 17". H1? = 0', then any matrix A will work, and dim{V) = n2. Now aesnme that f." 9'5 OT
and suppose that the it" component it; is nonzero. If we denote the columns of A by
131, . . . , on, then the condition Afr: ﬁmeans that mail +1 "f‘iJ'gtﬁgF +'e,1n'fR = (T. We
can solve this relation ﬁor 131 and express nig in terms of the other as — 1 column vectors
of A. This means that we can choose it — 1 columns {or nfn — 1) entries) of A ﬁ‘eely; the
column 131 is then determined by these choices. Thus dim{V} = nﬁt — l) in this case. In
summary, We have din1(V) = n2 if 17" = 5, and dimﬂf} = n{n — 1) otherwise. 58. a. Let g{$} be a function in V. Thus, 9’12} r —g(:r)_ New, if ﬁx) 2 g{:t')2 +g’fs)2,
thGIl fl?) = QCQU‘NE’WJ 1' Qiﬂiliiigull'i = glglﬂiﬂifﬂii " 2(s{m))s'($) = 3‘ 30:
f {1*} = ms}2 + g’{x)2 is a oonstsnt function. It). Let 9(3) be a function in V such that gm} '._'. 910) = 0, 11mm part a we know that
9(3)? + 31332 = k, a constant. Now 5{0}2 +910}? = 03 + [)2 —_— I], so that g; 2 u_ The
equation 9(2er + g’(:r)2 —— 0 means that gm) 2 g’[;;) = {J for all 2:. as claimed. c. First we note that ﬂat} 2 ﬁx) — ﬂﬂ} cost*2) — f’w) sight] is in V, since the functions
ﬂashcoslfx} and sinfs) are all in V1 and V is a subspace of F{R,R}. Note that 9(0) :.
H0} — HUNG9(0)  ﬂ“) 5511(0) = flOi  flU] = 0 A130, s’ls) = f’lxl + foJ Sinlﬂi) —
f’ﬂi) cos(:t}, so that 9"(0) = f’m} — f’m] = I]. By part b, we can conclude that gl’l‘} 1 D for all s, so that f(x) = f{0jeos(;c} +
f’(0}sin(:e), as claimed; ' 3:: + s: 3t+ es "“ 4. Fails to be linear, Since T(2Ig) = det(2hj = 4 does not equal
21112} = 2 detfl’g) = 2. 16. Lirrear,sinceT(M+NJ=(.M+N) 2]{M+N)=M[2 0],”? a} i§,§lMl3 new: 3H3 lime 5H: filmmm _ I 2 n' 3 u 2 o 3 o
A . I: _ _ _.1 ,
lso.T[ M) (IaMJI [0 3J [a 4J{kM)—£(M[O 3J~fﬂ 4] M) =kF(M).
To determine whether T is an isomorphism, WC ﬁnd the kernel of T. Now, T[M} =
T([ab]):{a bj20_3ti ob]__2e 3b 3:: 3:!) —o 0
on? gene o4eeJ"2c3d“ ]=[ l 4:: 4d ~2e —d_ ‘
We see that M = [a a] is in the kernel of T if a = c C d = of = 0. Time. the kernel consists
. 0 e . . .
of all 111atr1eee [0 DJ , where b is arbitrary. Since the kernel is nonzero, T fails to be an isomorphism. 'TU. Since the dimensions of the domain and endomejn are equal, it sufﬁces to examine when
the kernel is zero, Now ﬁt) is in the kernel of?“ if ﬂog} = ﬂel) = = f{e,,} = 0. Recall that a nonzero polynorrﬁel of degree 5 n has at most It zeros. If the n+ 1 numbers
(:0, c1, . . . ,en are all different, then the only polynomial in P“ with ﬂog} — f { r31} = =
f(o,,} = O is the zero polynomial, so that the kernel of T is zero, and T is an isomorphism. HoweVer, if some of the numbers are equal, for example, ('1, = (2,3, then there Will be
nonzero polynomials in the kernel (for example, the product of all {t  Ci) where 1 7E q),
so that '1' fails to be an isomorphism. ...
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 Spring '08
 lee
 Math, 4j, 3j, 0 3J

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