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WmmwﬁmﬁLikﬂm7 777 . ..... 7 ........................................................ “"‘”‘""“""””””“S"A”L‘VA‘QE : ”W M, 75‘ #56 ”is" i:(’\ (1+ ' " ”‘ ' ' "‘ '" 7.7. """ W1 MATH 223 PROBLEM SET 3 , ' SOLUTIONS 1. Prove one of the following statements.
(a) If (V1, . , . ,vn) spans V, then so does the list (V1 — V2,V2 ~ V3,  . . ,vn—1 — vmvn) obtained by subtracting from each vector (except the last) the following vector.
PROOF. We need to show that for any v E V, we can write v as a linear combination of (v1 ._ V2, vz — v3, . . . ,vn_1 — V“, V“). We know that (w, . . . ,vn) spans V, so we may write
(1) v : am +   ' +‘anvn.
We are looking for coefficients X1,Xz, . ,xn such that v = X1(V1—V2)+X2(V2—VS)+'~+XnVn. Let’s set
X] — (11
X2 — C12 + 01
X3 — as + Cl2 + CH
xn : an+an_1+~+a1+a1. Then the linear combination x : x1021 — V2) + X2(V2 — V3) + ~ ' ' + xnvn becomes x = al(V1_V2)+(a2“al)(V2—V3)+"‘+(an'—an—1)Vn
= GM—C11V2l(dzlClilvz—((12+C11)V3+~'—(an—1+~~+aiivn+lan+m+a1ivn
(1]V]+"‘+C111VTL V
= v by equation (1). Thus, this linear combination is v, and so this list spans all of V. El (b) If (V1, ~ . . ,vn) is linearly independent in V, then so is the list
(V1 — V2, V2 ~ V3, . . . ,vn_1 —~ v“, V“) obtained by subtracting from each vector (except the last) the following vector. PROOF. To prove that the list is independent, we must take a linear combination adding
11p to 0, and prove that all the coefficients are zero. 80 suppose 0 = bilvi —V2) + b2("2 —V3) + ' ' ' + anw
Then we may rearrange the right hand side to get 0 : blvl — MW ‘1' bZVZ — bZVB “l” ' 'p' "l‘ bn—an—l _ bn—lvn "l’ bnvn
= b1V1Hbz — b1W2 + ' ' ' + (b'n — 1311—1)an But this is a linear combination of the list (v1, . . . ,vn), which we have assumed is linearly
independent. Thus, the coefficients of this linear combination must all be zero. In other MATH 223 PROBLEM SET 3 SOLUTIONS words,
by = 0
b2~b1 _ o
b3b2 — 0
bn "“ th—l : 0.
In other words, b1 : 0, b2 2 b1, b3 2 b2, and so on until we conclude that bn : bn_1 :
= b1 : 0. Thus, we have shown" that the list (v1 — vz,v2 —— V3,. ..,vn_1 — umvn) is linearly independent. [I (c) Suppose that (w, . . .vn) is linearly independent in V, and w E V. If the list (w + w, . . . ,vn —I— w) is linearly dependent, then M) E span(v1, . . . ,vn).
PROOF. By assumption, the list (w + w, . , . ,vn + w) is linearly dependent, which means
that there exist scalars a1, . . . , an E R not all zero so that 0 : CL1(V1+W)+C12(V2+W)+“'+ (Ian—IW)
= mw+aWys~+awW+wvs~+aaw. Because the list (v1 , . . . ,an is linearly independent, the coefficient on w must be non~zero,.
In other words, a1 +  ' ' + an 7; 0. Thus, we can rearrange the equation to get —Mrs~+adw=ast~+awm and since —(a1——~+ an) 75 O, we can divide by—(cn + ~ ~ . + an) 7é Oto get _—g]_—v .I. ..+—___EL_T"—
—(al+"'+an)l _(al+"'+an) In other words, w is a linear combination of (v1, . . . ,vn), so W 6—: sp an(v1, . . . ,vn), which
is what we wanted to show. [:1 W : V“. 2. Find a function f : R2 ~+ R that satisfies f(a  v) = a . f(v) for every a E R and
v E R2, but which is not linear. EXAMPLE. Consider the function ' mm) = v x3 +133 Then “CL. (x,y)) = «3/ a3x3 + a3y3 : av \3/x3 + y3, so this satisfies homogeneity. However,
“(0’2” : {/51 “(290” : {7—8—1 bUt “(0:2) + (2)0» : f((2>2)) : W 7é ‘73 + % Thus! this function does not satisfy the additivity property, and hence is not linear. <> 3. For V and W subspaces of IR“, pfOVe 3that
(a) V D Wis a subspace of R“; and
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‘ every v E V and w E W. Given an m X n matrix A, let R(A) denote the subspace of
JR“ spanned by the rows of A (thought of, of course, as column vectors!). Prove that
RM) is orthogonal to the kernel of A. Deduce that the image of A is orthogonal to
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V+W={v+wlv€V, WEW}. (a) Prove that V + W is a subspace of R“. b What is V + V?
((c; Prove or give a counterexample to the following statement: If V1 , V2, W Q R“ are subspaces such that
V1 + W = V2 + W then V1 2 V2. ‘ ‘ ‘
(d) What can you say about dim(Vl1‘ + V2)? l
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