§3.5
Hoff man man & Kunze Solutions p. 105 # 2 Let B = {α1 , α2 , α3 } be a basis for Find the dual basis of B B .
3
defined by: α1 = (1, (1, 0, 1), α2 = (1, (1, 1, 1), and α3 = (2, (2, 2, 0).
C
−
Let B = { g1 , g2 , g3 } be the dual basis for B and {f 1 , f 2 , f 3 } be the dual basis for the standard basis. Then: g1 = a 1 f 1 + b1 f 2 + c1 f 3 . Hence we have the following system of equations: ∗
g1 (α1 ) = g1 (α2 ) = g1 (α3 ) =
a1 c1 a1 + b1 + c1 2a1 + 2b 2 b1
=1 =0 =0
−
The matrix for this system is:
−1 1
1 A = 1
1 1 0 −→ R = 0
0 1 1 1
Now we have g have g 1 = f = f 1
0
− f . 2
A similar solution shows g 2 = f 1
0 0 1 0 0 0 1
0
− f + f 2
and g and g 3 =
3
−
1 2
f 1 + f 2
−
1 −1 . 0
1 2
f 3 .
p. 105 #3 If A and B are n n matrices over the field F , F , show that trace(AB trace( AB)) = trace(BA trace( BA). ). No Now w show show that similar matrices have the same trace.
×
n k=1
ab a trace(AB trace(AB)) =
Let C Let C = AB = AB = (cij ) Then c Then c ij =
n
ik kj .
n
n
ik bki
i=1 k=1
n k=1 aik bki .
b =
So c So c ii =
n
ki aik =
−1
∃
AP . AP . trace(B trace(B ) = trace(P trace( P 1 AP ) AP ) = trace(AP trace(AP
p. 105 # 4 Let V Let V be the vector vector space of all polynomial polynomial functions functions from c0 + c1 x + c2 x2 . Define three linear functionals on V by 1
f 1 ( p) p) =
trace(BA trace(BA)).
k=1 i=1
Now if A is A is similar to B then B then P such such that B that B = = P P trace(A trace(A).
Now:
−
R to R which
2
p( p(x) dx,
f2 ( p) p) =
0
0
p( p(x) dx,
p( p(x) dx.
0
Show that { f 1, f 2 , f 3 } is a basis for V by exibiting a basis for V of V of which it is the dual. ∗
We wish to find a basis B = { p1 , p2, p3 } V V such that { f 1, f 2 , f 3} = B . p1 (x) = c 10 + c11x + c12x2 Let p2 (x) = c 20 + c21x + c22x2 then we have: 2 p3 (x) = c 30 + c31x + c32x
∗
⊂
1
P ) P ) =
have degree 2 or less: p(x) = −1
f3 ( p) p) =
−1
§3.5
Hoff man man & Kunze Solutions f 1 ( p1 )
= c10x +
c11
f 2 ( p1 )
= c10x +
c11
f 3 ( p1 )
= c10x +
2
2
c11 2
x2 +
c12
x2 +
c12
x2 +
3
3
c12 3
This linear system yeilds the following matrix:
1 A = 2
1/2 2 1 1/2
−
1
x = x = x3 0 = 3 2 0
3
−1
0
c10 +
c12
+
2
3
2c10 + 2c 2 c11 +
−c
10
1 −→ R = 0 0
1/3 1 8/3 0 1/3 0
−
c11
+
c11 2
0 0 1 0 0 1
−
=1
8c12 3
c12 3
=0 =0
1 1 . −3/2
Now we have p have p 1 (x) = 1 + x 23 x2 . Similar solutions show that: p2 (x) = 16 + 21 x2 and p and p 3 (x) = 13 + x 21 x2 . A simple check will show that f i ( pj ) = δ ij required. ed. It remains remains to be shown shown that { p1 , p2 , p3 } is a ij as requir basis for V for V .. c1 c2/6 c3 /3 = 0 Suppose that c that c 1 p1 + c2 p2 + c3 p3 = 0. This gives: c1 + c3 = 0 . This homogeneous linear 3c1 /2 + c2 /2 c3 /2 = 0 system yields the following matrix:
−
A =
−
1 1 3/2
−
−1/6 0 1 /2
−
−
− − − − 1 0 −1/3 0 1 0 −→ R = 0 1 −1/2 0 0 0
.
0 0 0 0 1 0
Hence this system has only the trivial solution. Now { p1 , p2 , p3 } is linearly independent, and since dim V = 3, it must be a basis.
p. 105 # 5 If A A and B are n are n
× n complex matrices, show that AB − BA = BA = I I is is impossible.
Evedently trace(A trace(A + B ) = traceA trace A + traceB traceB , and by # 3, we have trace( AB) AB ) = traceA traceAtraceB traceB . Suppose now that AB that AB BA = BA = I I ,, Then trace(AB trace( AB)) trace(BA trace(BA)) = traceI traceI therefore 0 = traceI traceI = n
−
−
→←
p. 105 # 6 Let n Let n and and m m be positive integers and F a F a field. Let f Let f 1, . . . , fm be linear functionals on F n . For α F n , define T ( T (α) = (f 1 (α), . . . , fm (α)). Show Show that T is T is a linear transformation from F n into F m . Then Then show show n m that every linear transformation from F into F into F is of the above form, for some f 1 , . . . , fm .
∈
Linearity is trivial. Let L Let L : : F F n F m be linear. Let { π1 , . . . πm } be the dual to the standard basis on F m . For 1 j m n and α F , define f j (α) = πj L(α). Then Then if if L(α) = (y1 , . . . ym ), we have f j (α) = yj . Henc Hencee T ( T (α) = (f 1 (α), . . . , fm (α)).
∈
−→
≤ ≤
◦
2
§3.5
Hoff man man & Kunze Solutions
p. 105 # 7 Let α1 = (1, (1, 0, 1, 2) and α2 = (2, (2, 3, 1, 1), and let W be W be the subspace of R4 spanned by α1 and α2 . Which linear functionals f : f :
−
f (x1 , x2 , x3 , x4 ) = c 1 x1 + c2 x2 + c3 x3 + c4 x4 are in the annihilator of W of W ?? ◦
If f
∈ ∈ W
then we have the following homogeneous linear system:
f (f (
α1 ) = c 1 c3 + 2c 2 c4 = 0 f ( f (α2 ) = 2c1 + 3c 3 c2 + c3 + c4 = 0
−
which gives the following matrix: A =
1
0 2 3
−1
1 −→ R = 0
2 0 1 0
1
0 1
− .
−1
2 0 1 0
1
Now we have c have c 1 = c 3 2c4 and c and c 2 = c 3 c4 . So if f f W , then f then f ((x) = (a 2b)x1 + (a ( a b)x2 + ax3 + bx4 for some real numbers a and b. b . By theorem 16, dim W + + dim W = 4. Since the space of functions functions defined above has dimension 2 in R , it is equal to W .
−
−
∈ ∈
∗
◦
−
◦
−
◦
p. 106 # 9 Let V Let V be be the vector space of all n
× n matrices over R, and let
B =
2
−1
−2
C = =
1
0 0 0 1
.
Let W be W be the subspace of V of V consisting consisting of all A suct that AB = AB = 0. 0. Let Let f be f be a linear functional on V V with f W . Suppose that f ( f (I ) = 0 and f and f ((C ) = 3, where I is I is the 2 2 identity matrix. Find f ( f (B ). ◦
∈ ∈
×
Let E 1 = Now I Now I = E 1 + C
Now 0 = f = f 0 = f
1 0 0 0
ther theref efor oree
E 2 =
E 3 =
0 0
0 0 1 0
.
0 = f ( f (E 1) + f ( f (C ) = f ( f (E 1 ) + 3 theref therefore ore f (E 1 ) =
1 2 2 −2 0 0 −1 1 = 0 1 2 0 0
0 1
= f ( f (E 1 ) + 2f 2 f (E 2 ) =
and
0 0 2 1 2
−3 + 2f 2f ((E )
−1
−3. Observe that:
−2 = 0. 0. 1
ther theref efor oree f ( f (E 2 ) = 3/2.
2
0 0
= f ( f (E 3 ) + 2f 2 f (C ) = f ( f (E 3 ) + 6 theref therefore ore f ( f (E 3 ) = 6. 1 2 Now B = 2E 1 2E 2 E 3 + E + E 4. Hence f Hence f ((B ) = 2f ( f (E 1 ) 2f (E 2 ) f ( f (E 3 ) + f + f ((C ) = 2( 3) ( 6) + 3 = 6.
−
−
−
−
−
− − 2(3/ 2(3/2) −
−
p. 106 #11 Let W Let W 1 and W and W 2 be subspaces of a finite dimensional vector space V . V . a) Prove that (W (W 1 + W 2 ) = W 1 W 2 . b) Prove that (W ( W 1 W 2 ) = W 1 + W 2 . ◦
◦
◦
◦
◦
◦
∩
∩ ∈ (W +W ) and w ∈ W and v ∈ W . Then w, v ∈ W +W . Hence f ( a) Suppose that f ∈ f (w) = f ( f (v) = 0. So f So f ∈ ∈ W ∩ W . Now (W (W + W ) ⊆ W ∩ W . 1
◦
1
◦
2
1
2
2
◦
◦
1
◦
1
2
◦
2
3
1
2
§3.5
Hoff man man & Kunze Solutions Now suppose that f W 1 W 2 and u W 1 + W + W 2 , then w W 1 , v f ( f (u) = f ( f (w) + f ( f (v) = 0. Hence Hence f f (W 1 + W 2 ) and W and W 1 W 2 (W 1 + W 2 ) .
∩ ∈ ∃ ∈ ∈ √ W such that u = w + v + v.. ∈ ∈ ∩ ⊆ b) Suppose that f ∈ (W ∩ W ) . Let α = {α , . . . , α } be a basis for W ∩ W , and choose β = {β , . . . , β }, γ = {γ , . . . , γ } so that α ∪ β is a basis for W and α ∪ γ is a basis for W . No Note te tha thatt ∪ γ is linearly independent and hence can be extended to a basis of V , the set α ∪ β ∪ V , say by the vectors δ = = {δ . . . , δ }. Now define f define f by f by f (α ) = f ( f (α ) = 0, f 0, f (β ) = 0, f 0, f (γ ) = f ( f (γ ), and f and f (δ ) = f ( f (δ ). Define f by f by f (α ) = f ( f (α ) = 0, f (β ) = f ( f (β ), f ), f (γ ) = 0, and f (δ ) = 0. f ∈ W , f ∈ W , and f and f = f + f + f , hence (W (W ∩ W ) ⊆ W + W . Now suppose that f ∈ ∈ W + W and u ∈ W ∩ W . f = g + h + h where g√ ∈ W and h ∈ W . Henc Hencee f ( f (u) = g( g (u) + h(u) = 0. Now f Now f ∈ ∈ (W ∩ W ) and W and W + W ⊆ (W ∩ W ) . ◦
∈
◦
◦
1
m
1
i
1
2
◦
r
1
i
1
i
2
◦
2
◦
1
2
1
1
i
2
◦
2
◦
s
1
t
1
2
◦
i
i
i
1
i
2
2
1
1
i
i
i
◦
1
1
i
1
i
◦
2
1
2
2
◦
1
2
◦
1
◦
1
2
1
2
◦
◦
2
◦
1
◦
1
◦
2
1
2
2
◦
p. 106 # 12 Let V Let V be be a finite dimensional vector space over the field F and F and let W let W be be a subspace of V . V . If f is f is a linear functional on W on W ,, prove that there is a linear functional g on V on V such that g that g((α) = f ( f (α) for each α W . W .
∈
Let { α1, · · · , αm} be a basis for W . W . Choose αm+1, · · · , αn so that { α1 , · · · , αm , αm+1, . . . αn } is a basis for V . V . Let {f 1 , . . . , fm , f m+1 . . . , fn } be the dual basis basis.. No Now w f = f (α1 )f 1 + · · · + f + f ((αm )f m . Defin Definee g = f ( f (α1 )f 1 + · · · + f ( f (αm )f m + 0f 0 f m+1 + · · · + 0f 0f n .
p. 106 # 13 Let F F be a subfield of C, and let V V be any vectorspace over F . F . Suppo Suppose se tha thatt f and g are linear functionals on V V such that the function h defined by h(α) = f ( f (α)g (α) is also a linear on V . V . Prove Prove that that either f either f = = 0 or g = 0. Suppose that neither f nor g is 0, and let dim V = n. Then Then dimker dimker f f = dim dim ker g = n ker f ker g = V . V . ‡ Let α V \(ker f ker g). Now suppose that h that h is linear. Then
∪ ∪
∈
∪ ∪
− 1.
Henc Hencee
2h(α) = h(2 h (2α) = f (2 f (2α)g(2α) = 4f ( f (α)g (α) = 4h(α). But this implies that h( h (α) = 0 (F ( F must must have characteristic 0), contradicting the fact that f ( f (α) = 0 and g (α) = 0. Hence h Hence h is not linear.
p. 106 # 14 Let F Let F be be a field of characteristic 0, and let V be V be a finite dimensional vector space over F . F . If α α 1 , . . . , αm are finitely many vectors in V , V , each diff erent erent from the zero vector, prove that there is a linear functional f on V on V such that f that f ((αi ) = 0, 1 i m.
≤ ≤
Let α = {α1 , . . . , αm }, and define an equivalence relation on α by αi αj i ff λ F such such that λαi = α j . Now α is a disjoint union of equivalence classes, say α 1, . . . , αk . For 1 i k choose β i αi and extend β i to a basis for V for V ,, say { β i , β i2 , . . . , β in i k define the functional f i by f by f i (β i ) = 1, f i (β ij in }. For 1 ij ) = 0 for 2 j n. Now f Now f i (αj ) = 0 if and only if α j αi . The functional f = in=1 f i has the required property.
≤ ≤
≤ ≤ ∈
W and W and W are subspaces of V , V , then W then W ‡ If W
∼ ≤ ≤
∃ ∈
∈
∪ ∪ W
is a subspace 4
⇔ W ⊆ W
or W or W
⊆ W . W .
§3.5
Hoff man man & Kunze Solutions p. 106 # 15 Let V Let V be be the space of 2 2 matrices over the field F and F and let P be P be a fixed 2 linear linear operator operator on V on V defined by T by T ((A) = P A. Prove that trace(T trace( T )) = 2traceP 2traceP ..
×
× 2 matrix. Let T be T be the
The matrices E ij i, j 2 where the i, j entry is 1, and all other entries are 0 form a basis for V . ij 1 The matrix for T in T in this basis is:
≤ ≤
p p0 B =
11 21
0
p12 p22 0 0
0 0 p11 p12
0 0 p12 p22
.
Now traceT traceT = = traceB traceB = 2traceP 2traceP ..
p. 107 # 16 Show that the trace functional on the n n matrices is unique in the following sense. If W If W is is the space of n that f ((AB) each A n n matrices over the field F and F and if f is f is a linear functional on W such W such that f AB ) = f ( f (BA) BA) for each A and B and B in W in W ,, then f then f is is a scaler multiple of the trace function. If, in addition, f ( f (I ) = n, n , then f then f is is the trace function.
×
×
The matrices E ij i, j n where the i, j entry is 1, and all other entries are 0 form a basis for W . W . ij 1 It is easy to verify the following facts: (1) For 1 i n, E i1 E 1i = E ii and E 1i E i1 = E = E 11 11 ii and E (2) For 1 i n and 1 k n i, E i,i i,i+k E i,i i,i k = 0 and E i,i i,i k E i,i i,i+k = E i,i i,i+k n Now (1) i f ( f (E ii f (E 11 if i if i = j, then f then f ((E ij 0. If A If A = (aij ) then A = i,j a E ij 11 ) and (2) ii ) = f ( ij ) = 0. =1 ij ij n n Hence f Hence f ((A) = i,j=1 aij f ( f (E ij f (E 11 f (E 11 )trace(A). 11 ) 11 )trace(A ij ) = f ( k=1 akk = f ( Suppose now that f that f ((I ) = n. necessarily true that f = n . It is not necessarily f = trace, but if we assume further that it is not the case that k | n where k where k is the characteristic of F of F then then we have: n = f = f ((E 11 )trace(I ) = nf ( nf (E 11 11 )trace(I 11 ) f ( f (E 11 ) = 1. So f = f ( f ( E )trace = trace. 11 11 11
≤ ≤
⇒ ∀
≤ ≤ ≤ ≤
≤ ≤ −
−
−
⇒
⇒
To see that the extra assumption was necessary let F = Z2 and f and f
≡ ≡ 0, then
1 0
0 = f but since trace
1 0 0 0
= 1, 1,
0 1
= 1 + 1 = trace
1 0 0 1
,
f = trace.
p. 107 #17 Let W Let W be be the vector space of n the matrices C of C of the form C = AB trace 0.
× n matrices over the field F , F , and let W let W be the subspace spanned by − BA. BA . Prove Prove that W is exactly the subspace of matrices which have 0
0
By #3 we see that if C if C W 0 , then trace(C trace(C ) = 0 so W 0 ker(trace). In # 16 we defined the matrix ‘units’ E ij i, j n, and observed that for 1 i n and ij for 1 1 k n i, E i,i and E i,i Hence E i,i E i,i i,i+k E i,i i,i k = 0 and E i,i k E i,i i,i+k = E i,i i,i+k . Hence E i,i+k = E i,i i,i k E i,i i,i+k i,i+k E i,i i,i k = 0, so we have that if i = j , then E ij W 0 . In # 16 we also also observ observed ed that that for 1 i n, E i1 E 1i = E ii ij ii and E 1i E i1 = E 11 . No Now w for fo r 2 k n define A to be the matrix with the 1,1 entry equal to 1, the k, k 11 k entry equal to 1 and all other entries equal to 0, then trace Ak = 0, and by the above observation we have Ak = E 1i E i1 E i1 E 1i , hence A hence A k W 0 . It’s clear that the set {E ij k n} is linearly independent and there are n2 1 {Ak : 2 ij : i = j } elements in this set. Now W 0 = ker(trace), as required.
∈ ∈
≤ ≤ −
−
− −
≤
−
∈ ≤ ≤ ∈ ∪
≤ ≤ 5
⊆ ≤
−
− ≤ ≤
≤ ≤
−
−
§3.7
Hoff man man & Kunze Solutions p. 115 # 1
Let F Let F be be a field and let f be f be the linear functional on F 2 defined by f by f ((x1 , x2 ) = ax 1 + bx2 . For each of the following operators, T , T , let g let g = T = T t f and f and find g find g((x1 , x2 ). a) T a) T ((x1 , x2 ) = (x1 , 0). g (x1 , x2 ) = (T t f )( f )(x x1 , x2 ) = f ( f (T ( T (x1 , x2 )) = ax 1. b) T b) T ((x1 , x2 ) = ( x2 , x1 ).
−
g (x1 , x2 ) = (T t f )( f )(x x1 , x2 ) = f ( f (T ( T (x1 , x2 )) = c) T c) T ((x1 , x2 ) = (x1 t
−ax + bx . 2
− x , x + x ). 2
1
1
2
g (x1 , x2 ) = (T f )( f )(x x1 , x2 ) = f ( f (T ( T (x1 , x2 )) = a( a (x1
− x ) + b(x + x ). 2
1
2
p. 115 # 2 Let V Let V be the vector space of polynomial polynomial functions functions over be the linear functional on V V definde by:
R.
Let a Let a and b be fixed real numbers and let f
b
f ( f ( p) p) =
p( p(x)dx.
a
If D D is the di ff erentiation erentiation operator on V , V , what is D t f ? f ? (Dt f )( f )( p) p) = f ( f (D( p)) p)) =
b a
p (x)dx = dx = p p((b) − p( p(a).
p. 115 # 3 Let V Let V be be the vector space of all n n matrices over the field F , F , and let B be a fixed n fixed n n matrix. If T is T is the linear operator on V V defined by T by T ((A) = AB BA, BA, and if f is f is the trace function, what if T if T t f ? f ?
×
(T t f ) f )A = f = f ((T A) = trace(AB trace( AB
×
−
t
− BA) BA) = 0. Hence T f = 0.
p. 115 # 4 Let V Let V be a finite–dim finite–dimensio ensional nal vector vector space over over the field F and F and let T be T be a linear operator on V . V . Let c Let c be a scaler and suppose there is a non-zero vector α in V V such that T that T α = c = c α. Prove that there is a non-zero linear functional f on V on V such that T that T t f = cf . cf . Let dim V = n. n . Extend α to a basis B = {α, α2 , . . . , αn } and let B = {f, f 2 , . . . , fn } be the dual basis. Let x Let x V . V . Then x Then x = = f f ((x)α + in=2 f i (x)αi so (T t f )( f )(x x) = f ( f (T x) = f ( f (x)f (α) + in=2 f i (x)f ( f (αi ) = cf ( cf (x). ∗
∈
p. 116 # 5 Let A Let A be an m an m
t
× n matrix with real entries. Prove that A = 0 if and only if A if A A = 0.
( ) Trivial.
⇒ (⇐)
Let B Let B = A = A t A then b then b ii =
m t k=1 aik aki =
m k=1
aki aki =
6
m 2 k=1 aki =
0
⇔a
ki =
0 k i.
∀ ∀
§3.7
Hoff man man & Kunze Solutions
p. 116 # 6 Let n be a positive integer and let V be V be the space of all polynomial functions over the field R which have degree at most n. n . Let D be the diff erentiation erentiation operator on V . V . Find Find a basis basis for the null null space space of the t transpose operator D operator D .
{1, x , . . . , x n } is a basis for V . V . Let {f 0 , f 1 , . . . , fn } be the dual dual basis. basis. The image image of D of D is the collection of polynomials with degree at most n 1, hence deg(D deg(D(V )) V )) = n. n . If f is f is a functional, then dim(ker f ) = n or f or f = = 0. Now f Now f ker Dt D(V ) V ) ker f . f .
∈ ∈
⇔
⊆
−
p. 116 # 7 Let V be V be a finite-dimensional vector space over the field F . F . Show Show that that T L(V, V ) V ) onto L onto L((V , V ). ∗
t
→ T
∗
is an isomorphism of
Suppose that T L(V, V ) V ) and T t = 0, then for every functional f , f , T t (f ) f ) = 0, hence f T T = 0. We claim that this implies that T T = 0. Suppose Suppose that that α such that T α = 0, then we may choose β 2 , . . . , β n such that {T α, β 2 , . . . , β n } is a basis for V (n = dim V ). V ). Let Let {f 1 , . . . , fn } be the dual basis. T t (f 1 )(α) = t f ( f (T ( T (α)) = 1 = 0, contradicting the assumption that T = 0. 0. Now T T t is an inject injection ion.. dim V = dim V dim L(V, V ) V ) = dim L(V , V ). Hence, T Hence, T T t is an isomorphis isomorphism. m.
∈ ∈
∗
⇒
∃
∗
∗
◦ ◦
→
→
p. 116 # 8 Let V Let V be be the vector space of n of n n matrices over the field F . F . a) If B on V by f B (A) = trace(B trace( B t A). Show that that f B is a fixed n n matrix, define a function f B on V f B is a linear functional on V . V . b) Show that every linear functional on V is V is of the above form. c) Show that B that B f B is an isomorphism of V of V onto V onto V .
×
×
∗
→
a) f a) f B (A + λC ) = trace(B trace(B t (A + λC )) )) = trace(B trace(B t A + λB t C ) = traceB traceB t A + λtraceB traceB t C .
√
b) For 1 i, k n let E ik V be V be the matrix with i, j entry 1 and all other entries entries 0. These matrices matrices ik form a basis for V . V . If f is f is a functional on V and M and M = (m ( mik ) V then V then we have:
≤ ≤
∈
∈
n
n ik ik
ik
11 11
ik ik
ik
k=1 i=1
i,k=1
f (E ) . . . ... Let B Let B =
n
f (f (E )m = f (E )m . f ( f (M ) = (†) f ( E ) .. . Then (B ( B M ) = f ( f (E )m . A comparison with ( †) shows that . 1n
t
n i=1
kk
ik ik
ik
f ( f (E n1 ) . . . f ( E nn nn ) t f ( f (M ) M ) = trace(B trace( B M ). ).
√
th c) If A = A = (aij ) is any matrix, then AE ij column is equal to the i th column of A A ij is the matrix whose j and all other entries are 0, hence the only non-zero entry on the diagonal is a ji . t Suppose now that for a matrix B , f B = 0. Then Then 0 = f = f B (E ij Now w the ij ) = B E ij ij = b ij , hence B = 0. No map B map B f B is an injection and hence an isomorphism since dim V = V = dim V . ∗
→
7
√
