Solutions to Chapter 2
2.1 (a) / H-> / ' does not define a linear functional since / ' ^ IR in general.
(b),(c),(d) These are linear functional.
(e) $ : / — i > Jo f2 is not a linear mapping; for example, we have
0 = &[f + (—/)] whereas in general
= 2
2.2 That £> is linear follows from the fact that
= f fo(t)[af(t) + Pg(t)}dt
Jo
[ fo(t)f(t)dt + (3 f fo(t)g(t)dt
Jo Jo
= a<p{f) + f3<p{g).
2.S The transition matrix from the given basis to the standard basis is
1 -1 0
p = 0 1 1
-1 0 1
The inverse of this is readily seen to be
2 2 2
1 i __I
"2 2 2
II I
2 2 2
,Book 4 Linear algebra
Hence the dual basis is
U2'2' 2 J' L 2 ' 2 ' 2 J ' 1 2 » 2 ' 2 JJ '
2.4 The transition matrix from the basis A1 to the basis A is
Its inverse is
Consequently, (A') d = {-2^>i + § |
2.5 The transition matrix from the given basis to the standard basis is
Its inverse is
so the dual basis is (b), namely {[-1,0], [2,1]}.
2.6 (i) {[2, -1,1,0], [7, - 3 , 1 , - 1 ] , [-10, 5, -2,1], [-8,3, -3,1]};
(ii) {(4,5, -2,11), (3,4, -2,6), (2,3, -1,4), (0,0,0,1)}.
2.7 Since V = A 0 B we have
i H B ^ f i n s)1 = {o}1 = vd
A±nB± = {A + B)1 = VL = {0}.
Consequently, Vd = A1 0 i?-1.
The answer to the question is eno' : Ad is the set of linear functional
/ : A -> F so if A ^ V we have that Ad is not a subset of Vd. What is
true is : if V = A 0 B then Vd = A1 0 5 ' where A',B' are subspaces
of F d with A1 ~jLd and £ ' - £ d . To see this, let f e Ad and define
/ : 7 -> F by /(v) = f{a) where t; = a + b. Then p : Ad -* 7 d
given by £>(/) = / is an injective linear transformation and <p(Ad) is a
68
, Solutions to Chapter 2
subspace of Vd that is isomorphic to Ad. Define similarly fi : Bd —• Vd
by fi(g) = # where 'g(v) = g(b). Then we have
{/i) /2J /s} is linearly independent. For, if + A + A3/3 = 0 then
we have
0 = {Xifi + A + A )(l) = Ax + A2 + A3;
0 = (Ai/i + A + A3/3X-X) = Xih + A2*2 +
0 = {Xifi + A + A )(X 2 ) = X1t21 + A2*l
Since the coefficient matrix is the Vandermonde matrix and since the t{
are given to be distinct, the only solution is Ai = A2 = A3 = 0. Hence
{/15/25/3} is linearly independent and so forms a basis for (IR3[X])d.
If {pi,p 2 ) p 3 } is a basis of V of which {/1, / 2 ) / 3 } is the dual then we
must have f%(p3-) = 6,-y; i.e.
t) = Sty-
is now easily seen that
(X-t2)(X-t3)
=ll;
= 3 4
+3
+3 = 17a+ 226.
69
2.1 (a) / H-> / ' does not define a linear functional since / ' ^ IR in general.
(b),(c),(d) These are linear functional.
(e) $ : / — i > Jo f2 is not a linear mapping; for example, we have
0 = &[f + (—/)] whereas in general
= 2
2.2 That £> is linear follows from the fact that
= f fo(t)[af(t) + Pg(t)}dt
Jo
[ fo(t)f(t)dt + (3 f fo(t)g(t)dt
Jo Jo
= a<p{f) + f3<p{g).
2.S The transition matrix from the given basis to the standard basis is
1 -1 0
p = 0 1 1
-1 0 1
The inverse of this is readily seen to be
2 2 2
1 i __I
"2 2 2
II I
2 2 2
,Book 4 Linear algebra
Hence the dual basis is
U2'2' 2 J' L 2 ' 2 ' 2 J ' 1 2 » 2 ' 2 JJ '
2.4 The transition matrix from the basis A1 to the basis A is
Its inverse is
Consequently, (A') d = {-2^>i + § |
2.5 The transition matrix from the given basis to the standard basis is
Its inverse is
so the dual basis is (b), namely {[-1,0], [2,1]}.
2.6 (i) {[2, -1,1,0], [7, - 3 , 1 , - 1 ] , [-10, 5, -2,1], [-8,3, -3,1]};
(ii) {(4,5, -2,11), (3,4, -2,6), (2,3, -1,4), (0,0,0,1)}.
2.7 Since V = A 0 B we have
i H B ^ f i n s)1 = {o}1 = vd
A±nB± = {A + B)1 = VL = {0}.
Consequently, Vd = A1 0 i?-1.
The answer to the question is eno' : Ad is the set of linear functional
/ : A -> F so if A ^ V we have that Ad is not a subset of Vd. What is
true is : if V = A 0 B then Vd = A1 0 5 ' where A',B' are subspaces
of F d with A1 ~jLd and £ ' - £ d . To see this, let f e Ad and define
/ : 7 -> F by /(v) = f{a) where t; = a + b. Then p : Ad -* 7 d
given by £>(/) = / is an injective linear transformation and <p(Ad) is a
68
, Solutions to Chapter 2
subspace of Vd that is isomorphic to Ad. Define similarly fi : Bd —• Vd
by fi(g) = # where 'g(v) = g(b). Then we have
{/i) /2J /s} is linearly independent. For, if + A + A3/3 = 0 then
we have
0 = {Xifi + A + A )(l) = Ax + A2 + A3;
0 = (Ai/i + A + A3/3X-X) = Xih + A2*2 +
0 = {Xifi + A + A )(X 2 ) = X1t21 + A2*l
Since the coefficient matrix is the Vandermonde matrix and since the t{
are given to be distinct, the only solution is Ai = A2 = A3 = 0. Hence
{/15/25/3} is linearly independent and so forms a basis for (IR3[X])d.
If {pi,p 2 ) p 3 } is a basis of V of which {/1, / 2 ) / 3 } is the dual then we
must have f%(p3-) = 6,-y; i.e.
t) = Sty-
is now easily seen that
(X-t2)(X-t3)
=ll;
= 3 4
+3
+3 = 17a+ 226.
69
