Basic Algebra

Solutions to Chapter 1




1.1
(i)False. For example, take 6 = — a\.
(ii)True.
(iii)False. {(1,1,1)} is a basis, so the dimension is 1.
(iv) False. For example, take A = {0} or A = {v, 2v}.
(v) True.
(vi) False. For example, take A = IRn.
(vii) True.
(viii) False, {(x, Xx) | x G IR} is a subspace of IR2 for every A G IR.
(ix) True,
(x) True,
(xi) False. An isomorphism is always represented by a non-singular
matrix.
(xii) False. Consider, for example, IR2 and C 2 . The statement is true,
however, if the vector spaces have the same ground field.
(xiii) False. is a counter-example,
(xiv) False. For example,

1 0 1 0 1 3" 0
0 0 1 0 0 4 0

(xv) True,
(xvi) True.
(xvii) False. Take, for example, /, g : IRn -> IR* given by f(x, y) = (0,0)
and g(x,y) = (x, y). Relative to the standard basis of IR* we see

,Book 4 Linear algebra

that / is represented by the zero matrix and g is represented by
the identity matrix; and there is no invertible matrix P such that
P ~ P = 0.
(xviii) True,
(xix) False. The transformation t is non-singular (an isomorphism), but
[l 2~|. . .
is singular.
L1 2 J
(xx) False. The matrix is not diagonalisable.

1.2 We have that

(a,6,c) G Ker*i <=> (a + 6,6 + c,c + a) = (0,0,0)



and so Kerti = {0}. It follows from the dimension theorem that Imti =
IR3.
As for t2, we have

(a,6,c) € Kert 2 <=> a - 6 = = O , 6 - c = O
= b =c

and so Ker£2 = {(^? a ) a ) I a G IR}. It is clear from the definition of £2
that Imt 2 = {(a, 6,0) | a,6<ElR}.
Likewise, it is readily seen that
Ker £3 = {0}, Im*3 = IR3,
Ker£4 = {(0,0,a) | a e IR}, lmt4 = {(a,6,6) | a,6GlR}.
If Kerti nlmti = {0} then by the dimension theorem we have

dim(Ker*t- 4- Imt,-) = dimIR3

and so Ker£t- + Im£,- = IR3. Now for i = 1,2,3,4 we have from the above
that Ker*t- n Imt,- = {0}. Thus we see that IR3 = Kert,- 0 Imt,- holds in
all cases.
Im<2 is t3-invariant. For, if v G Imt2 then v = (a, 6,0) and so

h{v) = £3(^,6,0) = (-6,a,0) Glm£ 2 .

However, Ker£2 is not £3-invariant. For (1,1,1) G Ker£2 but £3(1,1,1) =


32

, Solutions to Chapter 1

For the last part, we have that (t3 o £4)(a,6,c) = (-6,a, 6) and that
4 ° *3)(^j &5 c) = (-6, a, a). Consequently,

a e IR};


Ker(*4o*3) = {(0,0, a € IR};
Im(t4ot3) = {(-b,a a,6elR}.

Reducing the matrix to row-echelon form we obtain

3 - 1 1 1 _^ 3~ 1 - 1 3
-1 5-1 -1 5 -1 — • 0 4 2
1 - 1 3 3 -1 1 0 2-8
1 1
3 -1 3
0 2 - 8 —> 2-8
0 4 2 0 18

Note that we have been careful not to divide by any number that is
divisible by either 2 or 3 (since these will be zero in Z 2 and Z3 respec-
tively).
(i) When F = IR the rank of the row echelon matrix is 3, in which case
dim Im t = 3 and hence dim Ker t = 0.
(ii) When F = Z 2 we have that 2,18, —8 are zero so that the rank is 1,
in which case dimlmtf = 1 and dim Ker t = 2.
(iii) When F = Z3 we have that 18 is zero so that the rank is 2, in
which case dimlmi = 2 and dim Ker t = 1.
V = Kertf® Imi holds in cases (i) and (ii), but not in case (iii); for in
case (iii) we have that (1,1,1) belongs to both Kert and Imt.
1.4 If s o t = idy then s is surjective, hence bijective (since V is of finite
dimension). Then t = s~x and so t o s = idy.
Suppose that W is ^-invariant, so that t(W) C W. Since t is an
isomorphism we must have dimi(VK) = dim W and so t(W) = W. Hence
W = s[t{W)} = s{W) and W is s-invariant.
The result is false for infinite-dimensional spaces. For example, con-
sider the real vector space \R[X] of polynomials over IR. Let s be the
differentiation map and t the integration map. We have s o t = id but
t o s ^ id.
33

, Book 4 Linear algebra

1.5 KerD = {a \ a G F} and imD = {p{X) | degp(X) < n - 2}. Clearly,
ImD is isomorphic to Vn_i and Ker£> is isomorphic to F. Now Ker Dn
ImD ^ {0} since if a G F with a ^ 0 then the constant polynomial a
belongs to both.
The same results do not hold when the ground field is Z 2 . For exam-
ple, in this case we see that the polynomial X2 belongs to the kernel of
D.
1.6 Let 5, t € £{V, V). Then if w G Im(s o t) we have w = s[t(u)} for some
u G V which shows that w G Iras. Thus Im(s o tf) C Ims. The first
chain now follows by taking s = tn.
Similarly, if u G Kerf 1 then s[tn{u)\ = s(0) = 0 gives u G Ker(s o i n )
and so Kerf 1 C Ker(s o f ) . The second chain now follows by taking
s = t.
Now we cannot have an infinite number of strict inclusions in the first
chain since X cY implies that dimX < dimF, and the dimension of V
is finite. Hence the chain is finite. It follows that there exists a positive
integer p such that Im£p = Imi p+fc for all positive integers k. Since
dimlmi p + dim Ker tp = dimV the corresponding results for the kernel
chain are easily deduced.
To show that V = Im£ p ©Kerf it suffices, by the dimension argument,
to prove that Im*p n Kerf = {0}. Now li x e lmtp n Kerf then
tp(x) = 0 and there exists v G V such that x = tp(v). Consequently



and so v G Ker*2p = Kerf whence x = tp{v) = 0.
For the last part, observe that if x G I m f then x = tp(v) gives
t(x) = tp+1(v) G I m f + 1 C I m f and so I m f is ^-invariant. Also, if
x G K e r f then f (x) = 0 gives f + 1 ( x ) = 0 so tp[t{x)} = 0 whence
t(x) G Kerf and so Kerf is ^-invariant.
1.7 If / o / = 0 then (Vx G V) f{x) G Ker/ and so I m / C Ker/.
We know that

n = dim V — dim Im / + dim Ker / = r + dim Ker /

and, by the above, dim Ker / > dim Im / = r. Hence 2r < n.
If W is a subspace such that V = Ker/ 0 W then we have that
dim V = dim Ker / + dim W and so

= dimV — dim Ker/ = d i m l m / = r.
34