Linear Algebra

54 Pages·2008·0.377 MB·English

by  J. Saxl, Sebastian Pancratz

Linear Algebra
Prof. J. Saxl
Michaelmas 2005
LATEXed by Sebastian Pancratz
ii
These notes are based on a course of lectures given by Prof. J. Saxl in Part IB of the
Mathematical Tripos at the University of Cambridge in the academic year 2005(cid:21)2006.
ThesenoteshavenotbeencheckedbyProf.J.Saxlandshouldnotberegardedaso(cid:30)cial
notes for the course. In particular, the responsibility for any errors is mine (cid:22) please
email Sebastian Pancratz (sfp25) with any comments or corrections.
Contents
1 Vector spaces 1
2 Linear maps, matrices 7
2.1 Basic de(cid:28)nitions and properties . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 The space of linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.4 Change of bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.5 Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.6 Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Determinant and trace 15
4 Endomorphisms, matrices, eigenvectors 23
5 Dual spaces 33
6 Bilinear forms 37
7 Inner product spaces 45

Chapter 1
Vector spaces
Remark. Write F for the (cid:28)eld R or C. The crucial properties of F are as follows.
• F is an abelian group under +, with additive identity 0.
• F\{0} is an abelian group under ·, with multiplicative identity 1.
• Multiplication is distributive over addition, a(b+c) = ab+ac for all a,b,c ∈ F.
De(cid:28)nition. The set V is a vector space over F if the following holds.
(A) V is an abelian group under an operation +.
(A0) + : V → V, (v ,v ) 7→ v +v ∈ V;
1 2 1 2
(A1) For all v ,v ,v ∈ V, (v +v )+v = v +(v +v );
1 2 3 1 2 3 1 2 3
(A2) For all v ,v ∈ V, v +v = v +v ;
1 2 1 2 2 1
(A3) There exists 0 ∈ V such that v+0 = v for all v ∈ V.
(A4) For each v ∈ V there exists −v ∈ V such that −v+v = 0.
(B) There exists a multiplication · by scalars on V.
(B0) · : F×V → V, (λ,v) 7→ λv;
(B1) For all λ ∈ F,v ,v ∈ V, λ(v +v ) = λv +λv ;
1 2 1 2 1 2
(B2) For all λ ,λ ∈ F,v ∈ V, (λ +λ )v = λ v+λ v;
1 2 1 2 1 2
(B3) For all λ ,λ ∈ F,v ∈ V, (λ λ )v = λ (λ v);
1 2 1 2 1 2
(B4) For all v ∈ V, 1v = v.
Lemma 1.1. Let V be a vector space over F and let v ∈ V,λ ∈ F. Then
(i) 0·v = 0, λ·0 = 0;
(ii) −v = (−1)·v;
(iii) if λv = 0 then either λ = 0 or v = 0.
Proof. (i) 0·v = (0+0)·v = 0·v+0·v, so 0 = 0·v.
(ii) −v+v = 0 = 0v = (−1+1)v = (−1)v+1v = (−1)v+v, so −v = (−1)v.
2 Vector spaces
(iii) Let λv = 0 and suppose λ 6= 0. Then λ−1 exists and λ−1(λv) = λ−10 = 0 but
λ−1(λv) = (λ−1λ)v = 1v = v so v = 0.
Example. (i) The space Fn of n-tuples with entries in F.
(ii) Let X be any set. The set FX of all functions X → F is a vector space over F
addition and multiplication de(cid:28)ned pointwise.
De(cid:28)nition. LetV beavectorspaceoverF. AsubsetU ⊂ V isasubspaceofV, written
U ≤ V if
• 0 ∈ U;
• u ,u ∈ U =⇒ u +u ∈ U;
1 2 1 2
• λ ∈ F,u ∈ U =⇒ λu ∈ U.
Equivalently, U 6= ∅ and U is closed under linear combinations.
Lemma 1.2. If V is a vector space over F and U ≤ V then U is a vector space over F
under the restriction of the operations + and · on V to U.
Example. (i) RR is a vector space over R. The set C(R) of continuous real functions
is a subspace, and hence a real vector space.
(ii) Similarly, one can also consider the subspaces D(R) and P(R)) of di(cid:27)erentiable
and polynomial functions, respectively.
Remark. If n ∈ N , λ ,...,λ ∈ F, v ,...,v ∈ V write Pn λ v for the linear
0 1 n 1 n i=1 i i
combination λ v +···+λ v . By convention, P0 λ v = 0. Also note that all linear
1 1 n n i=1 i i
P
combinations are (cid:28)nite. If S ⊂ V, the linear combination λ v has only (cid:28)nitely
v∈S v
many v such that λ 6= 0.
v
De(cid:28)nition. The vector v ,...,v in V span (or generate) V over F if any v ∈ V is a
1 n
linear combination of v ,...,v . Write V = hv ,...,v i.
1 n 1 n
More generally, if S ⊂ V then S spans V if
n
X
∀v ∈ V ∃n ∈ N ∃v ,...,v ∈ V ∃λ ,...,λ ∈ F v = λ v .
0 1 n 1 n i i
i=1
Example. (i) P (R) is spanned by 1,x,x2.
2
(ii) P(R) is not (cid:28)nite-dimensional.
De(cid:28)nition. The vectors v ,...,v are linearly independent in V over F if whenever
1 n
λ v + ···λ v = 0 then λ = ··· = λ = 0. Otherwise, the vectors are linearly
1 1 n n 1 n
dependent.
More generally, if S ⊂ V then S is linearly independent precisely if every (cid:28)nite subset
of S is linearly independent.
Remark. 0 is never contained in a linearly independent set as 1·0 = 0.
Remark. (i) V = C is a vector space over R and 1,i are linearly independent.
3
(ii) V = C is also a vector space over C. In this case, 1,i are linearly dependent.
De(cid:28)nition. The vectors v ,...,v form a basis of V if they both span V over F and
1 n
are linearly independent.
Example. (i) P (R) has standard basis 1,x,x2.
2
(ii) Fn has standard basis e ,...,e where e = (0,...,1,...,0)T.
1 n i
(iii) {0} has basis ∅.
Lemma 1.3. v ,...,v ∈ V form a basis of V over F if and only if each element v ∈ V
1 n
can be written uniquely as v = Pn λ v with λ ∈ F.
i=1 i i i
Proof. Let v ∈ V. v ,...,v span V so v = Pn λ v for some λ ∈ F. If also
1 n i=1 i i i
v = Pn µ v then 0 = Pn (λ −µ )v , so as v ,...,v are linearly independent, we
i=1 i i i=1 i i i 1 n
have λ = µ for i = 1,...,n.
i i
Conversely, since each v ∈ V is a linear combination of v ,...,v , we have v ,...,v
1 n 1 n
span V over F. They are linearly dependent since if Pn λ v = 0 = Pn 0v then
i=1 i i i=1 i
λ = 0 for i = 1,...,n by uniqueness.
i
Lemma 1.4. if v ,...,v span V over F, some subset of {v ,...,v } is a basis of V.
1 n 1 n
Proof. If v ,...,v are linearly independent, we are done. Otherwise, for some k, there
1 n
exist α ,...,α ∈ F with v = α v + ··· + α v . (If λ v + ··· + λ v = 0
1 k−1 k 1 1 k−1 k−1 1 1 n n
with not all λ = 0, take k maximal with λ 6= 0 and let α = −λi.) Then
i k i λ
k
v ,...,v ,v ,...,v stillspanV sincewheneverv = Pn λ v thenv = Pk−1(α +
1 k−1 k+1 n i=1 i i i=1 i
λ )v +Pn λ v . Continue deleting until we have a basis.
i i i=k+1 i i
Theorem 1.5 (Steinitz Exchange Lemma). Let V be a (cid:28)nite dimensional vector space
over F. Let v ,...,v be linearly independent and let w ,...,w span V over F. Then
1 m 1 n
m ≤ n, and reordering the w if necessary, the vectors v ,...,v ,w ,...,w span V.
i 1 m m+1 n
Proof. Supposewehavereplacedr ≥ 0ofthew already,renumberingthew ifnecessary,
i i
wehavev ,...,v ,w ,...,w spanV. Ifr = m, wearedone. Soassumer < m. Then
1 r r+1 n
r n
X X
v = α v + β w
r+1 i i i i
i=1 i=r+1
for some α ,β ∈ F. Note that β 6= 0 for some i since v ,...,v ,v are linearly
i i i 1 r r+1
independent. After reordering w ,...,w we have β 6= 0. Then
r+1 n r+1
r
X −αi 1 X −βi
w = v + v + w .
r+1 i r+1 i
β β β
r+1 r+1 r+1
i=1 i=r+2
It follows that V is spanned by v ,...,v ,v ,w ,...,w . After m steps, we shall
1 r r+1 r+2 n
have replaced m of the w by the v , retaining the spanning property. It follows that
i i
m ≤ n.
Theorem 1.6. If V is a (cid:28)nite dimensional vector space over F, any two bases have the
same size, the dimension of V over F, dimFV.
4 Vector spaces
Proof. Let v ,...,v and w ,...,w be two bases. Then m ≤ n since the v are linearly
1 m 1 n i
independent and the w span V. Also n ≤ m since the w are linearly independent and
i i
the v span V.
i
Example. (i) dimFFn = n;
(ii) dimRP2(R) = 3;
(iii) dimRC = 2;
(iv) dimFF = 1.
Lemma 1.7. IfV isa(cid:28)nitedimensionalvectorspaceoverF,andifthevectorsv ,...,v
1 k
are linearly independent for some k ≥ 0, there exists a basis v ,...,v ,v ,...,v of
1 k k+1 n
V.
Proof. If v ,...,v span V, we are done. Otherwise, take v ∈ V \hv ,...,v i. Then
1 k k+1 1 k
v ,...,v are linearly independent. We shall obtain a basis after dimV −k steps.
1 k+1
Remark. If V is (cid:28)nite dimensional then whenever U ≤ V then dimU ≤ dimV with
equality if and only if U = V.
Lemma 1.8. Let V be a vector space over F with dimV = n.
(i) An independent set has at most n vectors, with equality if and only if it as basis.
(ii) A spanning set has at least n vectors, with equality if and only if it a basis.
Proof. (i) This follows from Lemma 1.7 and Theorem 1.6.
(ii) This follows from Lemma 1.4 and Theorem 1.6.
Lemma 1.9. Let dimFV = n. The following are equivalent.
(i) v ,...,v form a basis.
1 n
(ii) v ,...,v are linearly independent.
1 n
(iii) v ,...,v span V.
1 n
Lemma 1.10. Let S ⊂ V. There is a unique smallest subspace U of V containing
S, denoted by U = hSi, the subspace generated by S. In fact, U consists of linear
combinations of elements of S.
Proof. If we write U for the set of linear combinations of elements of S then U is a
subspace of V. On the other hand, U as de(cid:28)ned above has to be in any subspace
containing S.
Example. Let V = RR and S = {1,x,x2,...}. Then hSi = P(R), the subspace of
polynomial functions.
Remark. (i) The intersection of any collection of subspaces is a subspace.
(ii) If U,W ≤ V de(cid:28)ne U +W = {u+w : u ∈ U,w ∈ W}. Then U +W ≤ V.
5
(iii) Note that U ∪W is a subspace if and only if U ⊂ W or W ⊂ U.
Theorem 1.11. If U, W are (cid:28)nite dimensional subspaces of V then U+W is also (cid:28)nite
dimensional and dimU +W = dimU +dimW −dimU ∩W.
Proof. Let v ,...,v be a basis for U ∩ W. Extend this to v ,...,v ,u ,...,u
1 k 1 k 1 l
a basis for U and to v ,...,v ,w ,...,w a basis for W. We claim
1 k 1 m
v ,...,v ,u ,...,u ,w ,...,w is a basis for U +W.
1 k 1 l 1 m
P P
• If v ∈ U +W, then v = u+w for some u ∈ U, w ∈ W. Now u = α v + β u
i i i i
for some α ,β ∈ F and w = Pα0v +Pγ w for some α0,γ ∈ F, and therefore
i i i i i i i i
X X X
v = (α +α0)v + β u + γ w .
i i i i i i i
P P P
• Suppose α v + β u + γ w = 0. Then
i i i i i i
X X X
α v + β u = − γ w
i i i i i i
X
= δ v
i i
for some δ ∈ F, using that the LHS is in U and the RHS is in W, so both vectors
i
are in U ∩W. Then
X X
(α −δ )v + β u = 0,
i i i i i
and as v ,...,v ,u ,...,u form a basis of U, we have that all β are 0. But then
1 k 1 l i
X X
α v + γ w = 0,
i i i i
and as v ,...,v ,w ,...,w form a basis of W, we have that all α ,γ are 0.
1 k 1 m i i
De(cid:28)nition. Let V be a vector space over F and suppose U,W ≤ V. Then
V = U ⊕W
if every element v of V can be written uniquely as v = u+w with u ∈ U and w ∈ W.
If so, we say W is the complement (or complementary subspace) of U in V.
Lemma 1.12. Suppose U,W ≤ V. Then V = U ⊕W if and only if U +W = V and
U ∩W = {0}.
Lemma 1.13. If V is a (cid:28)nite dimensional vector space over F and U ≤ V, then U has
a complement in V. (Note this is not at all unique unless U = {0} or U = V.)
Proof. Take v ,...,v a basis for U and extend to a basis u ,...,u ,w ,...,w for
1 k 1 k k+1 n
V. Then W = hw ,...,w i is a complement of U in V.
k+1 n
P P
Lemma 1.14. Suppose V ,...,V ≤ V and let V = { v : v ∈ V } ≤ V. The sum
1 k i i i i
L P
is direct, written as V , if each v ∈ V is uniquely expressible as v with v ∈ V .
i i i i
Lemma 1.15. Let V ,...,V ≤ V. The following are equivalent.
1 k
P
(i) V is direct.
i
6 Vector spaces
(ii) If B is a basis for V then B = Sk B is a basis for PV .
i i i=1 i i
P
(iii) For each i, V ∩ V = {0}.
i j6=i j
Note that in (iii), if k > 2 it is not enough to assume V ∩V = {0} for all i 6= j.
i j
Proof. We show (i) =⇒ (ii). Let B be a basis for V , let B = Sk B . If v ∈ PV
i i i=1 i i
then v = Pk v , and v can be written as a linear combination of vectors in B , so
i=1 i i i
substitute for each v and obtain v as a linear combination of B. If we have a linear
i
combination of vectors in B equal to 0, collect together terms in each V , let v denote
i i
the part in V . Then v +···+v = 0. By uniqueness of expression for 0, we have v = 0.
i 1 k i
Now B is linear independent, so all coe(cid:30)cients are 0.
i