The document discusses topics in linear algebra, including direct complements of subspaces in finite dimensional spaces and the properties of linear maps. It provides proofs for the existence and uniqueness of complements, as well as the index of a linear map. Additionally, it explores arrangements of lines in three-dimensional space and their classification based on linear automorphisms.

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2. (1) (a) A direct complement to a subspace L1 in a finite dimensional space L is
a subspace L2 ⊂ L such that L = L1 ⊕ L2. Prove that for any subspace L1 a
direct complement exists, and the dimensions of any two direct
complements of L1 in L coincide.
(b) For a linear map F : L M → , let cokerF be a direct complement of ImF
in M. Define the index of the map F by indF = dim(cokerF) − dim(kerF).
Check using (a) that indF is welldefined. Prove that if L and M are finite
dimensional, indF depends only on the dimensions of L and M: indF =
dim(M) − dim(L).
(c) Set dim(M) = dim(L) = n. What can you deduce from (b) about systems of n
linear equations in n variables?
(2) Two ordered ntuples of subspaces �L1, L2, . . . , Ln� and �L1 � , L2 � ,
. . . L� n� in a finite dimensional L are identically arranged if there exists
a linear automorphism (bijecitve linear map from a space to itself) F : L L
such that F(Li) = for all L i = 1, . . . n. Show that � → i all triples of non-
coplanar, pairwise distinct lines through zero in R3 are identically arranged.
Classify the arrangements of quadruples of such lines in R3 .
Problems
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3. Solution
1)
(a) Given a finite dimensional linear space L and a subspace L1 ⊂ L we want to prove that
there exists a subspace L2 ⊂ L such that L1 ⊕ L2 = L. Furthermore, we want to prove that
the dimensions of all such direct complements to L1 coincide. Let dim L = l. We pick a
basis {ei}l1 for L1 and note that i=1 dim L1 = l1. We then use the basis extension theorem
and extend the basis of L1 to the basis of L. Thus, {e1, . . . , el1 , el1+1 . . . , el} spans l
L. Let L2 = span {ei}i=l1+1. Since, L1∩L2 = 0 and L1+L2 = L we note that L2 is a direct
complement to L1. Furthermore, dim L2 = l − l1. 2 Let L2 � be a direct complement to L1
and {ei � }i l � =1 be a basis for L� 2. We extend the basis of L� to L by using the basis
vectors of L1. Thus, 2 1, . . . , e� , e1, . . . , el1 } spans L. Hence, it must be that dim L
{e� l � � 2 = 2 l − l1. However, this is the same as dim L2. Thus, dimensions of all direct
complements of L1 coincide.
(b) Given F : L �→ M, we first show that ind F = dim(coker F) − dim(ker F) is well defined.
We note that since part (a) defined direct complement only for finite dimensional spaces
we restrict our attention to a finite dimensional M. Using the result from part (a) we find
that coker F, a direct complement to Im F ⊂ M, always exists and has a finite dimension.
Furthermore, ker F also always exists and has a well defined dimension. We can then take
dim coker F = c and dim ker F = k. Hence, ind F = c − k. We note that c is always a non-
negative integer and k can be either a nonnegative integer or infinity depending on the
dimension of L. Thus, ind F is well defined. For finite dimensional M and L. Let dim M = m
and dim Im F = i. Then dim coker F = m − i. Further, let dim L = l. Then dim ker F = dim
L−dim Im F = l−i. Thus, ind F = (m−i)−(l−i) = m − l = dim M − dim L
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4. (c) If dim M = dim L = n. Then ind F = 0 and also dim coker F = dim ker F. If ker F = 0 then
also coker F = 0 and the system of linear equations always has a solution, while the system
with a zero r.h.s has no nontrivial solution.
Problem 2.
Solution
We want to show that all triples of noncoplanar, pairwise distinct lines through zero in R3
are identically arranged. Let {e1, e2, e3} be a basis for R3 and let vi = a1ie1 + a2ie2 +
a3ie3 for i = 1, 2, 3 be direction vectors for three noncoplanar pairwise distinct lines in
R3. Further, let v� = a1 � ie1 + a� i 2ie2 + a3 � ie3 for i = 1, 2, 3 be the direction
vectors for a second set of noncoplanar pairwise distinct lines in R3. For vi and vi � to be
identically arranged we must find a linear map f such that f(vi) = v� for i = 1, 2, 3.
This is equivalent to i finding a matrix T such that T(aij ) = (aij � ). Since the three lines
are linearly independent we can invert the matrix of coefficients. Thus, T = (a� ij )(aij
)−1 and three such lines are identically arranged. To consider the arrangements of four
such lines we note that direction vectors for three such lines span R3 and thus we express
the direction vector for the fourth line as a linear combination of the first three. Namely,
v4 = 1v� + b� 3v3 b1v1 + b2v2 + b3v3 and v4 � = b� 1 2v2 � + b� � . Further, T(v4) =
b1T(v1) + b2T(v2)+b3T(v3). Hence, if we add a scaling factor to the first three direction
vectors T(vi) = b b i � i vi � for i = 1, 2, 3 it follows that T(v4) = v� and thus all 4
quadruples of such lines are identically arranged.
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