Download Chapter 3 Subspaces

Chapter 3
Subspaces
3.1
Definition and First Properties
A vector space V can contain many smaller vector spaces in a natural way. This leads to
the following definition:
Definition 3.1. A subset U of a vector space V is called a subspace if:
(S1) U is non-empty;
(S2) if u, v ∈ U , then u + v ∈ U (U is closed under addition);
(S3) if u ∈ U and α ∈ K, then αu ∈ U (U is closed under scalar multiplication).
Note that conditions (S1)–(S3) imply that U is itself a vector space over K, with the
addition and scalar multiplication operations inherited from V (this is an alternative
definition of a subspace).
Remark. Please, please, please note that a subspace must be non-empty, and it inherits the
addition and scalar multiplication from the bigger vector space. This is very important!
Examples 3.2. Here are some basic, but important, examples of subspaces:
(i) In any vector space V , the subset {0} consisting of the zero vector is a subspace,
called the trivial subspace.
(ii) Let V = Kn , and let 1 ≤ i ≤ n. Then the set Ui of column vectors with 0 in the ith
place is a subspace of V .
(iii) Let V = K[x], and let n be a non-negative integer. Then the set Kn [x] of polynomials of degree less than or equal to n is a subspace of V . Note that Kn [x] ⊂ Kn+1 [x]
for all non-negative integers n, so we have an infinite sequence of subspaces in K[x].
(iv) Let V = R2 . Then the subspaces of V are the trivial subspace, all straight lines
through the origin, and V itself. Similarly, the subspaces of V = R3 are the trivial
subspace, straight lines through the origin, planes through the origin and V itself.
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Let’s verify that (ii) gives a subspace. You should do (i), (iii) and (iv) yourself. So set
V = Kn and let 1 ≤ i ≤ n. Then certainly the column vector with 0s in every position
lies in Ui , so Ui is non-empty. Also, if you add two vectors with a zero in the ith position,
the result still has a zero in the ith position, so Ui is closed under addition. Finally, if you
multiply a vector with a zero in the ith position by any scalar, the result still has a zero
in the ith position, so Ui is closed under scalar multiplication. Hence Ui is a subspace.
Looking at the examples above, we might conjecture that the dimension of a subspace
is less than the dimension of the whole space. Notice in particular in part (iii), the
subspace Kn [x] has basis (1, x, x2 , . . . , xn ), so this subspace has dimension n + 1. Here we
have an infinite sequence of finite-dimensional subspaces in an infinite-dimensional space.
We now show that our naı̈ve conjecture is true:
Lemma 3.3. Suppose V is a finite-dimensional vector space of dimension n and U is a
subspace of V . Then U is finite-dimensional and dim U ≤ n. Moreover, if dim U = n,
then U = V .
Proof. Any linearly independent subset of U is also a linearly independent subset of V ,
so has size less than or equal to n. Let B be a maximal linearly independent subset of
U . Then B must span U , else we can add an element of U to B and get a larger linearly
independent subset, as in the proof of Lemma 2.13(iii). So B is a finite basis for U of size
less than or equal to n. If B has size n, then it must in fact be a basis for V by Lemma
2.13(iii), and we get U = V as required.
The following result shows how to build a subspace from a collection of vectors in a
vector space.
Theorem 3.4. For any subset S of a vector space V , Sp(S) is a subspace of V . Moreover,
Sp(S) is the smallest subspace of V containing S.
Proof. If S = ∅, then Sp(S) = {0} is the trivial subspace, so assume now that S is nonempty. Then S ⊆ Sp(S), so Sp(S) is non-empty. Now suppose u, v ∈ Sp(S). Then u =
α1 v1 + · · · + αn vn and v = β1 w1 + · · · + βm wm for some vectors v1 , . . . , vn , w1 , . . . , wm ∈ S
and scalars α1 , . . . , αn β1 , . . . , βm ∈ K. Thus
u + v = α1 v1 + · · · + αn vn + β1 w1 + · · · + βm wm
which is again a linear combination of vectors in S, so Sp(S) is closed under addition. If
α ∈ K, then
αu = αα1 v1 + · · · + ααn vn
is a linear combination of vectors in S, so Sp(S) is closed under scalar multiplication.
Hence Sp(S) is a subspace of V .
Finally, let U be any subspace of V containing S. Then, since U is closed under
addition and scalar multiplication, U must also contain all linear combinations of elements
from S, i.e., Sp(S) ⊆ U .
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As an example of how to calculate the span of a subset in practice, let’s consider the
vector space V = R4 and the subset
      
−2
1
0 


      



2  3   7 
S =  , ,  .

3  4  10






0
1
1
To find the dimension of Sp(S), we can put the vectors of S as rows of a matrix and
row reduce (I do it this way because I am happier row-reducing; if you want to put the
vectors as columns and do column reduction, then that is good too).






1 2 3 0
1 2 3 0
1 2 3 0





1 
 0 1 10
 0 7 10 1 
 −2 3 4 1 
7
7 
0 0 0 0
0 7 10 1
0 7 10 1
Then the number of nonzero rows tells us the dimension of Sp(S), and the nonzero rows
themselves tell us a basis for Sp(S). So, in this case, we have dim(Sp(S)) = 2, and Sp(S)
has a basis
    
0
1
 2   1  
    
  ,  10  .
 3   7  
1
0
7
3.2
Intersections, Sums and Direct Sums
Given two subspaces U, W of a vector space V , there are two very important subspaces
we can build out of U and W :
Theorem/Definition 3.5. Suppose U and W are subspaces of V . Then the subsets
U ∩ W := {v ∈ V | v ∈ U and v ∈ W }
and
U + W := {u + w | u ∈ U, w ∈ W }
are subspaces of V .
Proof. We prove that U + W is a subspace. (The proof for U ∩ W is similar, is on an
exercise sheet, and is examinable!). Since U and W are subspaces, they both contain
0. Hence 0 + 0 ∈ U + W , and we see that U + W is non-empty. Now suppose that
v1 , v2 ∈ U + W . Then there exist u1 , u2 ∈ U and w1 , w2 ∈ W such that v1 = u1 + w1 and
v2 = u2 + w2 . This means that
v1 + v2 = (u1 + w1 ) + (u2 + w2 ) = (u1 + u2 ) + (w1 + w2 ).
But u1 + u2 ∈ U , as U is a subspace, and similarly w1 + w2 ∈ W . Hence v1 + v2 ∈ U + W ,
and so U + W is closed under addition. Finally, suppose v ∈ U + W and α ∈ K. Then
v = u + w for some u ∈ U and w ∈ W , and we have
αv = α(u + w) = αu + αw.
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But αu ∈ U since U is a subspace, and similarly αw ∈ W . Thus αv ∈ U + W and we see
that U + W is closed under scalar multiplication. Thus U + W is a subspace.
Remark. It is not true in general that the union of two subspaces is a" subspace.
To see
#
1
this, think of V = R2 , let U be the x-axis, and W be the y-axis. Then
∈ U ⊂ U ∪W,
0
" #
0
∈ W ⊂ U ∪ W , but
1
" # " # " #
1
0
1
+
=
∈
/ U ∪ W,
0
1
1
so U ∪ W is not closed under addition, hence is not a subspace.
We now look at a special case of the sum of two subspaces.
Definition 3.6. Suppose U and W are two subspaces such that
V =U +W
and
U ∩ W = {0}.
Then we use the notation V = U ⊕ W , and we say that V is the direct sum of U and
W.
Theorem 3.7. If V = U ⊕ W , then every v ∈ V can be written uniquely as v = u + w
with u ∈ U , w ∈ W . Moreover, if B is a basis for U and B ′ is a basis for W , then B ∪ B ′
is a basis for V .
Proof. We know by definition of the direct sum that V = U + W , so for any v ∈ V ,
v = u + w for some u ∈ U and w ∈ W . To show this is unique, suppose that u′ ∈ U and
w′ ∈ W are some other vectors such that v = u′ + w′ . Then
u + w = u′ + w
=⇒ u − u′ = w′ − w
But u − u′ ∈ U and w′ − w ∈ W , since U and W are subspaces. Hence u − u′ ∈ U ∩ W
and w′ − w ∈ U ∩ W . But U ∩ W = {0}, so u = u′ and w = w′ , as required.
For the final assertion, note that everything in U can be written in terms of B,
everything in W can be written in terms of B ′ , so everything in V can be written in terms
of the union. Hence the union spans V . Now note that by the uniqueness statement, the
only way to write 0 as a sum of something in U and something in W is 0 = 0 + 0. Hence
if we have
X
X
αw w,
0=
αu u +
u∈B
w∈B′
each individual sum must be zero, and then linear independence of B and B ′ forces all
coefficients to be zero, hence the union B ∪ B ′ is linearly independent.
Example 3.8. Let V = Kn for some positive integer n, and let 1 ≤ i ≤ n. Let U denote
the subspace of vectors with 0s in positions i + 1, i + 2, . . . , n, and W denote the subspace
of vectors with 0s in positions 1, 2, . . . , i. Then U ∩ W = {0}, and in this case we have
V = U ⊕ W.
A special case of this is that R2 is the direct sum of the x-axis and the y-axis.
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The proof of the following useful dimension formula for sums is on a problem sheet:
Theorem 3.9. If U and W are finite-dimensional subspaces of the vector space V , then
dim(U + W ) = dim U + dim W − dim(U ∩ W ).
In particular, V = U ⊕ W if and only if U ∩ W = {0} and dim U + dim W = dim V .
We finish with a useful result which shows how to construct direct sums in practice.
Theorem 3.10. Suppose V is a finite-dimensional vector space and U is a subspace of
V . Then there exists a subspace W of V with V = U ⊕ W (such a subspace is called a
complement to U in V ).
Proof. Since V is finite dimensional, and U is a subspace of V , U is finite-dimensional.
Let (e1 , . . . , ek ) be a basis for U . By Theorem 2.14, we can extend the basis for U to
a basis (e1 , . . . , en ) for V . Now let W = Sp(ek+1 , . . . , en ) be the span of the new basis
vectors. Then it is easy to see that U ∩ W = {0} and dim U + dim W = dim V , so
V = U ⊕ W by Theorem 3.9.
Example 3.11. Let V = R3 and let
 



x


 
U = y  | x + y + z = 0 .


 z

Then, U has basis, for example,
  
1
1
    
−1 ,  0  .
−1
0

To extendtoa basis for all of V we just need to add one vector outside this subspace. For
1
 
example, 2 will do. Then if W is the span of this single vector, we have V = U ⊕ W .
3
Note that there are infinitely many choices for W : geometrically, we are saying that R3
is the direct sum of the plane U together with any line through the origin not contained
in the plane.
3.3
Quotient Spaces
Fix a vector space V and a subspace U of V . There is a natural way of building a new
vector space by “dividing by U ”. We do this by way of an equivalence relation on the
elements of V :
v ∼ w ⇐⇒ v − w ∈ U.
Let’s check this is an equivalence relation on V :
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(R) For any v ∈ V , v − v = 0 ∈ U , so v ∼ v.
(S) If v ∼ w, then v − w ∈ U . Since U is a subspace, we have −1(v − w) = w − v ∈ U ,
so w ∼ v.
(T) If v ∼ w and w ∼ x, then v − w ∈ U and w − x ∈ U . But then (v − w) + (w − x) =
v − x ∈ U , so v ∼ x.
Given v ∈ V , let [v] denote the equivalence class of v under the relation ∼. Then
[v] = {v + u | u ∈ U } (check!), so we sometimes also use the notation v + U for this class.
Note that for any v ∈ V and u ∈ U , [v] = [v + u].
Definition 3.12. The quotient space V /U is defined as follows:
(i) The elements of V /U are the equivalence classes [v] for v ∈ V .
(ii) For [v], [w] ∈ V /U , define [v] + [w] = [v + w].
(iii) For [v] ∈ V /U and α ∈ K, define α[v] = [αv].
Theorem 3.13. The set V /U with addition and scalar multiplication as defined above
forms a vector space over K.
Proof. We first need to show that the addition and multiplication are well-defined. Let’s
do addition in detail: Suppose v1 ∼ v2 and w1 ∼ w2 (so that [v1 ] = [v2 ] and [w1 ] = [w2 ]).
Then there exist u, u′ ∈ U such that v1 − v2 = u and w1 − w2 = u′ . Now calculate
[v1 ] + [w1 ]:
[v1 ] + [w1 ] = [v1 + w1 ]
= [(v2 + u) + (w2 + u′ )]
= [v2 + w2 + (u + u′ )]
= [v2 + w2 ]
since u + u′ ∈ U
= [v2 ] + [w2 ].
This shows that the result of the addition doesn’t depend on the choice of representatives
v1 , v2 , w1 , w2 . A similar calculation shows that multiplication is well-defined.
Now we need to check the axioms for a vector space. Axioms (A1), (A2) and (M1)–
(M4) follow from the corresponding properties in V . The zero element in V /U is [0],
which gives (A3). We have −[v] = [−v] for any v ∈ V , giving (A4).
Let’s have an easy example.
Example 3.14. Let V = K[x], and let U = {f (x) ∈ K[x] | f (0) = 0}. Then U is a
subspace of V : saying f (0) = 0 is the same as saying f has constant term equal to 0,
and if you add two such polynomials, or multiply by a scalar, you still get a polynomial
with 0 constant term. Now what are the elements of the quotient V /U ? By definition,
they are the equivalence classes under the relation f ∼ g ⇐⇒ f − g ∈ U . In words,
this says f and g are equivalent if f − g has zero constant term. Therefore, [f ] consists
of all polynomials with the same constant term as f . Thus we get one equivalence class
for each constant, i.e. one element of V /U for each element of the field K. In this case,
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we see that the quotient V /U is “the same” as the field K (viewed as a vector space over
itself).
We’ll make the notion of “the same” more concrete later on (isomorphic vector spaces).
We finish this section with a useful dimension formula for quotient spaces:
Theorem 3.15. Suppose V is a finite-dimensional vector space, and suppose U is a
subspace of V . Then dim(V /U ) = dim V − dim U .
Proof. Recall that V /U is defined by way of the equivalence relation v ∼ w ⇐⇒ v − w ∈
U , and we denote the equivalence class of v ∈ V by [v].
Suppose dim V = n and dim U = k. Let (e1 , . . . , ek ) be a basis for U and extend to a
basis (e1 , . . . , en ) for V , using Theorem 2.14. Then we claim that B = ([ek+1 ], . . . , [en ]) is
a basis for V /U .
Firstly, since V is spanned by e1 , . . . , en , the quotient is spanned by [e1 ], . . . , [en ]. But
ei ∈ U for 1 ≤ i ≤ k, so [ei ] = [0] ∈ V /U for 1 ≤ i ≤ k. Hence V /U is spanned by B.
Now suppose there exist αk+1 , . . . , αn ∈ K such that
[0] =
n
X
αi [ei ].
i=k+1
P
Then we get P
[0] = [ ni=k+1 αi ei ], by the definition of addition and scalar multiplication in
V /U . Hence ni=k+1 αi ei ∈ U . Since (e1 , . . . , ek ) is a basis for U , there exist β1 , . . . , βk ∈ K
such that
k
X
βi ei =
i=1
so that
n
X
αi ei ,
i=k+1
0=
n
X
αi ei ,
i=1
where we let αi = −βi for 1 ≤ i ≤ k. But (e1 , . . . , en ) is a basis for V , and hence is linearly
independent, so all the coefficients must be zero. In particular, αk+1 = · · · = αn = 0, and
we conclude that B is linearly independent.
Thus B is a basis for V /U of size n − k, giving the result.
3.4
Tensor Products
You’ve now come across several different ways of making new vector spaces from old ones:
we can take the sum of subspaces in a vector space to get a new subspace, we can take
the quotient of a vector space by a subspace to get a new space, and on Sheet 1 you saw
how to take the direct product of two vector spaces. In the final part of this chapter
we’re going to look very briefly at a third way of making new vector spaces, the tensor
product. There are two (equivalent) ways of defining the tensor product. I’ll begin with
the easier one, which has the advantage of making it clearer what is going on, but the
disadvantage of brushing quite a lot of important stuff under the carpet.
For the rest of this chapter, let V and W be vector spaces over a field K. Essentially,
to define the tensor product, one can begin by defining the tensor product of two
20
vectors v ⊗ w for each v ∈ V and w ∈ W . To avoid the space becoming too big, we want
to identify some of these products with each other, so we allow elements of the field to
pass freely past the ⊗ sign: that is, we decree that
α(v ⊗ w) = (αv) ⊗ w = v ⊗ (αw)
for all α ∈ K, v ∈ V and w ∈ W . Note that this also defines a scalar multiplication on
tensors. Finally, we define an addition on tensors, by allowing
v1 ⊗ w + v2 ⊗ w = (v1 + v2 ) ⊗ w
and
v ⊗ w1 + v ⊗ w2 = v ⊗ (w1 + w2 )
i.e., we can add two things as long as one of the arguments agrees. Note that these
definitions in reverse tell you how to expand of such expressions (think of expanding
brackets and pulling all scalars to the front): if v = α1 v1 + α2 v2 and w = β1 w1 + β2 w2 ,
then
v ⊗ w = (α1 v1 + α2 v2 ) ⊗ (β1 w1 + β2 w2 )
= (α1 v1 + α2 v2 ) ⊗ (β1 w1 ) + (α1 v1 + α2 v2 ) ⊗ (β2 w2 )
= (α1 v1 ) ⊗ (β1 w1 ) + (α2 v2 ) ⊗ (β1 w1 ) + (α1 v1 ) ⊗ (β2 w2 ) + (α2 v2 ) ⊗ (β2 w2 )
= α1 β1 (v1 ⊗ w1 ) + α2 β1 (v2 ⊗ w1 ) + α1 β2 (v1 ⊗ w2 ) + α2 β2 (v2 ⊗ w2 ).
With these preliminaries in hand:
Definition 3.16. The tensor product of V and W over K, denoted V ⊗K W or just
V ⊗ W , is the vector space spanned by all the objects v ⊗ w, with the identifications and
rules given above.
Remark. Note (and this is important!) that a general vector in the space V ⊗ W is a sum
of things of the form v ⊗ w where v ∈ V and w ∈ W : not everything can be expressed as
a “pure tensor” v ⊗ w.
Before giving a couple of examples, we collect some basic properties.
Lemma 3.17. Properties of the tensor product V ⊗ W :
(i) We have v ⊗ 0W = 0V ⊗W = 0V ⊗ w for all v ∈ V and w ∈ W .
(ii) If X is a subspace of V and Y is a subspace of W , then X ⊗ Y is a subspace of
V ⊗ W (identified in the obvious way).
(iii) If V has basis (e1 , . . . , em ) and W has basis (f1 , . . . , fn ), then (ei ⊗ fj )1≤i≤m,1≤j≤n is
a basis for V ⊗W . In particular, if dim V = m and dim W = n, then dim(V ⊗W ) =
mn.
Proof. (i). See problem sheet.
(ii). This is obvious from the definitions: X ⊗ Y is the span of all objects x ⊗ y with
x ∈ X and y ∈ Y inside V ⊗ W .
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(iii). Suppose
v ∈ V and w ∈ W . Then there exist αi , βj ∈ K such that v =
P
and w = nj=1 βj fj . Hence, by the properties defined above, we have
v⊗w =(
=
=
m
X
αi ei ) ⊗ (
i=1
n
m
XX
i=1 j=1
n
m X
X
n
X
Pm
i=1
αi ei
βj fj )
j=1
(αi ei ) ⊗ (βj fj )
αi βj (ei ⊗ fj )
i=1 j=1
so every object v⊗w can be written as a linear combination of the ei ⊗fj . Since everything
in V ⊗ W is a linear combination of objects v ⊗ w, this shows that the given set spans
V ⊗ W.
Linear independence is harder, mainly because we haven’t been very careful in our
definition. I’ll give you a hint how to rectify this below.
Examples 3.18. Here are two basic examples of tensor products:
(i) Suppose V = Kn is the space of column vectors with the standard basis (e1 , . . . , en )
and W is the space of row vectors of length n with the standard basis (f1 , . . . , fn )
(so fi is the row vector with a 1 in the ith position and 0s elsewhere). Then V ⊗ W
has dimension n2 and has basis (ei ⊗ fj )1≤i,j≤n . If we relabel by setting Eij = ei ⊗ fj
for each 1 ≤ i, j ≤ n, then we see that the vector space V ⊗ W looks very much
like the space Mn (K) of n × n matrices. This is no accident!
(ii) Suppose V = K[x] is the space of polynomials in x with coefficients in K, and
W = K[y] is the space of polynomials in y with coefficients in K. Then V ⊗ W has
basis (xi ⊗ y j )i,j∈N∪{0} . If we forget about the symbol ⊗ and identify each xi ⊗ y j
with the polynomial xi y j , then we see that the tensor product V ⊗ W looks just
like the space K[x, y] of polynomials in x and y with coefficients in K.
We finish this chapter with the “proper” definition of the tensor product V ⊗ W . It is
fairly technical, which is why I chose to introduce this idea more informally first. To start
with, we define the free vector space F (V × W ) to be the vector space over K with
basis (e(v,w) )v∈V,w∈W . I.e., F (V × W ) has one basis vector for each pair (v, w) ∈ V × W ;
in particular, because the e(v,w) are defined to be basis vectors, they are automatically
linearly independent. Note that F (V × W ) is a truly enormous vector space in general.
In set notation,
F (V ⊗ W ) = {
n
X
αi e(vi ,wi ) | n ∈ N, αi ∈ K, vi ∈ V, wi ∈ W }.
i=1
BE CAREFUL HERE! The above notation means “take any positive integer n, then take
any n elements of K and any n distinct pairs of elements from V × W , then write down
the formal sum shown. Repeat for all possible integers and field elements and pairs, and
you get F (V × W )”.
Now we define some subsets of F (V × W ):
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1. S1 := {e(v1 ,w) + e(v2 ,w) − e(v1 +v2 ,w) | v1 , v2 ∈ V, w ∈ W };
2. S2 := {e(v,w1 ) + e(v,w2 ) − e(v,w1 +w2 ) | v ∈ V, w1 , w2 ∈ W };
3. S3 := {αe(v,w) − e(αv,w) | α ∈ K, v ∈ V, w ∈ W };
4. S4 := {αe(v,w) − e(v,αw) | α ∈ K, v ∈ V, w ∈ W }.
Let R = Sp(S1 ∪ S2 ∪ S3 ∪ S4 ), so R is a subspace of F (V × W ).
Definition 3.19. With notation as above, define V ⊗K W to be the quotient F (V ×W )/R.
If we denote the equivalence class of ev,w in the quotient space by v ⊗ w, then we get
the previous construction back. In particular, quotienting out by R turns the defining
conditions of the subsets Si in F (V × W ) into equations in V ⊗ W . For example, the
condition
e(v1 ,w) + e(v2 ,w) − e(v1 +v2 ,w) | v1 , v2 ∈ V, w ∈ W
defining S1 becomes the equation
v1 ⊗ w + v2 ⊗ w = (v1 + v2 ) ⊗ w for all v1 , v2 ∈ V, w ∈ W.
Now the linear independence part of the proof in Lemma 3.17(iv) above becomes possible:
if there’s a linear dependence amongst the ei ⊗ fj in V ⊗ W , then this translates to a
linear combination of the e(ei ,fj ) lying inside R in F (V × W ), and one can show that this
is impossible.
A note to reassure you
Some of the constructions in this chapter are rather technical, and can seem quite daunting the first time round. This is especially true of the quotient and tensor product
constructions. The point of showing you them now is really the point of most of the
first half of this course: these objects will be extremely useful to you in later courses,
and seeing them now will help you and your lecturers later. Those of you who have
already done some group theory will have seen quotient constructions already; they are
immensely important across the spectrum of mathematics, and they always have the same
basic form. Tensor products are equally important, especially in representation theory,
which is used across pure mathematics, applied maths, physics, chemistry and beyond.
These ideas may seem rather abstract and “pure” now, but you will find a whole range of
applications as your degree progresses (remember when Calculus was part of Pure Maths
in A-level?).
Anyway, for the purposes of this course, I will expect you to know the basic constructions which lead to quotients and tensor products, and some of the important basic
properties like the dimension formulae, but I will not be asking you to do anything very
complicated with them.
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