Multilinear algebra

1 Universal sums and products
Multilinear algebra
Waffle
Mathcamp 2010
The basic idea of multilinear algebra is to study vector spaces with some kind of multiplication. In this
section, we will introduce the most basic and “universal” case of such a multiplication. Given two vector
spaces V and W , we will construct a vector space V ⊗ W (the tensor product of V and W ) such that we can
multiply a vector in V by a vector in W to get a vector in V ⊗ W .
First, as a warm-up, we will give a reinterpretation of the direct sum operation on vector spaces. Recall
that if V and W are vector spaces, their direct sum is the vector space V ⊕ W = V × W , with scalar
multiplication and addition given by c(v, w) = (cv, cw) and (v, w) + (v ′ , w ′ ) = (v + v ′ , w + w ′ ). We think of
V as sitting inside V ⊕ W by identifying v ∈ V with (v, 0) ∈ V ⊕ W , and similarly for W .
I’d now like to explain why one might come up with this construction of V ⊕ W . The idea is, we have
two different kind of vectors, v ∈ V and w ∈ W , and we can’t add them because they live in different vector
spaces. However, we’d like to build a new vector space in which it makes sense to add them together. That
is, given v ∈ V and w ∈ W , we want to have a “sum” v ⊕ w which lives in some vector space that we will call
V ⊕ W . Furthermore, this operation ⊕ should have the expected properties of addition. For example, we
should have that if c is a scalar, c(v⊕w) = cv⊕cw, and we should also have v⊕w+v ′ ⊕w ′ = (v+v ′ )⊕(w+w ′ ).
Based on this, we could make a definition.
Definition 1.1. Let V and W be vector spaces. Then a sum on V and W is another vector space E
together with a function α : V × W → E such that, writing α(v, w) = v ⊕ w, we have c(v ⊕ w) = cv ⊕ cw
and v ⊕ w + v ′ ⊕ w ′ = (v + v ′ ) ⊕ (w + w ′ ).
However, this notion of a “sum” does not quite give us what the direct sum should be. For example, for
any vector space E, α(v, w) = 0 defines a sum V × W → E. This is a bad sum because it throws away
information about V and W : elements of V and W are not all equal to 0, but this sum forgets about that.
On the other hand, note that if E is a sum of V and W and E is a subspace of F , F is also a sum of V and
W , with the same operation. This is also no good: F might have tons of other vectors in it that don’t come
from either V or W . That is, instead of losing information, this time we’ve gained extraneous information.
We want the direct sum to be the sum that is just right, and doesn’t gain or lose any information. The
following notion is how we can make this precise. Note that if α : V × W → E is a sum and T : E → F is
linear, then the composition T α : V × W → E → F is also a sum.
Definition 1.2. A sum α : V × W → E of V and W is universal (or initial) if for any sum β : V × W → F ,
there is a unique linear map T : E → F such that T (v ⊕E w) = v ⊕F w (i.e., T α = β).
Here’s why this is a reasonable definition. We have two different ways of getting vectors v ⊕ w, one from
α and one from β. The map T is such that T (v ⊕E w) = v ⊕F w. If E has no extra information besides being
a sum of V and W , this will be enough to define T : there are no other random vectors in E on which we
need to define T . Furthermore, if E has not lost any information, than T will be well-defined. For example,
if v ⊕w = 0 in E for all v and w, but this is not true in F , then T (v ⊕E w) = v ⊕F W will not be well-defined.
Furthermore, we can check that our original definition V ⊕W = V ×W is universal: indeed, if we consider
this as a sum, the map α is just the identity map 1 from V × W to itself, and there is a unique T such that
T ◦ 1 = β, namely T = β.
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However, while there should be a single definite direct sum of two vector spaces, it’s far from obvious
that there is only one “universal sum.” However, it turns out that there is.
Theorem 1.3. Let α : V × W → E and β : V × W → F be two universal sums. Then there is a unique
isomorphism T : E → F such that T (v ⊕E w) = v ⊕F w.
Proof. Since E is universal, there is a unique linear map T with the desired property; we just want to show
that T is an isomorphism. We show this by constructing an inverse map S : F → E. Namely, because F is
also universal, there is a unique map S : F → E such that S(v ⊕F w) = v ⊕E w.
We want to show that S is the inverse of T , i.e. that ST is the identity map 1 : E → E and T S is
1 : F → F . Note that ST (v⊕Ew) = S(v⊕F w) = v⊕Ew. That is, ST : E → E satisfies ST (v⊕Ew) = v⊕Ew.
By universality of E, there is a unique map E → E with this property. But the identity map also has this
property, so we must have ST = 1. The same argument shows that T S = 1.
Thus in fact, this highly abstract definition of a “universal sum” is in fact well-defined (up to isomorphism)
and gives a new definition of direct sum. It might seem silly to use such a complicated definition when we
can just define V ⊕ W = V × W . However, in the case of tensor product, which we now turn to, there is no
such simple concrete definition, at least without choosing a basis.
Tensor products will be like direct sums, except for the operation of multiplication rather than addition.
We need to write down some axioms that “multiplication” should satisfy, and then we can just copy the
definition we made above.
Definition 1.4. Let V and W be vector spaces. Then a product (or bilinear map) on V and W is another
vector space E together with a function µ : V × W → E such that, writing µ(v, w) = v ⊗ w, we have
c(v ⊗ w) = cv ⊗ w = v ⊗ cw, (v + v ′ ⊗ w = v ⊗ w + v ′ ⊗ w), and v ⊗ (w + w ′ ) = v ⊗ w + v ⊗ w ′ .
Example 1.5. The “dot product” k n × k n → k defined by v · w = v1w1 + v2w2 + · · · + vnwn is a product.
More generally, we can see that a function µ : k n × k m → k is a product if µ(v, w) is some polynomial in the
coordinates of v and the coordinates of w, each term of which is of the form cjkvjwk for cjk a scalar.
Definition 1.6. A product µ : V × W → E of V and W is universal (or initial) if for any product
λ : V × W → F , there is a unique linear map T : E → F such that T (v ⊗E w) = v ⊗F w (i.e., T µ = λ).
Theorem 1.7. Let µ : V × W → E and λ : V × W → F be two universal products. Then there is a unique
isomorphism T : E → F such that T (v ⊗E w) = v ⊗F w.
Proof. Exactly the same as Theorem 1.3
We can thus define tensor products.
Definition 1.8. The tensor product of two vector spaces V and W is the (unique up to isomorphism) vector
space V ⊗ W equipped with a universal product V × W → V ⊗ W .
But wait! There’s still a problem with this definition: how do we know that a universal product actually
exists? For sums, we knew it existed because we could explicitly construct it, as just the cartesian product.
For products, we will have to give an explicit construction to show existence as well, but it is much less
simple. We will do this at the beginning of the next section.
1.1 Exercises
Exercise 1.1. Show that a sum α : V × W → E is equivalent to a pair of linear maps, one V → E and the
other W → E. 1 (Hint: Consider α(v, 0) and α(0, w).)
Exercise 1.2. For V and W vector spaces, write Hom(V, W ) for the set of all linear maps from V to W .
1 By “equivalent”, we mean that there is a natural bijection between the set of sums V × W → E and the set of pairs of
maps V → E and W → E.
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(a): Show that Hom(V, W ) is naturally a vector space.
(b): Show that a product map V × W → E is equivalent to a linear map V → Hom(W, E).
(c): Find a natural nontrivial product map Hom(V, W ) × V → W .
Exercise 1.3. Show that 0 ⊗ 0 = 0, where 0 = {0} is the trivial vector space.
Exercise 1.4. Show that the map µ : k ×k → k given by µ(a, b) = ab is a product. Is it a universal product?
Exercise 1.5. We will do this in class tomorrow, but you can try to figure it out now if you want. Let
V = k n and W = k m , and define a product µ : V × W → k nm by
µ(v, w) = (v1w1, v1w2, . . . , v1wm, v2w1, . . . , v2wm, v3w1, . . . , vnwm).
Show that this is a universal product, so k n ⊗ k m ∼ = k nm .
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2 Tensor products
We begin with the result promised at the end of the previous section.
Theorem 2.1. Any two vector spaces have a tensor product.
Proof. Let V and W be vector spaces. Let E be the vector space of all formal linear combinations of symbols
v ⊗ w where v ∈ V and w ∈ W . That is, an element of E is a finite sum ci(vi ⊗ wi), where the ci are
scalars, vi ∈ V and wi ∈ W . Alternatively, E is a vector space with basis V × W , where we write the basis
element coming from (v, w) as v ⊗ w.
We now want to force the symbol ⊗ to represent a product. To do this, we let K be the subspace of E
spanned by all elements of the form c(v ⊗ w) − cv ⊗ w, c(v ⊗ w) − v ⊗ cw, (v + v ′ ) ⊗ w − v ⊗ w − v ′ ⊗ w, and
v ⊗ (w + w ′ ) − v ⊗ w − v ⊗ w ′ . That is, these are the elements such that if they were zero, that would say
that ⊗ is a product. We then define V ⊗ W to be the quotient vector space E/K.
We have a map µ : V × W → E → E/K = V ⊗ W where the first map sends (v, w) to v ⊗ w; we claim
that this map is a universal product. First of all, it is a product: that’s exactly what modding out by K
does. Now let λ : V × W → F be any other product. Note that since E and hence E/K is spanned by
elements of the form v ⊗ w, there is at most one linear map T : V ⊗ W preserving the product. We thus
want to show that one exists. We can first define a map ˜ T : E → F which sends v ⊗ w to v ⊗F w; this
is well defined since {v ⊗ w} forms a basis for E. To turn ˜ T into our desired map T : E/K → F , we just
need to show that K is contained in the kernel of ˜ T . But since ⊗F is a product, all the elements of K must
become zero when you replace ⊗ with ⊗F , which is exactly what ˜ T does. Thus ˜ T (K) = 0 so we obtain a
map T : E/K → F witnessing the universality of E/K.
This constructions is explicit enough to tell us what an element of V ⊗ W looks like: it is a linear
combination civi ⊗ wi. However, two such linear combinations can be equal, and they are equal exactly
when they are forced to be equal by the fact that ⊗ is a product.
Let’s now give a more concrete description of tensor products.
As a first example of how we can actually compute tensor products, let’s let W = 0, the trivial vector
space containing only 0. Then in V ⊗ 0, for any v ∈ V we have an element v ⊗ 0, and these will span all of
V ⊗ 0. Now note that for any scalar c, c(v ⊗ 0) = v ⊗ (c · 0) = v ⊗ 0. But the only way this can happen for
an element of a vector space is if v ⊗ 0 = 0! Indeed, if for example we let c = 2, we get 2(v ⊗ 0) = v ⊗ 0 so
subtracting v ⊗ 0 from both sides gives v ⊗ 0 = 0.
This means that every element of V ⊗ 0 is 0, i.e. V ⊗ 0 = 0.
OK, let’s look at a more interesting example. Let W = k, the field of scalars, considered as a 1-dimensional
vector space (if you are not comfortable thinking about vector spaces over general fields, you can assume
k = R). To understand V ⊗ k, we need to understand products µ : V × k → E. Given such a product, for
each v ∈ V we have an element µ(v, 1) ∈ E. Furthermore, for any other c ∈ k, by bilinearity (i.e., being
a product), µ(v, c) = cµ(v, 1), so these elements µ(v, 1) determine µ. Also, bilinearity says that µ(v, 1) is
linear as a function of v. Indeed, µ is bilinear iff µ(v, 1) is linear in v.
Thus we have found that a product on V × k is the same thing as a linear map on V . But by definition,
a product on V × k is the same thing as a linear map on V ⊗ k. Thus we should get the following.
Proposition 2.2. For any V , V ⊗ k ∼ = V .
Proof. To prove this rigorously, we have to give a product µ : V × k → V and show that it is universal. The
only natural product to write down is µ(v, c) = cv, where the right-hand side is scalar multiplication in V .
This is universal by the discussion above: given any other product λ : V × k → F , we get a map V → F by
just sending v to λ(v, 1), and this preserves the product since it sends µ(v, c) = cv to λ(cv, 1) = λ(v, c).
Finally, there is one more property that we need to compute general tensor products.
Proposition 2.3. For any U, V , and W , (U ⊕ V ) ⊗ W ∼ = (U ⊗ W ) ⊕ (V ⊗ W ). The isomorphism sends
(u ⊕ v) ⊗ w to (u ⊗ w) ⊕ (v ⊗ w).
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Proof. First, we define a map T : U ⊕ (V ) ⊗ W → (U ⊗ W ) ⊕ (V ⊗ W ) by saying
T ((u ⊕ v) ⊗ w) = (u ⊕ w) ⊗ (v ⊕ w).
This gives a well-defined linear map by the definition of the tensor product, since the right-hand side is
bilinear in u ⊕ v and w
On the other hand, we can also define a map S : (U ⊗ W ) ⊕ (V ⊗ W ) → U ⊕ (V ) ⊗ W by
S((u ⊕ w1) ⊗ (v ⊕ w2)) = (u ⊕ 0) ⊗ w1 + (0 ⊕ v) ⊗ w2.
This is a well-defined linear map by a similar analysis. Finally, we have T S = 1 and ST = 1, so they are
inverse isomorphisms.
With this, we can understand the tensor product of any two vector spaces. For example, let W = k ⊕k, a
2-dimensional vector space. Then for any V , V ⊗ W ∼ = (V ⊗ k) ⊕ (V ⊗ k) = V ⊕ V . More generally, iterating
this argument shows that V ⊗ W is a direct sum of copies of V , one for each dimension of W . Let’s make
this more precise.
Theorem 2.4. Let {ei} be a basis for V and {fj} be a basis for W . Then {ei ⊗ fj} is a basis for V ⊗ W .
In particular, dim(V ⊗ W ) = dim(V ) dim(W ).
Proof. We can verify this directly by showing explicitly that a bilinear map is uniquely determined by saying
where each (ei, fj) gets sent. Indeed, for any v ∈ V we can write v = ciei and for any w ∈ W we can write
w = djfj, and then we must have µ(v, w) = µ( ciei, djfj) =
i,j cidjµ(ei, fj). On the other hand,
given any values of µ(ei, fj), we can see that the expression above for µ(v, w) gives a well-defined bilinear
map, by uniqueness of the representation v = ciei and w = djfj. This says that a linear map out of
V ⊗ W is uniquely determined by saying where each ei ⊗ fj goes, i.e. {ei ⊗ fj} is a basis.
However, this also follows from the results we have proven already. Namely, giving a basis {ei} of V is
the same as giving an isomorphism kn → V which sends (1, 0, 0, . . . ) to e1, (0, 1, 0, . . . to e2, (0, 0, 1, . . . ) to
e3, and so on. Thus our bases give isomorphisms V ∼ = kn and W ∼ = km , and the “distributivity” property
(Proposition 2.3) together with k ⊗ k ∼ = k gives that kn ⊗ km ∼ = knm . Chasing through exactly what the map
in Proposition 2.3 is, we can see that the isomorphism V ⊗ W ∼ = kn ⊗ km ∼ = knm corresponds to picking the
basis {ei ⊗ fj} for V ⊗ W .
Note that we could have just taken Theorem 2.4 as the definition of the tensor product. However, this
would have been very awkward because it depends on a choice of a basis, and it’s not at all obvious that
you could make it independent of the basis. This is why we made our abstract definition.
2.1 Exercises
Exercise 2.1. Show that V ⊗ W ∼ = W ⊗ V naturally. (Hint: You can prove it picking a basis or using
the universal property. However, if your isomorphism involves a basis, you should show that if you picked a
different basis, you would still get the same isomorphism (this is part of what “naturally” should mean).)
Exercise 2.2. Show that (U ⊗ V ) ⊗ W ∼ = U ⊗ (V ⊗ W ).
Exercise 2.3. Let k = R, and consider C as a vector space over R. Show that for any R-vector space V ,
C ⊗ V can naturally be thought of as a vector space over C, whose dimension over C is the same as the
dimension of V over R. (C ⊗ V is called the complexification of V , and should be thought of as V but with
the scalars formally extended from R to C.)
Exercise 2.4. Let k[x] be the set of polynomials in a variable x with coeffients in k; this is a vector space
over k. Show that k[x] ⊗ k[y] ∼ = k[x, y] via f(x) ⊗ g(y) ↦→ f(x)g(y), where k[x, y] is the vector space of
polynomials in two variables x and y.
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Exercise 2.5. Give a definition of a trilinear map U × V × W → E (it should be something like “linear in
all three variables simultaneously”). Show that U ⊗ V ⊗ W (parenthesized in either way by Exercise 2.2)
is the universal vector space with a trilinear map from U × V × W (the same way V ⊗ W is the universal
vector space with a bilinear map from V × W ).
Exercise 2.6. Generalize Exercise 2.5 by replacing 3 by n: define an n-multilinear map V1×V2×· · ·×Vn → E,
and show that V1 ⊗V2 ⊗· · ·⊗Vn is the universal vector space with an n-multilinear map V1 ×V2 ×· · ·×Vn →
V1 ⊗ V2 ⊗ · · · ⊗ Vn.
Exercise 2.7. Define a product map of abelian groups to be a map µ : A × B → C satisfying µ(a + a ′ , b) =
µ(a, b) + µ(a ′ , b) and µ(a, b + b ′ ) = µ(a, b) + µ(a, b ′ ).
(a): Show that for any two abelian groups A and B, a universal product A × B → A ⊗ B exists. (Hint:
Imitate the proof of Theorem 2.1.)
(b): Show that Z/2 ⊗ Z/3 = 0. (Hint: Show that any product Z/2 × Z/3 → C is identically 0.)
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3 Exterior products and determinants
In this section, we will look at a specific notion of multiplication, namely one that relates to area and volume.
The basic idea is as follows. Given two vectors v and w, we can form the parallelogram that they span,
and write v ∧ w for something we think of as the “area” of the parallelogram. This is not quite the usual
notion of area, however, because we want to think of it not just as a single number but as also having a
“two-dimensional direction” (the same way a single vector v both has a size and a direction). That is, if
we had a parallelogram pointing in a “different direction”, i.e. in a different plane, we would think of it as
different.
Let’s write down some properties this v ∧ w ought to have. Scaling v or w scales the parallelogram, and
thus ought to scale its area. That is, for scalars c, we should have (cv) ∧ w = c(v ∧ w) = v ∧ (cw). This
suggests that the operation ∧ ought to be bilinear. Another property of ∧ is that for any vector v, v ∧ v
should be 0: if both vectors are pointing in the same direction, the “parallelogram” is just a line segment
and has no area.
As it turns out, these two properties are the only properties we really need.
Definition 3.1. Let V be a vector space. Then the exterior square 2 (V ) is the quotient of V ⊗ V by the
subspace spanned by the elements v ⊗ v for all v ∈ V . We write v ∧ w for the image of v ⊗ w under the
quotient map V ⊗ V → 2 (V ).
Let’s try to understand what this vector space 2 (V ) actually looks like. First, there is a very important
formal consequence of the fact that v ∧ v = 0 for all v. Namely, if we apply this to a sum v + w, we get
0 = (v + w) ∧ (v + w) = v ∧ v + v ∧ w + w ∧ v + w ∧ w = v ∧ w + w ∧ v.
Thus for any v and w, v ∧ w = −w ∧ v. That is, as a “multiplication”, this ∧ operation is “anticommutative”
(or “alternating”).
We can now see what 2 (V ) looks like if we pick a basis {ei} for V . We know that {ei ⊗ ej} is a basis
for V ⊗ V . However, in 2 (V ), ei ∧ ei = 0 and ei ∧ ej = −ej ∧ ei, so 2 (V ) can be spanned by just the
elements ei ∧ ej for which i < j. Seeing that all these are indeed linearly independent is a bit trickier.
Theorem 3.2. Suppose {ei} is a basis for V . Then {ei ∧ ej}i j, and
T (ei ⊗ ei) = 0. We want to show that T gives a map S : 2 (V ) → E; it suffices to show that T (v ⊗ v) = 0
for all v ∈ V . Let v = ciei; then
v ⊗ v = cicjei ⊗ ej =
c 2 i ei ⊗ ei +
cicj(ei ⊗ ej + ej ⊗ ei).
i
We thus see that T (v ⊗ v) = 0. Hence T gives a map S : 2 (V ) → E which sends ei ∧ ej to eij. Since
the eij are linearly independent in E by construction, this implies that the ei ∧ ej (for i < j) are linearly
independent, and hence a basis.
One thing to note about 2 (V ) is that not every element is of the form v ∧ w. That is, not every “area
vector” is just an area in some plane; it can also be a sum of areas in different planes. For example, if {ei}
is a basis, then e1 ∧ e2 + e3 ∧ e4 cannot be simplified to a single v ∧ w.
We can use the same idea to generalize from area to volume, or even n-dimensional volume.
7
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Definition 3.3. Let r ∈ N and V be a vector space. Then r (V ) is the quotient of V ⊗ · · · ⊗ V (with r
factors) by the subspace spanned by all tensors v1 ⊗ · · · ⊗ vn for which two of the vi are the equal. The
exterior algebra (V ) is the direct sum r(V r ).
We think of an element of r (V ) as some sort of “r-dimensional volume vector”. Note that 0(V ) = k
and 1 (V ) = V , the former because the “empty” tensor product is k (since V ⊗ k = V for any V ). The
same argument as for r = 2 shows the following.
Theorem 3.4. Suppose {ei} is a basis for V . Then {ei1 ∧ ei2 ∧ · · · ∧ eir}i1

4 More on Determinants; Duality
Let’s see that this definition of determinant we gave agrees with other definitions you may have seen. If
V = k2
a b
is 2-dimensional (with basis {e1, e2} and we represent T as a matrix , then det(T ) is supposed
c d
to be ad − bc. Now T (e1) = ae1 + ce2 and T (e2) = be1 + de2, so we have
2 T (e1∧e2) = T (e1)∧T (e2) = (ae1+ce2)∧(be1+de2) = ab(e1∧e1)+ad(e1∧e2)+bc(e2∧e1)+bd(e2∧e2) = (ad−bc)(e1∧e2).
In general, a similar calculation shows that the determinant of a matrix (aij) is given by a sum of products
± a iσ(i) where σ is a permutation of the numbers 1, . . . , n and the sign is given by the sign of the
permutation σ.
Many important properties of determinants are easy to understand from this definition. First of all, it is
manifestly independent of any choice of basis. Second, it is clear that det(ST ) = det(S) det(T ), since n (ST )
is just the composition n (S) n(T ), which is just multiplication by det(T ) followed by multiplication by
det(T ). We can then use this to prove the other extremely useful property of determinants:
Theorem 4.1. Let V be finite-dimensional and T : V → V be linear. Then T is invertible iff det(T ) = 0.
Proof. First, if T T −1 = I, then det(T ) det(T −1 ) = det(I) = 1 so det(T ) = 0. Conversely, suppose T is not
invertible. Then for some nonzero v, T (v) = 0. Pick a basis {ei} for V such that e1 = v. Then
so det(T ) = 0.
det(T )e1 ∧ · · · ∧ en =
n T (e1 ∧ · · · ∧ en) = T (e1) ∧ · · · ∧ T (en) = 0 ∧ · · · ∧ T (en) = 0,
Let’s now change gears and look at another construction, closely related to tensor products. We will
assume all vector spaces are finite-dimensional from now on.
Definition 4.2. Let V and W be vector spaces. Then we write Hom(V, W ) for the set of linear maps from
V to W . This is a vector space by defining addition and scalar multiplication pointwise.
More concretely, if we pick bases for V and W , we can think of Hom(V, W ) as a vector space of matrices:
if V is n-dimensional and W is m-dimensional, a linear map V → W can be represented as an n × m matrix
(assuming we pick bases). It follows that the dimension of Hom(V, W ) is nm.
We also know another way to form a vector space from V and W that is nm-dimensional, namely the
tensor product V ⊗W ! It follows that Hom(V, W ) is isomorphic to V ⊗W . However, we could ask whether it
is natural to think of them as being isomorphic–that is, whether we can write down an isomorphism between
them without choosing bases.
To understand this, we will specialize to the case W = k. In that case, V ⊗ W = V ⊗ k can naturally be
identified with V , and Hom(V, W ) = Hom(V, k) is called the dual of V and written V ∗ . An element of V ∗
is a linear function α that eats a vector and V and gives you a scalar.
Given a basis {ei} for V , we get a basis {α i } (the dual basis) for V ∗ as follows: let α i be the linear map
such that α i (ei) = 1 and α j (ei) = 0 if j = i. This completely defines αi since {ei} is a basis, and we can
easily see that these are linearly independent (because ( ciα i )(ej) = cj, so if ciα i = 0 then ci = 0 for
all i). Finally, the α i span all of V ∗ since given any α ∈ V ∗ , we can show that α = α(ei)αi.
Thus given a basis {ei} for V , we obtain an isomorphism T : V → V ∗ by defining T (ei) = α i . However,
for this to be a “natural” isomorphism, it should not depend on which basis we chose. Unfortunately, it
does!
Example 4.3. Let V = k 2 , and let {e1, e2} be the standard basis, with dual basis {α 1 , α 2 }. A different
basis for V is given by f1 = e1 and f2 = e1 +e2; call the dual basis to this {β 1 , β 2 }. Now β 1 (e1) = β 1 (f1) = 1
and β 1 (e2) = β 1 (f2 − f1) = −1, so β 1 = α 1 − α 2 . Similarly, β 2 = α 2 .
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Now the isomorphism T : V → V ∗ given by the basis {e1, e2} is T (e1) = α 1 , T (e2) = α 2 . However, the
isomorphism S : V → V ∗ given by the basis {f1, f2} is S(f1) = β 1 , S(f2)β 2 , and hence
S(e1) = S(f1) = β 1 = α 1 − α 2 ,
S(e2) = S(f2 − f1) = β 2 − β 1 = 2α 2 − α 1 .
Thus S and T are quite different, showing that the isomorphism V → V ∗ very much depends on what basis
you use!
Thus we should not think of V and V ∗ as being the same; we can only identify them if we’ve chosen a
basis. However, we can identify V with its double dual!
Theorem 4.4. Let V be a (finite-dimensional) vector space and v ∈ V . Let ˜v : V ∗ → k be defined by
˜v(α) = α(v). Then v ↦→ ˜v is an isomorphism from V to (V ∗ ) ∗ .
Proof. It is straightforward to check that ˜v : V ∗ → k is a linear map, so ˜v ∈ (V ∗ ) ∗ . Similarly, it is
straightforward to check that the map v → ˜v is linear. Thus we only need to check that it is an isomorphism.
If we pick a basis {ei} for V , let {α i } be the dual basis for V ∗ and let {fi} be the basis of (V ∗ ) ∗ dual to
{α i }. We claim that fi = ˜ei; it follows that our map is an isomorphism. Indeed, ˜ei(α i ) = αi(ei) = 1 and
˜ei(α j ) = α j (ei) = 0 if i = j, exactly the definition of the dual basis {fi} of {α i }.
Thus we should think of V ∗ as different from V (even though they have the same dimension), but V ∗∗
is the same as V .
Let’s now turn back to the general question of how Hom(V, W ) and V ⊗ W might be related. In the case
W = k, we found that Hom(V, k) = V ∗ and V ⊗ k = V were of the same dimension but not the same. For
general W , we have a similar situation.
Theorem 4.5. Define a map T : V ∗ ⊗ W → Hom(V, W ) by T (α ⊗ w)(v) = α(v)w (remember α(v) ∈ k is a
scalar, so we can multiply it with w). Then T is an isomorphism, and so V ∗ ⊗ W ∼ = Hom(V, W ) in a natural
(basis-independent) way.
Proof. Pick a basis {ei} for V , the dual basis {αi } for V ∗ , and a basis {fj} for W . A basis for Hom(V, W ) is
given by the matrices with all entries 0 except for the ijth entry 1; this is exactly the linear map βi j defined
by βi j (ei) = fj and βi j (ek) = 0 if k = i. We claim that T (αi ⊗ fj) = βi j , and hence T sends a basis to a basis
and is an isomorphism.
To check this, we must just evaluate T (αi ⊗ fj) on each ek and see that it agrees with βi j . But T (αi ⊗
fj)(ek) = αi (ek)fj, which is fj if i = k and 0 otherwise, which is exactly what we want.
Note that the coefficients of a map A : V → W with respect to the basis {βi j } is exactly the entries of
the matrix representing A. Thus if we think of a map A : V → W as an element of V ∗ ⊗ W , its coefficients
with respect to the basis {αi ⊗ fj} are exactly its matrix entries.
4.1 Exercises
Exercise 4.1. Given a linear map T : k 3 → k 3 given by a matrix
⎛
⎝
a11 a12 a13
a21 a22 a23
a31 a32 a33
⎞
⎠, show that
det(T ) = a11a22a33 + a12a23a31 + a13a21a32 − a11a23a32 − a13a22a31 − a12a21a33.
Exercise 4.2. Given a linear map T : k n → k n given by a matrix with entries aij (1 ≤ i, j ≤ n), show that
det(T ) =
ɛ(σ)
aiσ(i), σ∈Sn
where the product is over all permutations σ of {1, . . . , n} and ɛ(σ) = ±1 is the sign of the permutation.
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i

Exercise 4.3. Define a map T : V ∗ ⊗ W ∗ → (V ⊗ W ) ∗ by T (α ⊗ β)(v ⊗ w) = α(v)β(w) (so T (α ⊗ β) is a
map V ⊗ W → k). Show that T is an isomorphism. (Hint: Use bases.)
Exercise 4.4. Write the identity map I : V → V as an element of V ⊗ V ∗ , in terms of a basis {ei} for V
and the dual basis {α i } on V ∗ .
Exercise 4.5. Let µ : V × V → k be a bilinear map. We say µ is nondegenerate if for any v ∈ V , there
exists a w ∈ V such that µ(v, w) = 0. Show that a nondegenerate bilinear map µ gives an isomorphism
T : V → V ∗ defined by T (v)(w) = µ(v, w).
Exercise 4.6. Let {ei} be a basis for V and define µ(civi, divi) = cidi (the dot product with respect
to this basis). Show that µ is a nondegenerate bilinear map V × V → k, and show that the associated
isomorphism V → V ∗ sends the basis {ei} to its dual basis.
Exercise 4.7. Given a linear map T : V → W , you can get a linear map T ∗ : W ∗ → V ∗ by defining
T ∗ (α)(v) = α(T (v)) (so T ∗ (α) is a map V → k). We can think of T as an element of W ⊗ V ∗ and T ∗ as an
element of V ∗ ⊗ W ∗∗ = V ∗ ⊗ W . Show that if we identify W ⊗ V ∗ with V ∗ ⊗ W by just swapping the order
of tensors, then T is the same as T ∗ . (Hint: Use bases.)
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5 Traces and TQFT
Another way that Hom(V, W ) is related to tensor products is the following. Given f ∈ Hom(V, W ) and v ∈ V ,
we can evaluate f on v to get an element f(v) ∈ W . This is a bilinear map Hom(V, W )×V → W , so it gives a
linear map ev : Hom(V, W )⊗V → W . In particular, for W = k, we get a map tr : V ⊗V ∗ → k. Furthermore,
if we identify Hom(V, W ) with V ∗ ⊗W = W ⊗V ∗ as in Theorem 4.5, ev : Hom(V, W )⊗V = W ⊗V ∗ ⊗V → W
is just given by pairing up the V ∗ and V using tr. That is ev(w ⊗ v ∗ ⊗ v) = v ∗ (v)w (for v ∗ ∈ V ∗ ). This can
be seen by the map used in Theorem 4.5: an element of W ⊗ V ∗ maps V to W by pairing the element of v
with the V ∗ part of the tensor product.
More generally, there is a composition map Hom(U, V ) ⊗ Hom(V, W ) → Hom(U, W ) which composes two
linear maps. Writing these Hom’s in terms of tensor products and duals, this is a map U ∗ ⊗ V ⊗ V ∗ ⊗ W →
U ∗ ⊗ W . A similar analysis to the analysis above shows that this map is just given by pairing up the V and
V ∗ in the middle: we compose linear maps by sending u ∗ ⊗ v ⊗ v ∗ ⊗ w to v ∗ (v)u ∗ ⊗ w.
Let’s now see what happens when we let W = V . We then have Hom(V, V ) = V ⊗ V ∗ , and there is a
map tr : Hom(V, V ) = V ⊗ V ∗ → k, a map which takes a linear map T : V → V and gives a scalar. This is
called the trace tr(T ) of T .
You may have seen the trace of a matrix defined as the sum of its diagonal entries. This definition is really
quite mystifying: why on earth would you take the diagonal entries (as opposed to some other collection of
entries) and add them up? Why on earth would that be independent of a choice of basis?
On the other hand, our definition is manifestly independent of a choice of basis and is very naturally
defined: it’s just the natural evaluation map on V ⊗ V ∗ . We can furthermore check that this agrees with
the “sum of the diagonal entries” definition.
Example 5.1. Let V have basis {e1, e2} and let {α 1 , α 2 } be the dual basis of V ∗ . Let T : V → V have
a b
matrix
c d
matrix with ij entry 1 and all other entries 0. Thus T is given by
. We want to write down T as an element of V ⊗ V ∗ . Recall that ei ⊗ α j corresponds to the
ae1 ⊗ α 1 + be1 ⊗ α 2 + ce2 ⊗ α 1 + de2 ⊗ α 2 .
Now the trace just takes these tensors and evaluates them together, so ei ⊗ α j goes to 1 if i = j and 0
otherwise. Thus tr(T ) ends up being exactly a + d, the sum of the diagonal entries. The same argument
would generalize to n × n matrices for any n.
Besides basis-invariance, one of the most important properties of traces is the following.
Theorem 5.2. Let S, T : V → V be linear maps. Then tr(ST ) = tr(T S).
Proof. Write S and T as elements of V ∗ ⊗V , say S = v ∗ ⊗v and T = w ∗ ⊗w (actually, S and T will be linear
combinations of things of this form, but by linearity of everything we can treat each term separately). Then
recall that we compose linear maps by “pairing up the vectors in the middle”, so ST will be v ∗ (w)w ∗ ⊗ v, so
tr(ST ) = v ∗ (w)w ∗ (v). On the other hand, T S = w ∗ (v)v ∗ ⊗ w and tr(T S) = w ∗ (v)v ∗ (w). But multiplication
of scalars is commutative, so these are the same!
Let’s now look more closely at how duals and tensor products interact with linear maps. First, note that
an element u of a vector space U is just the same thing as a linear map T : k → U, namely the linear map
such that T (1) = u. Thus in showing that elements of V ∗ ⊗ W = Hom(V, W ) are the same as maps V → W ,
we’ve shown that maps V → W are the same as maps k → V ∗ ⊗ W . We can think of this as “moving V to
the other side” by dualizing it. In fact, this works more generally.
Theorem 5.3. Let U, V , and W be (finite-dimensional) vector spaces. Then linear maps U ⊗ V → W are
naturally in bijection with linear maps U → V ∗ ⊗ W = Hom(V, W ).
Proof. Given T : U ⊗ V → W , we can define ˜ T : U → Hom(V, W ) by ˜ (T )(u)(v) = T (u ⊗ v). Conversely,
given ˜ T : U → Hom(V, W ), we can define T : U ⊗ V → W by T (u ⊗ v) = ˜ T (u)(v). Clearly these two
operations are inverse to each other.
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That is, a V on one side of a linear map is equivalent to a V ∗ on the other side. Another way to see this
is that the set of maps U ⊗ V → W is
Hom(U ⊗ V, W ) = (U ⊗ V ) ∗ ⊗ W = U ∗ ⊗ V ∗ ⊗ W = U ∗ ⊗ (V ∗ ⊗ W ) = Hom(U, V ∗ ⊗ W )
(using Exercise 4.3 to say (U ⊗ V ) ∗ = U ∗ ⊗ V ∗ ).
As yet another way to understand this correspondence, we can take ˜ T : U → V ∗ ⊗ W and define a linear
map 1 ⊗ ˜ T : V ⊗ U → V ⊗ V ∗ ⊗ W by 1 ⊗ ˜ T (v ⊗ u) = v ⊗ ˜ (T )(u). We can then compose this with the map
V ⊗ V ∗ ⊗ W → k ⊗ W = W which just evaluates the V ∗ factor on the V factor (i.e., the trace map) to get a
map T : U ⊗ V = V ⊗ U → V ⊗ V ∗ ⊗ W → W . Chasing through all the definitions (perhaps using a basis)
shows that this is the same operation as the one used above.
We can understand this operation in another way by drawing pictures of lines. Unfortunately, I don’t
know how to make such pretty pictures in L ATEX, so I can’t provide these pictures in these notes.
5.1 Exercises
Exercise 5.1. Verify that if you think of maps U → V and V → W as elements of V ∗ ⊗ U and W ∗ ⊗ V ,
then you compose maps via the linear map W ∗ ⊗ V ⊗ V ∗ ⊗ U which sends w ∗ ⊗ v ∗ ⊗ v ⊗ u to v ∗ (v)w ∗ ⊗ u.
Exercise 5.2. Show that the dual of the trace map V ⊗ V ∗ → k is the map k ∗ = k → V ∗ ⊗ V = Hom(V, V )
which sends 1 to the identity I : V → V . (Here we are take the dual of a linear map as in Exercise 4.7.)
Exercise 5.3. Show that there is a natural (basis-free) isomorphism r (V ∗ ) ∼ = ( r(V )) ∗ . (Hint: Find a
way to “wedge together” r elements of V ∗ to get a map from r (V ) → k, and use a basis to show that this
gives an isomorphism r (V ∗ ) → ( r(V )) ∗ .
Exercise 5.4. Let V and W be vector spaces and let S : k → V ⊗ W and T : W ⊗ V → k be maps such that
the compositions V = k ⊗V S⊗1
→ V ⊗W ⊗V 1⊗T
→ V ⊗k = V and W = W ⊗k 1⊗S
T ⊗1
→ W ⊗V ⊗W → k ⊗W = W
are both the identity maps (here 1 ⊗ S, for example, means the map 1 ⊗ S(v ⊗ c) = v ⊗ S(c)). Show
that there is an isomorphism W ∼ = V ∗ such that when you identify W with V ∗ , S(1) is the identity map
I ∈ Hom(V, V ) = V ⊗ V ∗ and T : V ∗ ⊗ V → k is the trace map.
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