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im J|W = {0} — W .
g) There are unique linear transformations p : V ’’ V and p : V ’’ V for which
2
p2 = p —¦ p = p, p =p —¦p =p,
im p = W, im p = W ,
IdV = p + p .

We often write V = W • W whenever W is a linear complement of W . The maps p, p of
Theorem 1.5(g) are often called the (linear ) projections onto W and W . This can be extended
to the situation where there are r subspaces V1 , . . . , Vr ⊆ V for which
± 
r 
V = V1 + · · · + Vr = vj : vj ∈ Vj ,
 
j=1
2. CLASS FUNCTIONS AND THE CAYLEY-HAMILTON THEOREM 5

and we inductively have that Vk is a linear complement of (V1 • · · · • Vk’1 ) in (V1 + · · · + Vk ).
A linear complement for a subspace W ⊆ V always exists since we can extend a basis
{w1 , . . . , wr } of W to a basis {w1 , . . . , wr , wr+1 , . . . , wn } for V and then take W to be the
subspace spanned by {wr+1 , . . . , wn }. Theorem 1.5(b) implies that W is a linear complement.

2. Class functions and the Cayley-Hamilton Theorem
In this section k can be any ¬eld. Let A = [aij ] be an n — n matrix over k.

Definition 1.6. The characteristic polynomial of A is the polynomial (in the variable X)
n
ck (A)X k ∈ k[X],
charA (X) = det(XIn ’ [aij ]) =
k=0

where In is the n — n identity matrix.

This polynomial is monic and of degree n in X. The coe¬cients cr (A) ∈ k are functions of
the entries aij . The following is an important result about this polynomial.

Theorem 1.7 (Cayley-Hamilton Theorem: matrix version). The matrix A satis¬es
the polynomial identity
n
ck (A)Ak = 0.
charA (A) =
k=0

Example 1.8. Let
0 ’1
A= ∈ R[X].
10
Then
X1
= X 2 + 1.
charA (X) = det
’1 X
By calculation we ¬nd that A2 + I2 = O2 as claimed.

Lemma 1.9. Let A = [aij ] and P be an n — n matrix with coe¬cients in k. Then if P is
invertible,
charP AP ’1 (X) = charA (X).
Thus each of the coe¬cients ck (A) (0 k n) satis¬es
ck (P AP ’1 ) = ck (A).

Proof. We have
charP AP ’1 (X) = det(XIn ’ P AP ’1 )
= det(P (XIn )P ’1 ’ P AP ’1 )
= det(P (XIn ’ A)P ’1 )
= det P det(XIn ’ A) det P ’1
= charA (X).
Now comparing coe¬cients we obtain the result.

This result shows that as functions of A (hence of the aij ), the coe¬cients ck (A) are invariant
or class functions in the sense that they are invariant under conjugation,
cr (P AP ’1 ) = cr (A).
6 1. LINEAR AND MULTILINEAR ALGEBRA

Recall that for an n — n matrix A = [aij ], the trace of A, Tr A ∈ k, is de¬ned by
n
Tr A = ajj .
j=1

Proposition 1.10. For any n — n matrix over k we have

cn (A) = (’1)n det A.
cn’1 (A) = ’ Tr A and

Proof. Calculating the coe¬cient of X n’1 in det(XIn ’ [aij ]) we get
n
’ arr = ’ Tr[aij ].
r=1

Putting X = 0 gives
cn (A) = det([’aij ]) = (’1)n det[aij ].

Now let • : V ’’ V be a linear transformation on a ¬nite dimensional k-vector space with
a basis v = {v1 , . . . , vn }. Consider the matrix of • relative to v,

[•]v = [aij ],

where
n
•vj = arj vr .
r=1
Then the trace of • with respect to the basis v is

Trv • = Tr[•]v .

If we change to a second basis w say, there is an invertible n — n matrix P = [pij ] such that
n
wj = prj vr ,
r=1

and then
[•]w = P [•]v P ’1 .
Hence,
Trw • = Tr P [•]v P ’1 = Trv •.
Thus we see that the quantity
Tr • = Trv •
only depends on •, not the basis v. We call this the trace of •. We can similarly de¬ne
det • = det A.
More generally, we can consider the polynomial

char• (X) = char[•]v (X)

which by Lemma 1.9 is independant of the basis v. Thus all of the coe¬cients ck (A) are
functions of • and do not depend on the basis used, so we may write ck (•) in place of ck (A).
In particular, an alternative way to de¬ne Tr • and det • is as

det • = (’1)n det A.
Tr • = cn’1 (•) and

We also call char• (X) the characteristic polynomial of •. The following is a formulation of the
Cayley-Hamilton Theorem for a linear transformation.
2. CLASS FUNCTIONS AND THE CAYLEY-HAMILTON THEOREM 7

Theorem 1.11 (Cayley-Hamilton Theorem: linear transformation version).
If • : V ’’ V is a k-linear transformation on the ¬nite dimensional k-vector space V , then •
satis¬es the polynomial identity
char• (•) = 0.
More explicitly, if
n
cr (•)X r ,
char• (X) =
r=0
then
n
cr (•)•r = 0
r=0
where •0 = IdV .

There is an important connection between class functions of matrices such as the trace
and determinant and eigenvalues. It can be shown that for a complex square matrix A (or
more generally a matrix over an algebraically closed ¬eld k), the distinct eigenvalues are the
distinct roots of the characteristic polynomial charA (X) (thus there are at most n distinct
eigenvalues). However, on factoring charA (X) into linear factors we can get repeated linear
factors, corresponding to ˜repeated™ roots. If a linear factor (X ’ ») appears to degree d say, we
say that » is an eigenvalue of multiplicity d. If every eigenvalue of A has multiplicity 1 then A
is diagonalisable in the sense that there is an invertible matrix P satisfying
P AP ’1 = diag(»1 , . . . , »n ),
the diagonal matrix with the n distinct diagonal entries »k down the leading diagonal. More
generally, let
(2.1) charA (X) = (X ’ »1 ) · · · (X ’ »n )
where now we allow some of the »j to be repeated. Then we can describe Tr A and det A in
terms of the eigenvalues »j .

Proposition 1.12. We have
n
det A = (’1)n »1 · · · »n .
Tr A = »j and
j=1

Proof. This follows by considering the degree (n ’ 1) and constant terms in Equation (2.1)
and using Proposition 1.10.

We can also apply the above discussion to a linear transformation • : V ’’ V , where an
eigenvector for the eigenvalue » ∈ C is a non-zero vector v ∈ V satisfying •(v) = »v.
The characteristic polynomial may not be the smallest degree polynomial satis¬ed by a
matrix or a linear transformation. By de¬nition, a minimal polynomial of an n — n matrix A or
linear transformation • : V ’’ V is a (non-zero) monic polynomial f (X) of smallest possible
degree for which f (A) = 0 or f (•) = 0.

Lemma 1.13. For an n — n matrix A or a linear transformation • : V ’’ V , let f (X) be a
minimal polynomial and g(X) be any other polynomial for which g(A) = 0 or g(•) = 0. Then
f (X) | g(X). Hence f (X) is unique.

Proof. We give the proof for matrices, that for a linear transformation is similar.
Suppose that f (X) g(X). Then we have
g(X) = q(X)f (X) + r(X)
8 1. LINEAR AND MULTILINEAR ALGEBRA

where deg r(X) < deg f (X). Since r(A) = 0 and r(X) is of degree less than f (X), we have
a contradiction. Hence f (X) | g(X). In particular, if g(X) is of the same degree as f (X),
minimality of g(X) also gives g(X) | f (X), giving f (X) = g(X) as these are both monic.
We write minA (X) or min• (X) for the minimal polynomial of A or •. Note also that
minA (X) | charA (X) and min• (X) | char• (X).

3. Separability
Lemma 1.14. Let V be a ¬nite dimensional vector space over C and • : V ’’ V a linear
transformation. Suppose that
m
cr X r ∈ C[X]
0 = f (X) =
r=0
is a polynomial with no repeated linear factors over C and that • satis¬es the relation
m
cr •r = 0,
r=0

i.e., for every v ∈ V
m
cr •r (v) = 0.
,
r=0
Then there is a basis v = {v1 , . . . , vn } of V consisting of eigenvectors of the linear transformation
•.
Proof. By multiplying by the inverse of the leading coe¬cient of f (X) we can replace
f (X) by a monic polynomial with the same properties, so we will assume that f (X) is monic,
i.e., cm = 1. Factoring over C, we obtain
f (X) = fm (X) = (X ’ »1 ) · · · (X ’ »m ),
where the »j ∈ C are distinct. Put
fm’1 (X) = (X ’ »1 ) · · · (X ’ »m’1 ).
Notice that fm (X) = fm’1 (X)(X ’ »m ), hence (X ’ »m ) cannot divide fm’1 (X), since this
would lead to a contradiction to the assumption that fm (X) has no repeated linear factors.
Using long division of (X ’ »m ) into fm’1 (X), we see that
fm’1 (X) = qm (X)(X ’ »m ) + rm ,
where the remainder rm ∈ C cannot be 0 since then (X ’ »m ) would divide fm’1 (X). Dividing
by rm if necessary, we see that for some non-zero sm ∈ C,
sm fm’1 (X) ’ qm (X)(X ’ »m ) = 1.
Substituting X = •, we have for any v ∈ V ,
sm fm’1 (•)(v) ’ qm (•)(• ’ »m IdV )(v) = v.
Notice that we have
(• ’ »m IdV ) (sm fm’1 (•)(v)) = sm fm (•)(v) = 0
and
fm’1 (•) (qm (•)(• ’ »m IdV )(v)) = qm (•)fm (•)(v) = 0.
Thus we can decompose v into a sum v = vm + vm where
(• ’ »m IdV )(vm ) = 0 and fm’1 (•)(vm ) = 0.
4. BASIC NOTIONS OF MULTILINEAR ALGEBRA 9

Let
Vm = {v ∈ V : (• ’ »m IdV )(v) = 0},
Vm = {v ∈ V : fm’1 (•)(v) = 0}.
Thus we have shown that V = Vm + Vm . If v ∈ Vm © Vm , then from above we would have
v = sm fm’1 (•)(v) ’ qm (•)(• ’ »m IdV )(v) = 0.
So Vm © Vm = {0}, hence V = Vm • Vm . We can now consider Vm in place of V , noticing that
for v ∈ Vm , •(v) ∈ Vm , since
fm’1 (•)(•(v)) = • (fm’1 (•)(v)) = 0.
Continuing in this fashion, we eventually see that
V = V1 • · · · • Vm
where for v ∈ Vk ,
(• ’ »k )(v) = 0.
If we choose a basis v(k) of Vk , then the (disjoint) union
v = v(1) ∪ · · · ∪ v(m)
is a basis for V , consisting of eigenvectors of •.
The condition on • in this result is sometimes referred to as the separability or semisimplicity
of •. We will make use of this when discussing characters of representations.

4. Basic notions of multilinear algebra
In this section we will describe the tensor product of r vector spaces. We will most often
consider the case where r = 2, but give the general case for completeness. Multilinear algebra
is important in di¬erential geometry, relativity, electromagnetism, ¬‚uid mechanics and indeed
much of advanced applied mathematics where tensors play a rˆle.o
Let V1 , . . . , Vr and W be k-vector spaces. A function
F : V1 — · · · — Vr ’’ W
is k-multilinear if it satis¬es
(ML-1)
F (v1 , . . . , vk’1 , vk + vk , vk+1 , . . . , vr ) = F (v1 , . . . , vk , . . . , vr ) + F (v1 , . . . , vk’1 , vk , vk+1 , . . . , vr ),
(ML-2)
F (v1 , . . . , vk’1 , tvk , vk+1 , . . . , vr ) = tF (v1 , . . . , vk’1 , vk , vk+1 , . . . , vr )
for vj , vj ∈ V and t ∈ k. It is symmetric if for any permutation σ ∈ Sr (the permutation group
on r objects),
(ML-S) F (vσ(1) , . . . , vσ(k) , . . . , vσ(r) ) = F (v1 , . . . , vk , . . . , vr ),

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