Hermitian Spaces
13.1
Introduction
The goal of this chapter is to show that there are nice normal forms for symmetric matrices,
skew-symmetric matrices, orthogonal matrices, and normal matrices. The spectral theorem
for symmetric matrices states that symmetric matrices have real eigenvalues and that they
can be diagonalized over an orthonormal basis. The spectral theorem for Hermitian matrices
states that Hermitian matrices also have real eigenvalues and that they can be diagonalized
over a complex orthonormal basis. Normal real matrices can be block diagonalized over an
orthonormal basis with blocks having size at most two, and there are refinements of this
normal form for skew-symmetric and orthogonal matrices.
13.2
Normal Linear Maps
We begin by studying normal maps, to understand the structure of their eigenvalues and
eigenvectors. This section and the next two were inspired by Lang [65], Artin [3], Mac Lane
and Birkhoff [70], Berger [6], and Bertin [10].
Definition 13.1. Given a Euclidean space E, a linear map f : E → E is normal if
f ◦ f∗ = f∗ ◦ f.
A linear map f : E → E is self-adjoint if f = f∗, skew-self-adjoint if f = −f∗, and orthogonal
if f ◦ f∗ = f∗ ◦ f = id.
Obviously, a self-adjoint, skew-self-adjoint, or orthogonal linear map is a normal linear
map. Our first goal is to show that for every normal linear map f : E → E, there is an
orthonormal basis (w.r.t. −, − ) such that the matrix of f over this basis has an especially
331
332
CHAPTER 13. SPECTRAL THEOREMS
nice form: It is a block diagonal matrix in which the blocks are either one-dimensional
matrices (i.e., single entries) or two-dimensional matrices of the form
λ
µ .
−µ λ
This normal form can be further refined if f is self-adjoint, skew-self-adjoint, or orthog-
onal. As a first step, we show that f and f ∗ have the same kernel when f is normal.
Proposition 13.1. Given a Euclidean space E, if f : E → E is a normal linear map, then
Ker f = Ker f ∗.
Proof. First, let us prove that
f (u), f (v) = f ∗(u), f ∗(v)
for all u, v ∈ E. Since f∗ is the adjoint of f and f ◦ f∗ = f∗ ◦ f, we have
f (u), f (u) = u, (f ∗ ◦ f)(u) ,
= u, (f ◦ f∗)(u) ,
= f ∗(u), f ∗(u) .
Since −, − is positive definite,
f (u), f (u) = 0 iff f (u) = 0,
f ∗(u), f ∗(u) = 0 iff f ∗(u) = 0,
and since
f (u), f (u) = f ∗(u), f ∗(u) ,
we have
f (u) = 0 iff f ∗(u) = 0.
Consequently, Ker f = Ker f ∗.
The next step is to show that for every linear map f : E → E there is some subspace W
of dimension 1 or 2 such that f (W ) ⊆ W . When dim(W ) = 1, the subspace W is actually
an eigenspace for some real eigenvalue of f . Furthermore, when f is normal, there is a
subspace W of dimension 1 or 2 such that f (W ) ⊆ W and f∗(W ) ⊆ W . The difficulty is
that the eigenvalues of f are not necessarily real. One way to get around this problem is to
complexify both the vector space E and the inner product −, − .
Every real vector space E can be embedded into a complex vector space E , and every
C
linear map f : E → E can be extended to a linear map f : E → E .
C
C
C
13.2. NORMAL LINEAR MAPS
333
Definition 13.2. Given a real vector space E, let E be the structure E × E under the
C
addition operation
(u1, u2) + (v1, v2) = (u1 + v1, u2 + v2),
and let multiplication by a complex scalar z = x + iy be defined such that
(x + iy) · (u, v) = (xu − yv, yu + xv).
The space E is called the complexification of E.
C
It is easily shown that the structure E is a complex vector space. It is also immediate
C
that
(0, v) = i(v, 0),
and thus, identifying E with the subspace of E consisting of all vectors of the form (u, 0),
C
we can write
(u, v) = u + iv.
Observe that if (e1, . . . , en) is a basis of E (a real vector space), then (e1, . . . , en) is also
a basis of E (recall that e
C
i is an abreviation for (ei, 0)).
A linear map f : E → E is extended to the linear map f : E → E defined such that
C
C
C
f (u + iv) = f (u) + if (v).
C
For any basis (e1, . . . , en) of E, the matrix M(f ) representing f over (e1, . . . , en) is iden-
tical to the matrix M (f ) representing f over (e
C
C
1, . . . , en), where we view (e1, . . . , en) as a
basis of E . As a consequence, det(zI − M(f)) = det(zI − M(f )), which means that f
C
C
and f have the same characteristic polynomial (which has real coefficients). We know that
C
every polynomial of degree n with real (or complex) coefficients always has n complex roots
(counted with their multiplicity), and the roots of det(zI − M(f )) that are real (if any) are
C
the eigenvalues of f .
Next, we need to extend the inner product on E to an inner product on E .
C
The inner product −, − on a Euclidean space E is extended to the Hermitian positive
definite form −, −
on E as follows:
C
C
u1 + iv1, u2 + iv2
= u
C
1, u2 + v1, v2 + i( u2, v1 − u1, v2 ).
It is easily verified that −, −
is indeed a Hermitian form that is positive definite, and
C
it is clear that −, −
agrees with −, − on real vectors. Then, given any linear map
C
f : E → E, it is easily verified that the map f∗ defined such that
C
f ∗(u + iv) = f ∗(u) + if ∗(v)
C
for all u, v ∈ E is the adjoint of f w.r.t. −, − .
C
C
Assuming again that E is a Hermitian space, observe that Proposition 13.1 also holds.
We deduce the following corollary.
334
CHAPTER 13. SPECTRAL THEOREMS
Proposition 13.2. Given a Hermitian space E, for any normal linear map f : E → E, we
have Ker (f ) ∩ Im(f) = (0).
Proof. Assume v ∈ Ker (f) ∩ Im(f) = (0), which means that v = f(u) for some u ∈ E, and
f (v) = 0. By Proposition 13.1, Ker (f ) = Ker (f ∗), so f (v) = 0 implies that f ∗(v) = 0.
Consequently,
0 = f ∗(v), u
= v, f (u)
= v, v ,
and thus, v = 0.
We also have the following crucial proposition relating the eigenvalues of f and f ∗.
Proposition 13.3. Given a Hermitian space E, for any normal linear map f : E → E, a
vector u is an eigenvector of f for the eigenvalue λ (in C) iff u is an eigenvector of f ∗ for
the eigenvalue λ.
Proof. First, it is immediately verified that the adjoint of f − λ id is f∗ − λ id. Furthermore,
f − λ id is normal. Indeed,
(f − λ id) ◦ (f − λ id)∗ = (f − λ id) ◦ (f∗ − λ id),
= f ◦ f∗ − λf − λf∗ + λλ id,
= f ∗ ◦ f − λf∗ − λf + λλ id,
= (f ∗ − λ id) ◦ (f − λ id),
= (f − λ id)∗ ◦ (f − λ id).
Applying Proposition 13.1 to f − λ id, for every nonnull vector u, we see that
(f − λ id)(u) = 0 iff (f∗ − λ id)(u) = 0,
which is exactly the statement of the proposition.
The next proposition shows a very important property of normal linear maps: Eigenvec-
tors corresponding to distinct eigenvalues are orthogonal.
Proposition 13.4. Given a Hermitian space E, for any normal linear map f : E → E, if
u and v are eigenvectors of f associated with the eigenvalues λ and µ (in C) where λ = µ,
then u, v = 0.
Proof. Let us compute f (u), v in two different ways. Since v is an eigenvector of f for µ,
by Proposition 13.3, v is also an eigenvector of f ∗ for µ, and we have
f (u), v = λu, v = λ u, v
13.2. NORMAL LINEAR MAPS
335
and
f (u), v = u, f ∗(v) = u, µv = µ u, v ,
where the last identity holds because of the semilinearity in the second argument, and thus
λ u, v = µ u, v ,
that is,
(λ − µ) u, v = 0,
which implies that u, v = 0, since λ = µ.
We can also show easily that the eigenvalues of a self-adjoint linear map are real.
Proposition 13.5. Given a Hermitian space E, all the eigenvalues of any self-adjoint linear
map f : E → E are real.
Proof. Let z (in C) be an eigenvalue of f and let u be an eigenvector for z. We compute
f (u), u in two different ways. We have
f (u), u = zu, u = z u, u ,
and since f = f ∗, we also have
f (u), u = u, f ∗(u) = u, f (u) = u, zu = z u, u .
Thus,
z u, u = z u, u ,
which implies that z = z, since u = 0, and z is indeed real.
There is also a version of Proposition 13.5 for a (real) Euclidean space E and a self-adjoint
map f : E → E.
Proposition 13.6. Given a Euclidean space E, if f : E → E is any self-adjoint linear map,
then every eigenvalue λ of f is real and is actually an eigenvalue of f (which means that
C
there is some real eigenvector u ∈ E such that f(u) = λu). Therefore, all the eigenvalues of
f are real.
Proof. Let E be the complexification of E, −, −
the complexification of the inner product
C
C
−, − on E, and f : E → E the complexification of f : E → E. By definition of f and
C
C
C
C
−, − , if f is self-adjoint, we have
C
f (u
= f (u
C
1 + iv1), u2 + iv2 C
1) + if (v1), u2 + iv2 C
= f (u1), u2 + f (v1), v2 + i( u2, f(v1) − f(u1), v2 )
= u1, f(u2) + v1, f(v2) + i( f (u2), v1 − u1, f(v2) )
= u1 + iv1, f(u2) + if (v2) C
= u1 + iv1, f (u
,
C
2 + iv2) C
336
CHAPTER 13. SPECTRAL THEOREMS
which shows that f is also self-adjoint with respect to −, − .
C
C
As we pointed out earlier, f and f have the same characteristic polynomial det(zI−f ) =
C
C
det(zI − f), which is a polynomial with real coefficients. Proposition 13.5 shows that the
zeros of det(zI − f ) = det(zI − f) are all real, and for each real zero λ of det(zI − f), the
C
linear map λid − f is singular, which means that there is some nonzero u ∈ E such that
f (u) = λu. Therefore, all the eigenvalues of f are real.
Given any subspace W of a Euclidean space E, recall that the orthogonal complement
W ⊥ of W is the subspace defined such that
W ⊥ = {u ∈ E | u, w = 0, for all w ∈ W }.
Recall from Proposition 9.9 that E = W ⊕ W ⊥ (this can be easily shown, for example,
by constructing an orthonormal basis of E using the Gram–Schmidt orthonormalization
procedure). The same result also holds for Hermitian spaces; see Proposition 11.10.
As a warm up for the proof of Theorem 13.10, let us prove that every self-adjoint map on
a Euclidean space can be diagonalized with respect to an orthonormal basis of eigenvectors.
Theorem 13.7. (Spectral theorem for self-adjoint linear maps on a Euclidean space) Given
a Euclidean space E of dimension n, for every self-adjoint linear map f : E → E, there is
an orthonormal basis (e1, . . . , en) of eigenvectors of f such that the matrix of f w.r.t. this
basis is a diagonal matrix
λ
1
. . .
λ2 . . .
.
.
.
. ,
..
..
. .
..
. . . λn
with λi ∈ R.
Proof. We proceed by induction on the dimension n of E as follows. If n = 1, the result is
trivial. Assume now that n ≥ 2. From Proposition 13.6, all the eigenvalues of f are real, so
pick some eigenvalue λ ∈ R, and let w be some eigenvector for λ. By dividing w by its norm,
we may assume that w is a unit vector. Let W be the subspace of dimension 1 spanned by w.
Clearly, f (W ) ⊆ W . We claim that f(W ⊥) ⊆ W ⊥, where W ⊥ is the orthogonal complement
of W .
Indeed, for any v ∈ W ⊥, that is, if v, w = 0, because f is self-adjoint and f(w) = λw,
we have
f (v), w = v, f (w)
= v, λw
= λ v, w = 0
13.2. NORMAL LINEAR MAPS
337
since v, w = 0. Therefore,
f (W ⊥) ⊆ W ⊥.
Clearly, the restriction of f to W ⊥ is self-adjoint, and we conclude by applying the induction
hypothesis to W ⊥ (whose dimension is n − 1).
We now come back to normal linear maps. One of the key points in the proof of Theorem
13.7 is that we found a subspace W with the property that f (W ) ⊆ W implies that f(W ⊥) ⊆
W ⊥. In general, this does not happen, but normal maps satisfy a stronger property which
ensures that such a subspace exists.
The following proposition provides a condition that will allow us to show that a nor-
mal linear map can be diagonalized. It actually holds for any linear map. We found the
inspiration for this proposition in Berger [6].
Proposition 13.8. Given a Hermitian space E, for any linear map f : E → E and any
subspace W of E, if f (W ) ⊆ W , then f∗ W ⊥ ⊆ W ⊥. Consequently, if f(W ) ⊆ W and
f ∗(W ) ⊆ W , then f W ⊥ ⊆ W ⊥ and f∗ W ⊥ ⊆ W ⊥.
Proof. If u ∈ W ⊥, then
w, u = 0 for all w ∈ W .
However,
f (w), u = w, f ∗(u) ,
and f (W ) ⊆ W implies that f(w) ∈ W . Since u ∈ W ⊥, we get
0 = f (w), u = w, f ∗(u) ,
which shows that w, f ∗(u) = 0 for all w ∈ W , that is, f∗(u) ∈ W ⊥. Therefore, we have
f ∗(W ⊥) ⊆ W ⊥.
We just proved that if f (W ) ⊆ W , then f∗ W ⊥ ⊆ W ⊥. If we also have f∗(W ) ⊆ W ,
then by applying the above fact to f ∗, we get f ∗∗(W ⊥) ⊆ W ⊥, and since f∗∗ = f, this is
just f (W ⊥) ⊆ W ⊥, which proves the second statement of the proposition.
It is clear that the above proposition also holds for Euclidean spaces.
Although we are ready to prove that for every normal linear map f (over a Hermitian
space) there is an orthonormal basis of eigenvectors (see Theorem 13.11 below), we now
return to real Euclidean spaces.
If f : E → E is a linear map and w = u + iv is an eigenvector of f : E → E for the
C
C
C
eigenvalue z = λ + iµ, where u, v ∈ E and λ, µ ∈ R, since
f (u + iv) = f (u) + if (v)
C
and
f (u + iv) = (λ + iµ)(u + iv) = λu − µv + i(µu + λv),
C
338
CHAPTER 13. SPECTRAL THEOREMS
we have
f (u) = λu − µv and f(v) = µu + λv,
from which we immediately obtain
f (u − iv) = (λ − iµ)(u − iv),
C
which shows that w = u − iv is an eigenvector of f for z = λ − iµ. Using this fact, we can
C
prove the following proposition.
Proposition 13.9. Given a Euclidean space E, for any normal linear map f : E → E, if
w = u + iv is an eigenvector of f associated with the eigenvalue z = λ + iµ (where u, v ∈ E
C
and λ, µ ∈ R), if µ = 0 (i.e., z is not real) then u, v = 0 and u, u = v, v , which implies
that u and v are linearly independent, and if W is the subspace spanned by u and v, then
f (W ) = W and f ∗(W ) = W . Furthermore, with respect to the (orthogonal) basis (u, v), the
restriction of f to W has the matrix
λ
µ .
−µ λ
If µ = 0, then λ is a real eigenvalue of f , and either u or v is an eigenvector of f for λ. If
W is the subspace spanned by u if u = 0, or spanned by v = 0 if u = 0, then f (W ) ⊆ W and
f ∗(W ) ⊆ W .
Proof. Since w = u + iv is an eigenvector of f , by definition it is nonnull, and either u = 0
C
or v = 0. From the fact stated just before Proposition 13.9, u − iv is an eigenvector of f for
C
λ − iµ. It is easy to check that f is normal. However, if µ = 0, then λ + iµ = λ − iµ, and
C
from Proposition 13.4, the vectors u + iv and u − iv are orthogonal w.r.t. −, − , that is,
C
u + iv, u − iv
= u, u − v, v + 2i u, v = 0.
C
Thus, we get u, v = 0 and u, u = v, v , and since u = 0 or v = 0, u and v are linearly
independent. Since
f (u) = λu − µv and f(v) = µu + λv
and since by Proposition 13.3 u + iv is an eigenvector of f ∗ for λ − iµ, we have
C
f ∗(u) = λu + µv
and f ∗(v) = −µu + λv,
and thus f (W ) = W and f ∗(W ) = W , where W is the subspace spanned by u and v.
When µ = 0, we have
f (u) = λu and f (v) = λv,
and since u = 0 or v = 0, either u or v is an eigenvector of f for λ. If W is the subspace
spanned by u if u = 0, or spanned by v if u = 0, it is obvious that f (W ) ⊆ W and
f ∗(W ) ⊆ W . Note that λ = 0 is possible, and this is why ⊆ cannot be replaced by =.
13.2. NORMAL LINEAR MAPS
339
The beginning of the proof of Proposition 13.9 actually shows that for every linear map
f : E → E there is some subspace W such that f(W ) ⊆ W , where W has dimension 1 or
2. In general, it doesn’t seem possible to prove that W ⊥ is invariant under f . However, this
happens when f is normal.
We can finally prove our first main theorem.
Theorem 13.10. (Main spectral theorem) Given a Euclidean space E of dimension n, for
every normal linear map f : E → E, there is an orthonormal basis (e1, . . . , en) such that the
matrix of f w.r.t. this basis is a block diagonal matrix of the form
A
1
. . .
A2 . . .
.
.
.
.
..
..
. .
..
. . . Ap
such that each block Aj is either a one-dimensional matrix (i.e., a real scalar) or a two-
dimensional matrix of the form
λ
A
j
−µj
j =
,
µj
λj
where λj, µj ∈ R, with µj > 0.
Proof. We proceed by induction on the dimension n of E as follows. If n = 1, the result is
trivial. Assume now that n ≥ 2. First, since C is algebraically closed (i.e., every polynomial
has a root in C), the linear map f : E → E has some eigenvalue z = λ + iµ (where
C
C
C
λ, µ ∈ R). Let w = u + iv be some eigenvector of f for λ + iµ (where u, v ∈ E). We can
C
now apply Proposition 13.9.
If µ = 0, then either u or v is an eigenvector of f for λ ∈ R. Let W be the subspace
of dimension 1 spanned by e1 = u/ u if u = 0, or by e1 = v/ v otherwise. It is obvious
that f (W ) ⊆ W and f∗(W ) ⊆ W . The orthogonal W ⊥ of W has dimension n − 1, and by
Proposition 13.8, we have f W ⊥ ⊆ W ⊥. But the restriction of f to W ⊥ is also normal,
and we conclude by applying the induction hypothesis to W ⊥.
If µ = 0, then u, v = 0 and u, u = v, v , and if W is the subspace spanned by u/ u
and v/ v , then f (W ) = W and f ∗(W ) = W . We also know that the restriction of f to W
has the matrix
λ
µ
−µ λ
with respect to the basis (u/ u , v/ v ). If µ < 0, we let λ1 = λ, µ1 = −µ, e1 = u/ u , and
e2 = v/ v . If µ > 0, we let λ1 = λ, µ1 = µ, e1 = v/ v , and e2 = u/ u . In all cases, it
is easily verified that the matrix of the restriction of f to W w.r.t. the orthonormal basis
(e1, e2) is
λ
A
1
−µ1
1 =
,
µ1
λ1
340
CHAPTER 13. SPECTRAL THEOREMS
where λ1, µ1 ∈ R, with µ1 > 0. However, W ⊥ has dimension n − 2, and by Proposition 13.8,
f W ⊥ ⊆ W ⊥. Since the restriction of f to W ⊥ is also normal, we conclude by applying
the induction hypothesis to W ⊥.
After this relatively hard work, we can easily obtain some nice normal forms for the
matrices of self-adjoint, skew-self-adjoint, and orthogonal linear maps. However, for the sake
of completeness (and since we have all the tools to so do), we go back to the case of a
Hermitian space and show that normal linear maps can be diagonalized with respect to an
orthonormal basis. The proof is a slight generalization of the proof of Theorem 13.6.
Theorem 13.11. (Spectral theorem for normal linear maps on a Hermitian space) Given
a Hermitian space E of dimension n, for every normal linear map f : E → E there is an
orthonormal basis (e1, . . . , en) of eigenvectors of f such that the matrix of f w.r.t. this basis
is a diagonal matrix
λ
1
. . .
λ2 . . .
.
.
.
. ,
..
..
. .
..
. . . λn
where λj ∈ C.
Proof. We proceed by induction on the dimension n of E as follows. If n = 1, the result is
trivial. Assume now that n ≥ 2. Since C is algebraically closed (i.e., every polynomial has
a root in C), the linear map f : E → E has some eigenvalue λ ∈ C, and let w be some unit
eigenvector for λ. Let W be the subspace of dimension 1 spanned by w. Clearly, f (W ) ⊆ W .
By Proposition 13.3, w is an eigenvector of f ∗ for λ, and thus f ∗(W ) ⊆ W . By Proposition
13.8, we also have f (W ⊥) ⊆ W ⊥. The restriction of f to W ⊥ is still normal, and we conclude
by applying the induction hypothesis to W ⊥ (whose dimension is n − 1).
Thus, in particular, self-adjoint, skew-self-adjoint, and orthogonal linear maps can be
diagonalized with respect to an orthonormal basis of eigenvectors. In this latter case, though,
an orthogonal map is called a unitary map. Also, Proposition 13.5 shows that the eigenvalues
of a self-adjoint linear map are real. It is easily shown that skew-self-adjoint maps have
eigenvalues that are pure imaginary or null, and that unitary maps have eigenvalues of
absolute value 1.
Remark: There is a converse to Theorem 13.11, namely, if there is an orthonormal basis
(e1, . . . , en) of eigenvectors of f , then f is normal. We leave the easy proof as an exercise.
13.3
Self-Adjoint, Skew-Self-Adjoint, and Orthogonal
Linear Maps
We begin with self-adjoint maps.
13.3. SELF-ADJOINT AND OTHER SPECIAL LINEAR MAPS
341
Theorem 13.12. Given a Euclidean space E of dimension n, for every self-adjoint linear
map f : E → E, there is an orthonormal basis (e1, . . . , en) of eigenvectors of f such that the
matrix of f w.r.t. this basis is a diagonal matrix
λ
1
. . .
λ2 . . .
.
.
.
. ,
..
..
. .
..