Download Proposition 7.3 If α : V → V is self-adjoint, then 1) Every eigenvalue

Proposition 7.3 If α : V → V is self-adjoint, then
1) Every eigenvalue of α is real.
2) Eigenvectors with distinct eigenvalues are orthogonal.
Proof 1) Suppose that α(v) = λv with v 6= 0. Then
hα(v), vi = hv, α(v)i
⇓
hλv, vi = hv, λvi
⇓
λhv, vi = λhv, vi.
As hv, vi =
6 0 it follows that λ = λ and thus λ ∈ R.
2) If α(vi ) = λi vi where λ1 6= λ2 then
λ1 hv1 , v2 i = hα(v1 ), v2 i = hv1 , α(v2 )i = λ2 hv1 , v2 i
(notice that we have used the fact that λ1 ∈ R). It follows that hv1 , v2 i = 0 as λ1 6= λ2 . 2
Corollary 7.4 If A is a hermitian n × n matrix then
∆A (t) = (λ1 − t) · · · (λn − t)
with λ1 , . . . , λn ∈ R.
Proof The corresponding linear map α : Cn → Cn , given by α(v) = Av, is self-adjoint and it
follows from the last proposition that all the eigenvalues are real. 2.
Remark. (Version for self-adjoint maps). If we state last corollary in terms of linear maps, we
get the following: if α : V → is self-adjoint then
∆α (t) = (λ1 − t) · · · (λn − t)
with λ1 , . . . , λn ∈ R.
If V is an inner product space, then any subspace W is also an inner product space with respect
to the same inner product. Furthermore, if α : V → V is self-adjoint and W is α-invariant, then
α|W is self-adjoint as well.
Lemma 7.5 If α : V → V is self-adjoint and W is α-invariant, then
W ⊥ = {v ∈ V : hv, wi = 0 ∀w ∈ W }
is also α-invariant.
Proof Let v ∈ W ⊥ . We want to show that α(v) ∈ W ⊥ but for any w ∈ W we have
hα(v), wi = hv, α(w)i
which is zero as α(w) ∈ W and v ∈ W ⊥ . This shows that α(v) ∈ W ⊥ as required. 2.
The next result is the main result of this chapter and tells us that a self-adjoint operator
α : V → V on a finite dimensional inner product space is always diagonalisable. In fact more is
true as we will see that one can moreover choose the basis of eigenvectors to be orthonormal.
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Theorem 7.6 (Spectral Theorem) Let α : V → V be a self-adjoint linear operator on an ndimensional inner product space. Then there is an orthonormal basis v1 , . . . , vn of eigenvectors
of α.
Proof (Not examinable). We first prove by induction on n that we can find a orthonormal
basis of eigenvectors. For the case n = 1 pick a orthonormal vector (which is then ofcourse an
eigenvector).
Now suppose that n > 1 and the result is true for all V with dim V < n. We know from Corollary
7.4 that ∆α (t) has a real eigenvalue λ. Let W1 = Eα (λ) be the corresponding eigenspace and
choose any orthonormal basis for W1 . By Lemma 7.5 we know that W2 = W1⊥ is α-invariant.
The restriction α|W2 is then self-adjoint and by induction hypothesis, we can choose a orthonormal basis of or W2 consisting of eigenvectors of α|W2 , i.e. of α. Finally, V = W1 ⊕ W2 is
an orthogonal direct sum decomposition, so we can combine the two bases above to obtain an
orthonormal basis of V . 2
Corollary 7.7 If λ1 , . . . , λn are the corresponding eigenvalues with respect to the ON basis
v1 , . . . , vn , then α can be written explicitly by the formula
n
X
α(w) =
k=1
λk hvk , wivk .
Proof The explicit formula for α follows from the fact that
w=
n
X
k=1
hvk , wi vk .
It then follows that
α(w) =
n
X
hvk , wi α(vk )
k=1
n
X
=
hvk , wi λ α(vk ).
k=1
Theorem 7.8 (The matrix version of the Spectral Theorem)
(1) If A is a symmetric n × n matrix over R then there is an orthogonal invertible real matrix P (i.e. P −1 = P t ) such that P −1 AP is diagonal.
(2) If A is a hermitian n × n matrix over C then there is a unitary complex matrix P (i.e.
t
P −1 = P ) such that P −1 AP is diagonal.
Proof (Not examinable) In both cases the linear map α : Fn → Fn mapping v to Av is selfadjoint. By the Spectral Theorem we have an orthonormal basis v1 , . . . , vn of eigenvectors. Let
P be the base change matrix from the orthonormal basis to the standard basis e1 , . . . , en . The
columns of P are the coordinates of the orthonormal vectors v1 , . . . , vn expressed in e1 , . . . , en .
t
This translates into P t P = I when P is real and P P = I when P is complex. 2
Example. Consider the hermitian matrix
A=
1 i
−i 1
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We have ∆A (t) = (t − 1)2 − 1 = t(t − 2) and the eigenvalues are 0 and 2. We next determine
the eigenspaces. Firstly we solve Av = 0. That is
1 i
x
0
=
−i 1
y
0
that has the solution C
−i
1
.
Then we solve (A − 2I)v = 0 or
that has the solution C
i
1
−1
i
−i −1
x
y
=
0
0
.
In order to get an orthonormal basis we need to normalise these eigenvectors. This gives thus
basis vectors
1
1
−i
i
, v2 = √
.
v1 = √
1
2
2 1
So if we let
then
P −1 AP
1
P =√
2
=
0 0
0 2
−i i
1 1
.
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,