Si considerino le seguenti diagonalizzazioni :
A
=
U
Λ
U
H
B
=
V
M
V
H
A
+
B
=
W
Γ
W
H
{\displaystyle \mathrm {A} =\mathrm {U\Lambda U} ^{H}\qquad \mathrm {B} =\mathrm {VMV} ^{H}\qquad \mathrm {A} +\mathrm {B} =\mathrm {W\Gamma W} ^{H}}
dove
U
{\displaystyle \mathrm {U} }
V
{\displaystyle \mathrm {V} }
W
{\displaystyle \mathrm {W} }
unitarie . Dette
u
i
{\displaystyle \mathbf {u} _{i}}
v
i
{\displaystyle \mathbf {v} _{i}}
w
i
{\displaystyle \mathbf {w} _{i}}
U
{\displaystyle \mathrm {U} }
V
{\displaystyle \mathrm {V} }
W
{\displaystyle \mathrm {W} }
U
=
⟨
u
j
,
…
,
u
n
⟩
{\displaystyle {\mathcal {U}}=\langle \mathbf {u} _{j},\dots ,\mathbf {u} _{n}\rangle }
V
=
⟨
v
k
−
j
+
1
,
…
,
v
n
⟩
{\displaystyle {\mathcal {V}}=\langle \mathbf {v} _{k-j+1},\dots ,\mathbf {v} _{n}\rangle }
W
=
⟨
w
1
,
…
,
w
k
⟩
{\displaystyle {\mathcal {W}}=\langle \mathbf {w} _{1},\dots ,\mathbf {w} _{k}\rangle }
con
j
≤
k
≤
i
{\displaystyle j\leq k\leq i}
dim
(
U
∩
V
∩
W
)
=
1
{\displaystyle \dim \left({\mathcal {U}}\cap {\mathcal {V}}\cap {\mathcal {W}}\right)=1}
Allora esiste un vettore
z
∈
U
∩
V
∩
W
{\displaystyle \mathbf {z} \in {\mathcal {U}}\cap {\mathcal {V}}\cap {\mathcal {W}}}
norma euclidea :
‖
z
‖
2
=
1
z
∈
U
{\displaystyle \left\|\mathbf {z} \right\|_{2}=1\qquad \mathbf {z} \in \mathrm {U} }
perciò:
z
=
α
j
u
j
+
⋯
+
α
n
u
n
{\displaystyle \mathbf {z} =\alpha _{j}\mathbf {u} _{j}+\dots +\alpha _{n}\mathbf {u} _{n}}
con
α
i
∈
C
{\displaystyle \alpha _{i}\in \mathbb {C} }
U
{\displaystyle \mathrm {U} }
‖
z
‖
2
=
1
{\displaystyle \left\|\mathbf {z} \right\|_{2}=1}
z
H
A
z
≥
λ
j
{\displaystyle \mathbf {z} ^{H}\mathrm {Az} \geq \lambda _{j}}
usando la diagonalizzazione unitaria di
A
{\displaystyle \mathrm {A} }
z
H
B
z
≥
μ
k
−
j
+
1
{\displaystyle \mathbf {z} ^{H}\mathrm {Bz} \geq \mu _{k-j+1}}
z
H
(
A
+
B
)
z
=
γ
k
{\displaystyle \mathbf {z} ^{H}\mathrm {(A} +\mathrm {B)z} =\gamma _{k}}
Da queste ultime tre disuguaglianze si ricava la prima disuguaglianza del teorema:
γ
k
≥
z
H
(
A
+
B
)
z
=
z
H
A
z
+
z
H
B
z
≥
λ
j
+
μ
k
−
j
+
1
{\displaystyle \gamma _{k}\geq \mathbf {z} ^{H}\mathrm {(A} +\mathrm {B)} \mathbf {z} =\mathbf {z} ^{H}\mathrm {A} \mathbf {z} +\mathbf {z} ^{H}\mathrm {B} \mathbf {z} \geq \lambda _{j}+\mu _{k-j+1}}
Per la seconda disuguaglianza del teorema si procede in modo analogo.
