Linear Algebra, Theory And Applications, 2012a

178 SPECTRAL THEORY
Then
TT ∗ =
T ∗ T =
( )( ) (
t11 a ∗ t11 0 T |t11 | 2 )
+ a
0 T 1 a T1
∗ = ∗ a a ∗ T1
∗
T 1 a T 1 T1
∗
( )( ) (
t11 0 T t11 a ∗ |t11 | 2 )
t
a T1
∗ =
11 a ∗
0 T 1 at 11 aa ∗ + T1 ∗ T 1
Since these two matrices are equal, it follows a = 0. But now it follows that T1 ∗ T 1 = T 1 T1
∗
and so by induction T 1 is a diagonal matrix D 1 . Therefore,
( )
t11 0
T =
T
0 D 1
a diagonal matrix.
Now here is a proof which doesn’t involve block multiplication. Since T is normal,
T ∗ T = TT ∗ . Writing this in terms of components and using the description of the adjoint
as the transpose of the conjugate, yields the following for the ik th entry of T ∗ T = TT ∗ .
TT ∗
T ∗ T
{ ∑ }} {
t ij t ∗ jk = ∑ { ∑ }} {
t ij t kj = t ∗ ijt jk = ∑ t ji t jk .
j
j
j
j
Now use the fact that T is upper triangular and let i = k = 1 to obtain the following from
the above.
∑
|t 1j | 2 = ∑ |t j1 | 2 = |t 11 | 2
j
j
You see, t j1 = 0 unless j = 1 due to the assumption that T is upper triangular. This shows
T is of the form
⎛
⎞
∗ 0 ··· 0
0 ∗ ··· ∗
⎜
⎝
.
. .. . ..
. ⎟
⎠ .
0 ··· 0 ∗
Now do the same thing only this time take i = k = 2 and use the result just established.
Thus, from the above, ∑
|t 2j | 2 = ∑ |t j2 | 2 = |t 22 | 2 ,
j
j
showing that t 2j =0ifj>2 which means T has the form
⎛
⎞
∗ 0 0 ··· 0
0 ∗ 0 ··· 0
0 0 ∗ ··· ∗
.
⎜ . .
⎝
.
. . .. . ..
. ⎟
. ⎠
0 0 0 0 ∗
Next let i = k = 3 and obtain that T looks like a diagonal matrix in so far as the first 3
rows and columns are concerned. Continuing in this way it follows T is a diagonal matrix.
Theorem 7.4.11 Let A be a normal matrix. Then there exists a unitary matrix U such
that U ∗ AU is a diagonal matrix.