Diagonalisation of matrices

8.16 DIAGONALISATION OF MATRICES
orthonormal and the transformation matrix S is unitary then
'
&
Ski ek Srj er
ei |ej =
k
=
Ski∗
k
r
Srj ek |er r
k
=
Ski∗
Srj δkr =
r
Ski∗ Skj = (S† S)ij = δij ,
k
showing that the new basis is also orthonormal.
Furthermore, in addition to the properties of general similarity transformations,
for unitary transformations the following hold.
(i) If A is Hermitian (anti-Hermitian) then A is Hermitian (anti-Hermitian),
i.e. if A† = ±A then
(A )† = (S† AS)† = S† A† S = ±S† AS = ±A .
(8.99)
(ii) If A is unitary (so that A† = A−1 ) then A is unitary, since
(A )† A = (S† AS)† (S† AS) = S† A† SS† AS = S† A† AS
= S† IS = I.
(8.100)
8.16 Diagonalisation of matrices
Suppose that a linear operator A is represented in some basis ei , i = 1, 2, . . . , N,
by the matrix A. Consider a new basis xj given by
xj =
N
Sij ei ,
i=1
where the xj are chosen to be the eigenvectors of the linear operator A , i.e.
A xj = λj xj .
(8.101)
In the new basis, A is represented by the matrix A = S−1 AS, which has a
particularly simple form, as we shall see shortly. The element Sij of S is the ith
component, in the old (unprimed) basis, of the jth eigenvector xj of A, i.e. the
columns of S are the eigenvectors of the matrix A:


↑ ↑
↑
S =  x1 x2 · · · xN  ,
↓ ↓
↓
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MATRICES AND VECTOR SPACES
that is, Sij = (xj )i . Therefore A is given by
(S−1 AS)ij =
=
=
k
l
k
l
(S−1 )ik Akl Slj
(S−1 )ik Akl (xj )l
(S−1 )ik λj (xj )k
k
=
λj (S−1 )ik Skj = λj δij .
k
So the matrix A is diagonal with the eigenvalues of A as the diagonal elements,
i.e.


λ1 0 · · · 0

.. 
 0 λ2
. 
.
A = 
 .

.
.
.
 .
.
0 
0
···
λN
0
Therefore, given a matrix A, if we construct the matrix S that has the eigenvectors of A as its columns then the matrix A = S−1 AS is diagonal and has the
eigenvalues of A as its diagonal elements. Since we require S to be non-singular
(|S| = 0), the N eigenvectors of A must be linearly independent and form a basis
for the N-dimensional vector space. It may be shown that any matrix with distinct
eigenvalues can be diagonalised by this procedure. If, however, a general square
matrix has degenerate eigenvalues then it may, or may not, have N linearly
independent eigenvectors. If it does not then it cannot be diagonalised.
For normal matrices (which include Hermitian, anti-Hermitian and unitary
matrices) the N eigenvectors are indeed linearly independent. Moreover, when
normalised, these eigenvectors form an orthonormal set (or can be made to do
so). Therefore the matrix S with these normalised eigenvectors as columns, i.e.
whose elements are Sij = (xj )i , has the property
(S† S)ij =
k
(S† )ik (S)kj =
Ski∗ Skj =
k
†
(xi )∗k (xj )k = (xi ) xj = δij .
k
Hence S is unitary (S−1 = S† ) and the original matrix A can be diagonalised by
A = S−1 AS = S† AS.
Therefore, any normal matrix A can be diagonalised by a similarity transformation
using a unitary transformation matrix S.
286
8.16 DIAGONALISATION OF MATRICES
Diagonalise the matrix

1
A= 0
3

3
0 .
1
0
−2
0
The matrix A is symmetric and so may be diagonalised by a transformation of the form
A = S† AS, where S has the normalised eigenvectors of A as its columns. We have already
found these eigenvectors in subsection 8.14.1, and so we can write straightaway


1 √0
−1
1 
S= √
0
2
0 .
2
1
0
1
We note that although the eigenvalues of A are degenerate, its three eigenvectors are
linearly independent and so A can still be diagonalised. Thus, calculating S† AS we obtain




1 √0
−1
1
1
1
0
3
√0
1
†
S AS =
2 0   0 −2 0   0
2
0 
0
2
3
0
1
−1
0
1
1
0
1


4
0
0
=  0 −2 0  ,
0
0
−2
which is diagonal, as required, and has as its diagonal elements the eigenvalues of A. If a matrix A is diagonalised by the similarity transformation A = S−1 AS, so
that A = diag(λ1 , λ2 , . . . , λN ), then we have immediately
Tr A = Tr A =
N
λi ,
(8.102)
i=1
|A | = |A| =
N
λi ,
(8.103)
i=1
since the eigenvalues of the matrix are unchanged by the transformation. Moreover, these results may be used to prove the rather useful trace formula
| exp A| = exp(Tr A),
(8.104)
where the exponential of a matrix is as defined in (8.38).
Prove the trace formula (8.104).
At the outset, we note that for the similarity transformation A = S−1 AS, we have
(A )n = (S−1 AS)(S−1 AS) · · · (S−1 AS) = S−1 An S.
Thus, from (8.38), we obtain exp A = S−1 (exp A)S, from which it follows that | exp A | =
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