8.16 DIAGONALISATION OF MATRICES
Diagonalise the matrixA=
103
0 − 20
301
.
The matrixAis symmetric and so may be diagonalised by a transformation of the form
A′=S†AS,whereShas the normalised eigenvectors ofAas its columns. We have already
found these eigenvectors in subsection 8.14.1, and so we can write straightaway
S=
1
√
2
10 − 1
0
√
20
10 1
.
We note that although the eigenvalues ofAare degenerate, its three eigenvectors are
linearly independent and soAcan still be diagonalised. Thus, calculatingS†ASwe obtain
S†AS=
1
2
101
0
√
20
− 101
103
0 − 20
301
10 − 1
0
√
20
10 1
=
40 0
0 − 20
00 − 2
,
which is diagonal, as required, and has as its diagonal elements the eigenvalues ofA.
If a matrixAis diagonalised by the similarity transformationA′=S−^1 AS,sothatA′= diag(λ 1 ,λ 2 ,...,λN), then we have immediately
TrA′=TrA=∑Ni=1λi, (8.102)|A′|=|A|=∏Ni=1λi, (8.103)since the eigenvalues of the matrix are unchanged by the transformation. More-
over, these results may be used to prove the rather usefultrace formula
|expA|= exp(TrA), (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 transformationA′=S−^1 AS, we have
(A′)n=(S−^1 AS)(S−^1 AS)···(S−^1 AS)=S−^1 AnS.Thus, from (8.38), we obtain expA′=S−^1 (expA)S, from which it follows that|expA′|=