Mathematical Methods for Physics and Engineering - Matematica.NET

8.16 DIAGONALISATION OF MATRICES◮Diagonalise the matrixA =⎛⎝ 1 0 0 −2 30⎞⎠ .3 0 1The matrix A is symmetric and so may be diagonalised by a transformation of the formA ′ = S † AS, whereS has the normalised eigenvectors of A as its columns. We have alreadyfound these eigenvectors in subsection 8.14.1, and so we can write straightaway⎛⎞S = √ 1 1√0 −1⎝ 0 2 0 ⎠ .2 1 0 1We note that although the eigenvalues of A are degenerate, its three eigenvectors arelinearly independent and so A can still be diagonalised. Thus, calculating S † AS we obtain⎛⎞ ⎛S † AS = 1 1√0 1⎝ 0 2 0 ⎠ ⎝ 1 0 3⎞ ⎛⎞1√0 −10 −2 0 ⎠ ⎝ 0 2 0 ⎠2−1 0 1 3 0 1 1 0 1⎛= ⎝ 4 0 0⎞0 −2 0 ⎠ ,0 0 −2which 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, sothat A ′ = diag(λ 1 ,λ 2 ,...,λ N ), then we have immediatelyTr A ′ =TrA =N∑λ i , (8.102)i=1|A ′ | = |A| =N∏λ i , (8.103)since the eigenvalues of the matrix are unchanged by the transformation. Moreover,these results may be used to prove the rather useful trace formulai=1| 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 A n S.Thus, from (8.38), we obtain exp A ′ = S −1 (exp A)S, from which it follows that | exp A ′ | =287