Mathematical Methods for Physicists: A concise introduction - Site Map

DIAGONALIZATION OF A MATRIX
(2) The eigenvectors corresponding to distinct eigenvalues are orthogonal.
Let X 1 and X 2 be eigenvectors of ~H corresponding to the distinct eigenvalues 1
and 2 , respectively, so that
~HX 1 ˆ 1 X 1 ;
~HX 2 ˆ 2 X 2 :
…3:65†
…3:66†
Taking the hermitian conjugate of (3.66) and noting that * ˆ , we have
X y 2 ~ H ˆ 2 X y 2 :
…3:67†
Multiplying (3.65) from the left by X y 2
subtracting, we obtain
and (3.67) from the right by X 1, then
… 1 2 †X y 2 ‡ X 1 ˆ 0:
…3:68†
Since 1 ˆ 2 , it follows that X y 2 X 1 ˆ 0 or that X 1 and X 2 are orthogonal.
If X is an eigenvector of ~H, any multiple of X, X, is also an eigenvector of ~H.
Thus we can normalize the eigenvector X with a properly chosen scalar . This
means that the eigenvectors of ~H corresponding to distinct eigenvalues are orthonormal.
Just as the three orthogonal unit coordinate vectors ^e 1 ; ^e 2 ; and ^e 3 form
the basis of a three-dimensional vector space, the orthonormal eigenvectors of ~H
may serve as a basis for a function space.
Diagonalization of a matrix
Let A ~ ˆ…a ij † be a square matrix of order n, which has n linearly independent
eigenvectors X i with the corresponding eigenvalues i : ~AX i ˆ i X i . If we denote
the eigenvectors X i by column vectors with elements x 1i ; x 2i ; ...; x ni , then the
eigenvalue equation can be written in matrix form:
0
10
1 0 1
a 11 a 12 a 1n x 1i x 1i
a 21 a 22 a 2n
x 2i
x 2i
. . .
.
ˆ
B
@ . . . CB
A@
. C i
.
: …3:69†
B
A . C
@ A
a n1 a n2 a nn x ni x ni
From the above matrix equation we obtain
X n
kˆ1
a jk x ki ˆ i x ji :
…3:69b†
Now we want to diagonalize ~A. To this purpose, we can follow these steps. We
®rst form a matrix ~ S of order n n whose columns are the vector X i , that is,
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