# Change of basis and similarity transformations

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Change of basis and similarity transformations
```MATRICES AND VECTOR SPACES
Construct an orthonormal set of eigenvectors for the matrix


1
0
3
A =  0 −2 0  .
3
0
1
We ﬁrst determine the eigenvalues using |A − λI| = 0:
1−λ
0
3 −2 − λ
0 = −(1 − λ)2 (2 + λ) + 3(3)(2 + λ)
0= 0
3
0
1−λ = (4 − λ)(λ + 2)2 .
Thus λ1 = 4, λ2 = −2 = λ3 . The eigenvector




1
0
3
x1
 0 −2 0   x2  = 4 
x3
3
0
1
x1 = (x1 x2 x3 )T is found from



x1
1  1 
1

x2
0
⇒
x = √
.
2
x3
1
A general column vector that is orthogonal to x1 is
x = (a b − a)T ,
and it is easily shown that

1
Ax =  0
3
0
−2
0
(8.89)




3
a
a
0   b  = −2  b  = −2x.
1
−a
−a
Thus x is a eigenvector of A with associated eigenvalue −2. It is clear, however, that there
is an inﬁnite set of eigenvectors x all possessing the required property; the geometrical
analogue is that there are an inﬁnite number of corresponding vectors x lying in the
plane that has x1 as its normal. We do require that the two remaining eigenvectors are
orthogonal to one another, but this still leaves an inﬁnite number of possibilities. For x2 ,
therefore, let us choose a simple form of (8.89), suitably normalised, say,
x2 = (0 1 0)T .
The third eigenvector is then speciﬁed (to within an arbitrary multiplicative constant)
by the requirement that it must be orthogonal to x1 and x2 ; thus x3 may be found by
evaluating the vector product of x1 and x2 and normalising the result. This gives
1
x3 = √ (−1 0 1)T ,
2
to complete the construction of an orthonormal set of eigenvectors. 8.15 Change of basis and similarity transformations
Throughout this chapter we have considered the vector x as a geometrical quantity
that is independent of any basis (or coordinate system). If we introduce a basis
ei , i = 1, 2, . . . , N, into our N-dimensional vector space then we may write
x = x1 e1 + x2 e2 + · · · + xN eN ,
282
8.15 CHANGE OF BASIS AND SIMILARITY TRANSFORMATIONS
and represent x in this basis by the column matrix
x2 · · · xn )T ,
x = (x1
having components xi . We now consider how these components change as a result
of a prescribed change of basis. Let us introduce a new basis ei , i = 1, 2, . . . , N,
which is related to the old basis by
ej =
N
Sij ei ,
(8.90)
i=1
the coeﬃcient Sij being the ith component of ej with respect to the old (unprimed)
basis. For an arbitrary vector x it follows that
x=
N
xi ei =
i=1
N
xj ej =
j=1
N
xj
j=1
N
Sij ei .
i=1
From this we derive the relationship between the components of x in the two
coordinate systems as
xi =
N
Sij xj ,
j=1
which we can write in matrix form as
x = Sx
(8.91)
where S is the transformation matrix associated with the change of basis.
Furthermore, since the vectors ej are linearly independent, the matrix S is
non-singular and so possesses an inverse S−1 . Multiplying (8.91) on the left by
S−1 we ﬁnd
x = S−1 x,
(8.92)
which relates the components of x in the new basis to those in the old basis.
Comparing (8.92) and (8.90) we note that the components of x transform inversely
to the way in which the basis vectors ei themselves transform. This has to be so,
as the vector x itself must remain unchanged.
We may also ﬁnd the transformation law for the components of a linear
operator under the same change of basis. Now, the operator equation y = A x
(which is basis independent) can be written as a matrix equation in each of the
two bases as
y = A x .
y = Ax,
But, using (8.91), we may rewrite the ﬁrst equation as
Sy = ASx
⇒
283
y = S−1 ASx .
(8.93)
MATRICES AND VECTOR SPACES
Comparing this with the second equation in (8.93) we ﬁnd that the components
of the linear operator A transform as
A = S−1 AS.
(8.94)
Equation (8.94) is an example of a similarity transformation – a transformation
that can be particularly useful in converting matrices into convenient forms for
computation.
Given a square matrix A, we may interpret it as representing a linear operator
A in a given basis ei . From (8.94), however, we may also consider the matrix
A = S−1 AS, for any non-singular matrix S, as representing the same linear
operator A but in a new basis ej , related to the old basis by
ej =
Sij ei .
i
Therefore we would expect that any property of the matrix A that represents
some (basis-independent) property of the linear operator A will also be shared
by the matrix A . We list these properties below.
(i) If A = I then A = I, since, from (8.94),
A = S−1 IS = S−1 S = I.
(8.95)
(ii) The value of the determinant is unchanged:
|A | = |S−1 AS| = |S−1 ||A||S| = |A||S−1 ||S| = |A||S−1 S| = |A|.
(8.96)
(iii) The characteristic determinant and hence the eigenvalues of A are the
same as those of A: from (8.86),
|A − λI| = |S−1 AS − λI| = |S−1 (A − λI)S|
= |S−1 ||S||A − λI| = |A − λI|.
(8.97)
(iv) The value of the trace is unchanged: from (8.87),
Aii =
(S−1 )ij Ajk Ski
Tr A =
i
=
i
j
i
j
k
Ski (S−1 )ij Ajk =
j
k
= Tr A.
k
δkj Ajk =
Ajj
j
(8.98)
An important class of similarity transformations is that for which S is a unitary matrix; in this case A = S−1 AS = S† AS. Unitary transformation matrices
are particularly important, for the following reason. If the original basis ei is
284
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