A Linear
transformation ($\hat{T}$)
is an operation in a vector space that takes each vector and ‘transforms’ it
into another ‘form’ of vector, i.e. $|\alpha >\to |\alpha
'>=\hat{T}|\alpha >$. One of the important properties of this operation
is that, this operation is linear for
any vector: $|\alpha >,|\beta >$, and any scalars a, b:
$\hat{T}(a|\alpha >+b|\beta
>)=a(\hat{T}|\alpha >)+b(\hat{T}|\beta >)$ (16.1)
For instance, what a linear
transformation $\hat{T}$ does to a set of basis vectors $|{{e}_{1}}>,|{{e}_{2}}>,...,|{{e}_{n}}>$
is as follows:
$\hat{T}|{{e}_{1}}>={{T}_{11}}|{{e}_{1}}>+{{T}_{21}}|{{e}_{2}}>+...+{{T}_{n1}}|{{e}_{n}}>$
$\hat{T}|{{e}_{2}}>={{T}_{12}}|{{e}_{1}}>+{{T}_{22}}|{{e}_{2}}>+...+{{T}_{n2}}|{{e}_{n}}>$
…
$\hat{T}|{{e}_{n}}>={{T}_{1n}}|{{e}_{1}}>+{{T}_{2n}}|{{e}_{2}}>+...+{{T}_{nn}}|{{e}_{n}}>$
Or,
$\hat{T}|{{e}_{j}}>=\sum\limits_{i=1}^{n}{{{T}_{ij}}|{{e}_{ij}}>}$
For $j=1,2,...,n$ (16.2)
For an arbitrary vector:
$|\alpha
>={{a}_{1}}|{{e}_{1}}>+{{a}_{2}}|{{e}_{2}}>+...+{{a}_{n}}|{{e}_{n}}>=\sum\limits_{j=1}^{n}{{{a}_{j}}|{{e}_{j}}>}$ (16.3)
The linear transformation is:
$\hat{T}|\alpha
>=\sum\limits_{j=1}^{{}}{{{a}_{j}}(\hat{T}|{{e}_{j}}>)=\sum\limits_{j=1}^{n}{\sum\limits_{i=1}^{n}{{{a}_{j}}{{T}_{ij}}|{{e}_{i}}>=\sum\limits_{i=1}^{n}{\left(
\sum\limits_{j=1}^{n}{{{T}_{ij}}{{a}_{j}}} \right)|{{e}_{i}}>}}}}$
(16.4)
Evidently, the linear
transformation $\hat{T}$ transforms a vector with components ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}$
into a vector with components $a_{i}^{'}$
:
$a_{i}^{'}=\sum\limits_{j=1}^{n}{{{T}_{ij}}{{a}_{j}}}$ (16.5)
In the same fashion as n components are uniquely characterize vector $|\alpha >$, there
are n2 elements Tij those are uniquely characterize linear
transformation $\hat{T}$ with respect to the same given basis vector:
$\hat{T}\leftrightarrow
({{T}_{11}},{{T}_{12}},...,{{T}_{nn}})$
(16.6)
When the basis vector is
orthonormal, each element Tij
becomes:
${{T}_{ij}}=<{{e}_{i}}|\hat{T}|{{e}_{j}}>$
(16.7)
In the form of a matrix the linear transformation can be
expressed as follows:
$T = \begin{pmatrix}
T_{11} & T_{12} & ... & T_{1n} \\
T_{21} & T_{22} & ... & T_{2n} \\
... \\
T_{n1} & T_{n2} & ... & T_{nn}
\end{pmatrix}$
(16.8)
The
sum of two linear transformations is defined in the same way as the usual rule
for adding two matrices, as follows:
$\left( \hat{S}+\hat{T} \right)|\alpha
>=\hat{S}|\alpha >+\hat{T}|\alpha >$ (16.9)
i.e. by adding their
corresponding matrix elements:
U = S +
T ↔ Uij = Sij + Tij (16.10)
And the product of two linear
transformations $\left( \hat{S}\hat{T} \right)$ is the net effect of performing
the multiplication in succession: first $\hat{T}$ and then$\hat{S}$:
$|\alpha >\to |\alpha '>=\hat{T}|\alpha >\to
|\alpha ''>=\hat{S}|\alpha '>=\hat{S}(\hat{T}|\alpha
>)=\hat{S}\hat{T}|\alpha >$ (16.11)
Hence, the matrix U,
that represents the combined linear transformation $\hat{U}=\hat{S}\hat{T}$ can
be given:
${{{a}''}_{i}}=\sum\limits_{j=1}^{n}{{{S}_{ij}}{{{{a}'}}_{j}}=\sum\limits_{j=1}^{n}{{{S}_{ij}}\left(
\sum\limits_{k=1}^{n}{{{T}_{jk}}{{a}_{k}}} \right)=\sum\limits_{k=1}^{n}{\left(
\sum\limits_{j=1}^{n}{{{S}_{ij}}{{T}_{jk}}}
\right){{a}_{k}}=\sum\limits_{k=1}^{n}{{{U}_{ik}}{{a}_{k}}}}}}$
(16.12)
Then,
U = ST $\Leftrightarrow
{{U}_{ik}}=\sum\limits_{j=1}^{n}{{{S}_{ij}}{{T}_{jk}}}$ (16.13)
This is the standard rule for matrix multiplication, i.e.
the corresponding ikth
element of the product is found by multiplying
the ith row of S and the kth column of T then
add.
The n-tuple
components of $|\alpha >$ can be written as $n\times 1$ column matrix:
a $ = \begin{pmatrix}
a_{1} \\
a_{2} \\
... \\
a_{n}
\end{pmatrix}$
(16.14)
and the transformation rule in matrix form can be
written:
a’ = Ta (16.15)
Another useful terminology in the theory of matrix is the
transpose of a matrix: TT is the same
set of elements, but with rows and columns interchanged:
TT $ = \begin{pmatrix}
T_{11} & T_{21} & ... & T_{n1} \\
T_{12} & T_{22} & ... & T_{n2} \\
... \\
T_{1n} & T_{2n} & ... & T_{nn}
\end{pmatrix}$
(16.16)
And the transpose of column
matrix (16.14) is a row matrix aT :
aT = $\left(
{{a}_{1}}{{a}_{2}}...{{a}_{n}} \right)$ (16.17)
A square matrix is symmetric (TT = T) when the
matrix is equal to its transpose and anti-symmetric
(TT = -T) when its transpose is reversed the sign.
The complex conjugate T* of a matrix is constructed by taking
the complex conjugate of each element of the matrix:
\begin{pmatrix}
T^{*}_{11} & T^{*}_{12} & ... & T^{*}_{1n} \\
T^{*}_{21} & T^{*}_{22} & ... & T^{*}_{2n} \\
... \\
T^{*}_{n1} & T^{*}_{n2} & ... & T^{*}_{nn}
\end{pmatrix}
(16.18)
And also for column matrix:
a* $ = \begin{pmatrix}
a^{*}_{1} \\
a^{*}_{2} \\
... \\
a^{*}_{n}
\end{pmatrix}$
(16.19)
A matrix is real when the conjugate is equal to the matrix itself: T* = T and imaginary when: T* = -T.
The Hermitian conjugate of a matrix is the transposed conjugate of a
matrix, which is indicated by a dagger: T†
$\begin{pmatrix} T^{*}_{11} & T^{*}_{21} & ... & T^{*}_{n1} \\
T^{*}_{12} & T^{*}_{22} & ... & T^{*}_{n2} \\
... \\
T^{*}_{1n} & T^{*}_{2n} & ... & T^{*}_{nn}
\end{pmatrix}$
(16.20)
And the row matrix:
a†
=
$\left( {{a}^{*}_{1}}{{a}^{*}_{2}}...{{a}^{*}_{n}} \right)$ (16.21)
A square matrix is Hermitian if it is equal to its
Hermitian conjugate: T† = T and
skew Hermitian if: T† = -T. With this notation,
the inner product can be expressed in matrix form as follows:
$<\alpha |\beta >=$a†b
(16.22)
The commutator is defined as the difference between two ordering of
matrix multiplication, since matrix multiplication is not commutative (ST ≠ TS):
[S,T] ≡
ST – TS
(16.23)
The transpose of a product is the
product of transposes in reverse order:
(ST)T = TTST (16.24)
And for the Hermitian conjugates:
(ST)† = T†S†
(16.25)
An inverse of a matrix T-1
is defined:
T-1T
= TT-1 = 1 (16.26)
Here, 1 is a unit matrix, i.e.
a matrix consists of ones on the main diagonal and zeroes everywhere else. A
unit matrix represents every vector into itself. In other words:
1ij
= δij (16.27)
The
inverse of a matrix exists if and only if the determinant is non-zero:
T-1
= (1/det
T) CC (16.28)
CC is the cofactor of the matrix, the cofactor of
the element Tij can be
calculated by multiplying the determinant of the sub-matrix obtained from T by erasing the ith-row and jth-column.
A matrix without an inverse is said to be singular.
The inverse of a product is the product of the inverses in reverse order:
(ST)-1 = T-1S-1 (16.29)
A matrix is said to be unitary if its inverse is equal to its
Hermitian conjugate:
U†
= U-1 (16.30)
If the basis is orthonormal, the
columns and the rows of unitary matrix constitute also an orthonormal set.
The vector components depend on
the choice of basis as do the elements in the matrix representing a linear
transformation. When we switch to a different basis, the components will also
change. The old basis vectors are linear combinations of the new ones:
$|{{e}_{j}}>=\sum\limits_{i=1}^{n}{{{S}_{ij}}|{{f}_{i}}>}$ for $j=1,2,3,...,n$ (16.31)
And the components with respect
to new basis:
$a_{i}^{f}=\sum\limits_{j=1}^{n}{{{S}_{ij}}a_{j}^{e}}$
(16.32)
Here, the superscripts indicate
the basis vectors. This equation can be also expressed in matrix form:
af = Sae
(16.33)
The matrix representing a given
linear transformation is also modified by the change of basis. In the old basis
we had:
ae’ = Teae
(16.34)
Since ae = S-1af, by multiplying both
sides with S-1, we
obtain:
af’
= Sae’ = S(Teae) = STeS-1af
(16.35)
Hence,
Tf = STeS-1
(16.36)
Generally, two matrices T1 and T2 are similar,
if : T2 = ST1S-1,
given that the matrix S is
non-singular. If the first basis is orthonormal, the second will also be
orthonormal if and only if the matrix S
is unitary.
While the elements of a matrix change in the new basis, the determinant and trace of the matrix are unchanged. For
the determinant of a product is the product of the determinants:
det(Tf) = det(STeS-1)
= det(S)det(Te)det(S-1) = det(Te) (16.37)
And the trace is the sum of the
diagonal elements:
Tr(T) = $\sum\limits_{i=1}^{m}{{{T}_{ii}}}$
(16.38)
With the property:
Tr(T1T2) = Tr(T2T1)
(16.39)
Hence,
Tr(Tf) = Tr(STeS-1)
= Tr(Te) (16.40)
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