PX153 - K10 - special matrices

symmetric and anti-symmetric matrices

if $A^{T} = A : A$ is a symmetric matrix
if $A^{T} = - A : A$ is an anti-symmetric matrix
any $n \times n$ matrix is the sum of a symmetric and an anti-symmtric matrix
proof:

\begin{aligned} A & = \frac{(A + A^{T}) + (A - A^{T})}{2} \\ B & = \frac{A + A^{T}}{2} = B^{T} : s y m m e t r i c \\ C & = \frac{A - A^{T}}{2} \\ C^{T} & = \frac{A^{T} - A}{2} = - C : a n t i - s y m m e t r i c \end{aligned}

orthogonal matrices

if $A^{T} = A^{- 1} : A$ is an orthogonal matrix
eg: $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] : A^{T} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$
- $A A^{T} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I : A$ is orthogonal

singular matrices

if $det A = 0 : A$ is a singular matrix

hermitian conjugate matrix

the hermitian conjugate of a matrix is the transpose of the complex conjugate of the matrix:

\begin{aligned} A^{†} = (A^{*})^{T} & = (A^{T})^{*} \\ (a^{†})_{i j} & = a_{j i}^{*} \end{aligned}

hermitian or anti-hermitian matrices

if $A^{†} = A : A$ is a hermitian matrix
if $A^{†} = - A : A$ is an anti-hermitian matrix
- eg: $A = [\begin{matrix} 0 & - i \\ i & 0 \end{matrix}]$
  $A^{*} = [\begin{matrix} 0 & i \\ - i & 0 \end{matrix}]$
  $A^{†} = A : A$ is a hermitian matrix
any $n \times n$ matrix can be written as a sum of a hermitian and an anti-hermitian matrix
proof:

\begin{aligned} A & = \frac{1}{2} (A + A^{†}) + \frac{1}{2} (A - A^{†}) \\ B & = \frac{1}{2} (A + A^{†}) = B : h e r m i t i a n \\ C & = \frac{1}{2} (A - A^{†}) \\ C^{†} & = \frac{1}{2} (A^{†} - A) = - C : a n t i - h e r m i t i a n \end{aligned}

unitary matrices

$A$ is a unitary matrix if $A^{†} = A^{- 1}$
eg:

\begin{aligned} A & = [\begin{array}{c} 0 & - i \\ i & 0 \end{array}] \\ A^{†} & = [\begin{array}{c} 0 & i \\ - i & 0 \end{array}] \\ A^{†} A & = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] = I \\ ∴ & A i s u n i t a r y \end{aligned}

for unitary matrices:

\begin{aligned} | λ_{i} |^{2} & = 1 \\ λ_{i}^{*} λ_{i} & = 1 \end{aligned}

proof:

\begin{aligned} U {\vec{x}}^{i} & = λ_{i} {\vec{x}}^{i} \\ U^{- 1} U {\vec{x}}^{i} & = U^{- 1} λ_{i} {\vec{x}}^{i} \\ \frac{1}{λ_{i}} {\vec{x}}^{i} & = U^{- 1} {\vec{x}}^{i} \\ \frac{1}{λ_{i}} {\vec{x}}^{i †} {\vec{x}}^{i} & = {\vec{x}}^{i †} U^{†} {\vec{x}}^{i} \\ \frac{1}{λ_{i}} {\vec{x}}^{i †} {\vec{x}}^{i} & = (U {\vec{x}}^{i})^{†} {\vec{x}}^{i} \\ \frac{1}{λ_{i}} {\vec{x}}^{i †} {\vec{x}}^{i} & = λ_{i}^{*} {\vec{x}}^{i †} {\vec{x}}^{i} \\ (\frac{1}{λ_{i}} - λ_{i}^{*}) ({\vec{x}}^{i †} {\vec{x}}^{i}) & = 0 \end{aligned}

- $({\vec{x}}^{i †} {\vec{x}}^{i})$ cannot be $0 :$

\frac{1}{λ_{i}} - λ_{i}^{*} = 0 ⟹ λ_{i}^{*} λ_{i} = 1

- the modulus of eigenvalues is unity