Difference between orthogonal and unitary transformation
Answers
Step-by-step explanation:
A square matrix $A=[A_1\;A_2\;\cdots\;A_n]$ ($A_i$ for the ith column vector of $A$) is unitary if its inverse is equal to its conjugate transpose, i.e., $A^{-1}=A^{*T}$. In particular, if a unitary matrix is real $A=A^*$, then $A^{-1}=A^T$ and it is orthogonal. Both the column and row vectors ( $A_i, i=1,\cdots,n$) of a unitary or orthogonal matrix are orthogonal (perpendicular to each other) and normalized (of unit length), or orthonormal, i.e., their inner product satisfies:
\begin{displaymath}(A_i,A_j)=A_i^{*T} A_j=\delta_{i,j}=\left\{ \begin{array}{ll}
1 & \mbox{if } i=j 0 & \mbox{otherwise} \end{array} \right.
\end{displaymath}
These $n$ orthonormal vectors can be used as the basis vectors of the n-dimensional vector space.
Any unitary (orthogonal) matrix $A$ can define a unitary (orthogonal) transform of a vector $X=[x_1,\cdots,x_n]^T$:
\begin{displaymath}\left\{ \begin{array}{ll}
Y=A^{*T}X=A^{-1} X & \mbox{(forwa...
...\\
\par
X=AY & \mbox{(inverse transform)}
\end{array} \right. \end{displaymath}
or in tabular form:
\begin{displaymath}\left\{ \begin{array}{ll}
Y=\left[ \begin{array}{c} y_1 y...
...^n y_i \; A_i & \mbox{(inverse transform)}
\end{array} \right. \end{displaymath}