Linear algebra/Orthogonal matrix

Alternative notations

$Q^{-1}=Q^{\mathrm {T} }$	$Q^{\mathrm {T} }Q^{-1}=Q^{-1}Q^{\mathrm {T} }=I$
${\underline {\underline {Q}}}^{-1}={\underline {\underline {Q}}}^{\mathrm {T} }$	${\underline {\underline {Q}}}^{\mathrm {T} }\cdot {\underline {\underline {Q}}}^{-1}={\underline {\underline {Q}}}^{-1}\cdot {\underline {\underline {Q}}}^{\mathrm {T} }={\underline {\underline {I}}}$
$Q_{jk}^{-1}=Q_{kj}\equiv Q_{jk}^{\mathrm {T} }$	$\sum _{k}Q_{ki}Q_{kj}^{-1}=\sum _{k}Q_{ik}^{-1}Q_{jk}=\delta _{ij}$

A real-valued and square matrix $Q$ is orthogonal (or orthonormal) if $Q^{\mathrm {T} }=Q^{-1},$ where $Q^{\mathrm {T} }$ is the transpose and $Q^{-1}$ is its inverse. Equivalently, $Q$ is orthogonal when $Q^{\mathrm {T} }Q=QQ^{\mathrm {T} }=I,$ where $I$ is the identity matrix. It has a number of well-known properties:

• A real square matrix is orthogonal if and only if its columns form an orthonormal basis on the Euclidean space ℝⁿ, which is the case if and only if its rows form an orthonormal basis of ℝⁿ.
• The determinant of any orthogonal matrix is +1 or −1. But the converse is not true; having a determinant of ±1 is no guarantee of orthogonality. An orthogonal matrix with a determinant equal to +1 is called a special orthogonal matrix.
• An orthogonal matrix can always be diagonalized over the complex numbers to exhibit a full set of eigenvalues, all of which must have (complex) modulus 1.
• All permutation matrices are orthogonal (but the converse is not true.)
• All orthogonal matrices are unitary (but the converse is not true.)
• Under the operation of multiplication, the n × n orthogonal matrices form the orthogonal group known as O(n).

Example

Since orthogonal matrices form a group under multiplication, we can construct a non-trivial orthogonal matrix my multiplying two matrices that are easy to understand. Consider, for example, ${\underline {\underline {C}}}={\underline {\underline {A}}}\cdot {\underline {\underline {B}}}:$

$\overbrace {\begin{bmatrix}{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&0\\-{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&0\\0&0&-1\end{bmatrix}} ^{\underline {\underline {A}}}\cdot \overbrace {\begin{bmatrix}1&0&0\\0&{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}\\0&-{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}\end{bmatrix}} ^{\underline {\underline {B}}}=\overbrace {\begin{bmatrix}{\frac {1}{\sqrt {2}}}&{\frac {1}{2}}&{\frac {1}{2}}\\-{\frac {1}{\sqrt {2}}}&{\frac {1}{2}}&{\frac {1}{2}}\\0&{\frac {1}{\sqrt {2}}}&-{\frac {1}{\sqrt {2}}}\\\end{bmatrix}} ^{\underline {\underline {C}}}$

Both ${\underline {\underline {A}}}$ and ${\underline {\underline {B}}}$ posses a symmetry that could lead one to postulate a non-existent symmetry among off-diagonal elements. But no such symmetry exists for ${\underline {\underline {C}}}$. The upper-left 2x2 submatrix in ${\underline {\underline {A}}}$ represent a 45 degree rotation around the $x_{3}$ (or $z$) axis, plus and inversion through that same axis, i.e., $x_{3}\rightarrow -x_{3}$.