Orthogonal matrix: Difference between revisions

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One way to express this is
:<math display="block">Q^\mathrm{T} Q = Q Q^\mathrm{T} = I,</math>
 
:<math>Q^\mathrm{T} Q = Q Q^\mathrm{T} = I,</math>
 
where {{math|''Q''<sup>T</sup>}} is the [[transpose]] of {{mvar|Q}} and {{mvar|I}} is the [[identity matrix]].
 
This leads to the equivalent characterization: a matrix {{mvar|Q}} is orthogonal if its transpose is equal to its [[Invertible matrix|inverse]]:
:<math display="block">Q^\mathrm{T}=Q^{-1},</math>
 
:<math>Q^\mathrm{T}=Q^{-1},</math>
 
where {{math|''Q''<sup>−1</sup>}} is the inverse of {{mvar|Q}}.
 
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==Overview==
An orthogonal matrix is the real specialization of a unitary matrix, and thus always a [[normal matrix]]. Although we consider only real matrices here, the definition can be used for matrices with entries from any [[field (mathematics)|field]]. However, orthogonal matrices arise naturally from [[dot product]]s, and for matrices of complex numbers that leads instead to the unitary requirement. Orthogonal matrices preserve the dot product,<ref>[http://tutorial.math.lamar.edu/Classes/LinAlg/OrthogonalMatrix.aspx "Paul's online math notes"]{{Citation broken|date=January 2013|note=See talk page.}}, Paul Dawkins, [[Lamar University]], 2008. Theorem 3(c)</ref> so, for vectors {{math|'''u'''}} and {{math|'''v'''}} in an {{mvar|n}}-dimensional real [[Euclidean space]]
:<math display="block">{\mathbf u} \cdot {\mathbf v} = \left(Q {\mathbf u}\right) \cdot \left(Q {\mathbf v}\right) </math>
 
:<math>{\mathbf u} \cdot {\mathbf v} = \left(Q {\mathbf u}\right) \cdot \left(Q {\mathbf v}\right) </math>
 
where {{mvar|Q}} is an orthogonal matrix. To see the inner product connection, consider a vector {{math|'''v'''}} in an {{mvar|n}}-dimensional real [[Euclidean space]]. Written with respect to an orthonormal basis, the squared length of {{math|'''v'''}} is {{math|'''v'''<sup>T</sup>'''v'''}}. If a linear transformation, in matrix form {{math|''Q'''''v'''}}, preserves vector lengths, then
:<math display="block">{\mathbf v}^\mathrm{T}{\mathbf v} = (Q{\mathbf v})^\mathrm{T}(Q{\mathbf v}) = {\mathbf v}^\mathrm{T} Q^\mathrm{T} Q {\mathbf v} .</math>
 
:<math>{\mathbf v}^\mathrm{T}{\mathbf v} = (Q{\mathbf v})^\mathrm{T}(Q{\mathbf v}) = {\mathbf v}^\mathrm{T} Q^\mathrm{T} Q {\mathbf v} .</math>
 
Thus [[dimension (vector space)|finite-dimensional]] linear isometries—rotations, reflections, and their combinations—produce orthogonal matrices. The converse is also true: orthogonal matrices imply orthogonal transformations. However, linear algebra includes orthogonal transformations between spaces which may be neither finite-dimensional nor of the same dimension, and these have no orthogonal matrix equivalent.
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The {{nowrap|2 × 2}} matrices have the form
:<math display="block">\begin{bmatrix}
 
:<math>\begin{bmatrix}
p & t\\
q & u
\end{bmatrix},</math>
 
which orthogonality demands satisfy the three equations
<math display="block">\begin{align}
 
: <math>
\begin{align}
1 & = p^2+t^2, \\
1 & = q^2+u^2, \\
0 & = pq+tu.
\end{align}</math>
</math>
 
In consideration of the first equation, without loss of generality let {{math|1=''p'' = cos ''θ''}}, {{math|1=''q'' = sin ''θ''}}; then either {{math|1=''t'' = −''q''}}, {{math|1=''u'' = ''p''}} or {{math|1=''t'' = ''q''}}, {{math|1=''u'' = −''p''}}. We can interpret the first case as a rotation by {{mvar|θ}} (where {{math|1=''θ'' = 0}} is the identity), and the second as a reflection across a line at an angle of {{math|{{sfrac|''θ''|2}}}}.
 
:<math display="block">
\begin{bmatrix}
\cos \theta & -\sin \theta \\
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The special case of the reflection matrix with {{math|1=''θ'' = 90°}} generates a reflection about the line at 45° given by {{math|1=''y'' = ''x''}} and therefore exchanges {{mvar|x}} and {{mvar|y}}; it is a [[permutation matrix]], with a single 1 in each column and row (and otherwise 0):
:<math display="block">\begin{bmatrix}
 
:<math>\begin{bmatrix}
0 & 1\\
1 & 0
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===Higher dimensions===
Regardless of the dimension, it is always possible to classify orthogonal matrices as purely rotational or not, but for {{nowrap|3 × 3}} matrices and larger the non-rotational matrices can be more complicated than reflections. For example,
<math display="block">
 
:<math>
\begin{bmatrix}
-1 & 0 & 0\\
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A [[Householder reflection]] is constructed from a non-null vector {{math|'''v'''}} as
:<math display="block">Q = I - 2 \frac{{\mathbf v}{\mathbf v}^\mathrm{T}}{{\mathbf v}^\mathrm{T}{\mathbf v}} .</math>
 
:<math>Q = I - 2 \frac{{\mathbf v}{\mathbf v}^\mathrm{T}}{{\mathbf v}^\mathrm{T}{\mathbf v}} .</math>
 
Here the numerator is a symmetric matrix while the denominator is a number, the squared magnitude of {{math|'''v'''}}. This is a reflection in the hyperplane perpendicular to {{math|'''v'''}} (negating any vector component parallel to {{math|'''v'''}}). If {{math|'''v'''}} is a unit vector, then {{math|1=''Q'' = ''I'' − 2'''vv'''<sup>T</sup>}} suffices. A Householder reflection is typically used to simultaneously zero the lower part of a column. Any orthogonal matrix of size {{nowrap|''n'' × ''n''}} can be constructed as a product of at most {{mvar|n}} such reflections.
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===Matrix properties===
A real square matrix is orthogonal [[if and only if]] its columns form an [[orthonormal basis]] of the [[Euclidean space]] {{math|'''R'''<sup>''n''</sup>}} with the ordinary Euclidean [[dot product]], which is the case if and only if its rows form an orthonormal basis of {{math|'''R'''<sup>''n''</sup>}}. It might be tempting to suppose a matrix with orthogonal (not orthonormal) columns would be called an orthogonal matrix, but such matrices have no special interest and no special name; they only satisfy {{math|1=''M''<sup>T</sup>''M'' = ''D''}}, with {{mvar|D}} a [[diagonal matrix]].
 
The [[determinant]] of any orthogonal matrix is +1 or −1. This follows from basic facts about determinants, as follows:
:<math display="block">1=\det(I)=\det\left(Q^\mathrm{T}Q\right)=\det\left(Q^\mathrm{T}\right)\det(Q)=\bigl(\det(Q)\bigr)^2 .</math>
 
:<math>1=\det(I)=\det\left(Q^\mathrm{T}Q\right)=\det\left(Q^\mathrm{T}\right)\det(Q)=\bigl(\det(Q)\bigr)^2 .</math>
 
The converse is not true; having a determinant of ±1 is no guarantee of orthogonality, even with orthogonal columns, as shown by the following counterexample.
:<math display="block">\begin{bmatrix}
 
:<math>\begin{bmatrix}
2 & 0 \\
0 & \frac{1}{2}
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Now consider {{math|(''n'' + 1) × (''n'' + 1)}} orthogonal matrices with bottom right entry equal to 1. The remainder of the last column (and last row) must be zeros, and the product of any two such matrices has the same form. The rest of the matrix is an {{math|''n'' × ''n''}} orthogonal matrix; thus {{math|O(''n'')}} is a subgroup of {{math|O(''n'' + 1)}} (and of all higher groups).
 
:<math display="block">\begin{bmatrix}
& & & 0\\
& \mathrm{O}(n) & & \vdots\\
& & & 0\\
0 & \cdots & 0 & 1
\end{bmatrix}</math>
</math>
 
Since an elementary reflection in the form of a [[Householder matrix]] can reduce any orthogonal matrix to this constrained form, a series of such reflections can bring any orthogonal matrix to the identity; thus an orthogonal group is a [[reflection group]]. The last column can be fixed to any unit vector, and each choice gives a different copy of {{math|O(''n'')}} in {{math|O(''n'' + 1)}}; in this way {{math|O(''n'' + 1)}} is a [[fiber bundle|bundle]] over the unit sphere {{math|''S''<sup>''n''</sup>}} with fiber {{math|O(''n'')}}.
 
Similarly, {{math|SO(''n'')}} is a subgroup of {{math|SO(''n'' + 1)}}; and any special orthogonal matrix can be generated by [[Givens rotation|Givens plane rotations]] using an analogous procedure. The bundle structure persists: {{math|SO(''n'') ↪ SO(''n'' + 1) → ''S''<sup>''n''</sup>}}. A single rotation can produce a zero in the first row of the last column, and series of {{math|''n'' − 1}} rotations will zero all but the last row of the last column of an {{math|''n'' × ''n''}} rotation matrix. Since the planes are fixed, each rotation has only one degree of freedom, its angle. By induction, {{math|SO(''n'')}} therefore has
:<math display="block">(n-1) + (n-2) + \cdots + 1 = \frac{n(n-1)}{2}</math>
 
:<math>(n-1) + (n-2) + \cdots + 1 = \frac{n(n-1)}{2}</math>
 
degrees of freedom, and so does {{math|O(''n'')}}.
 
Permutation matrices are simpler still; they form, not a Lie group, but only a finite group, the order [[factorial|{{math|''n''[[factorial|!]]}}]] [[symmetric group]] {{math|S<sub>''n''</sub>}}. By the same kind of argument, {{math|S<sub>''n''</sub>}} is a subgroup of {{math|S<sub>''n'' + 1</sub>}}. The even permutations produce the subgroup of permutation matrices of determinant +1, the order {{math|{{sfrac|''n''!|2}}}} [[alternating group]].
 
===Canonical form===
More broadly, the effect of any orthogonal matrix separates into independent actions on orthogonal two-dimensional subspaces. That is, if {{mvar|Q}} is special orthogonal then one can always find an orthogonal matrix {{mvar|P}}, a (rotational) [[change of basis]], that brings {{mvar|Q}} into block diagonal form:
 
:<math display="block">P^\mathrm{T}QP = \begin{bmatrix}
R_1 & & \\ & \ddots & \\ & & R_k
\end{bmatrix}\ (n\text{ even}),
\ P^\mathrm{T}QP = \begin{bmatrix}
R_1 & & & \\ & \ddots & & \\ & & R_k & \\ & & & 1
\end{bmatrix}\ (n\text{ odd}).</math>
 
where the matrices {{math|''R''<sub>1</sub>, ..., ''R''<sub>''k''</sub>}} are {{nowrap|2 × 2}} rotation matrices, and with the remaining entries zero. Exceptionally, a rotation block may be diagonal, {{math|±''I''}}. Thus, negating one column if necessary, and noting that a {{nowrap|2 × 2}} reflection diagonalizes to a +1 and −1, any orthogonal matrix can be brought to the form
:<math display="block">P^\mathrm{T}QP = \begin{bmatrix}
 
:<math>P^\mathrm{T}QP = \begin{bmatrix}
\begin{matrix}R_1 & & \\ & \ddots & \\ & & R_k\end{matrix} & 0 \\
0 & \begin{matrix}\pm 1 & & \\ & \ddots & \\ & & \pm 1\end{matrix} \\
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===Lie algebra===
Suppose the entries of {{mvar|Q}} are differentiable functions of {{mvar|t}}, and that {{math|1=''t'' = 0}} gives {{math|1=''Q'' = ''I''}}. Differentiating the orthogonality condition
:<math display="block">Q^\mathrm{T} Q = I </math>
 
:<math>Q^\mathrm{T} Q = I </math>
 
yields
:<math display="block">\dot{Q}^\mathrm{T} Q + Q^\mathrm{T} \dot{Q} = 0</math>
 
:<math>\dot{Q}^\mathrm{T} Q + Q^\mathrm{T} \dot{Q} = 0</math>
 
Evaluation at {{math|1=''t'' = 0}} ({{math|1=''Q'' = ''I''}}) then implies
:<math display="block">\dot{Q}^\mathrm{T} = -\dot{Q} .</math>
 
:<math>\dot{Q}^\mathrm{T} = -\dot{Q} .</math>
 
In Lie group terms, this means that the [[Lie algebra]] of an orthogonal matrix group consists of [[skew-symmetric matrix|skew-symmetric matrices]]. Going the other direction, the [[matrix exponential]] of any skew-symmetric matrix is an orthogonal matrix (in fact, special orthogonal).
 
For example, the three-dimensional object physics calls [[angular velocity]] is a differential rotation, thus a vector in the Lie algebra {{math|<math>\mathfrak{so}(3)</math>(3)}} tangent to {{math|SO(3)}}. Given {{math|1='''ω''' = (''xθ'', ''yθ'', ''zθ'')}}, with {{math|1='''v''' = (''x'', ''y'', ''z'')}} being a unit vector, the correct skew-symmetric matrix form of {{mathmvar|'''ω'''}} is
<math display="block">
 
:<math>
\Omega = \begin{bmatrix}
0 & -z\theta & y\theta \\
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The exponential of this is the orthogonal matrix for rotation around axis {{math|'''v'''}} by angle {{mvar|θ}}; setting {{math|1=''c'' = cos {{sfrac|''θ''|2}}}}, {{math|1=''s'' = sin {{sfrac|''θ''|2}}}},
:<math display="block">R\exp(\Omega) = \begin{bmatrix}
 
:<math>
\exp(\Omega) =
\begin{bmatrix}
1 - 2s^2 + 2x^2 s^2 & 2xy s^2 - 2z sc & 2xz s^2 + 2y sc\\
2xy s^2 + 2z sc & 1 - 2s^2 + 2y^2 s^2 & 2yz s^2 - 2x sc\\
2xz s^2 - 2y sc & 2yz s^2 + 2x sc & 1 - 2s^2 + 2z^2 s^2
\end{bmatrix}.</math>
.</math>
 
==Numerical linear algebra==
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Consider an [[overdetermined system of linear equations]], as might occur with repeated measurements of a physical phenomenon to compensate for experimental errors. Write {{math|1=''A'''''x''' = '''b'''}}, where {{mvar|A}} is {{math|''m'' × ''n''}}, {{math|''m'' > ''n''}}.
A {{mvar|QR}} decomposition reduces {{mvar|A}} to upper triangular {{mvar|R}}. For example, if {{mvar|A}} is {{nowrap|5 × 3}} then {{mvar|R}} has the form
<math display="block">R = \begin{bmatrix}
 
:<math>R = \begin{bmatrix}
\cdot & \cdot & \cdot \\
0 & \cdot & \cdot \\
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For example, consider a non-orthogonal matrix for which the simple averaging algorithm takes seven steps
:<math display="block">\begin{bmatrix}3 & 1\\7 & 5\end{bmatrix}
 
:<math>\begin{bmatrix}3 & 1\\7 & 5\end{bmatrix}
\rightarrow
\begin{bmatrix}1.8125 & 0.0625\\3.4375 & 2.6875\end{bmatrix}
\rightarrow \cdots \rightarrow
\begin{bmatrix}0.8 & -0.6\\0.6 & 0.8\end{bmatrix}</math>
 
and which acceleration trims to two steps (with {{mvar|γ}} = 0.353553, 0.565685).
 
:<math display="block">\begin{bmatrix}3 & 1\\7 & 5\end{bmatrix}
\rightarrow
\begin{bmatrix}1.41421 & -1.06066\\1.06066 & 1.41421\end{bmatrix}
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Gram-Schmidt yields an inferior solution, shown by a Frobenius distance of 8.28659 instead of the minimum 8.12404.
 
:<math display="block">\begin{bmatrix}3 & 1\\7 & 5\end{bmatrix}
\rightarrow
\begin{bmatrix}0.393919 & -0.919145\\0.919145 & 0.393919\end{bmatrix}</math>
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The problem of finding the orthogonal matrix {{mvar|Q}} nearest a given matrix {{mvar|M}} is related to the [[Orthogonal Procrustes problem]]. There are several different ways to get the unique solution, the simplest of which is taking the [[singular value decomposition]] of {{mvar|M}} and replacing the singular values with ones. Another method expresses the {{mvar|R}} explicitly but requires the use of a [[matrix square root]]:<ref>[http://people.csail.mit.edu/bkph/articles/Nearest_Orthonormal_Matrix.pdf "Finding the Nearest Orthonormal Matrix"], [[Berthold K.P. Horn]], [[MIT]].</ref>
:<math display="block">Q = M \left(M^\mathrm{T} M\right)^{-\frac 1 2}</math>
 
:<math>Q = M \left(M^\mathrm{T} M\right)^{-\frac 1 2}</math>
 
This may be combined with the Babylonian method for extracting the square root of a matrix to give a recurrence which converges to an orthogonal matrix quadratically:
:<math display="block">Q_{n + 1} = 2 M \left(Q_n^{-1} M + M^\mathrm{T} Q_n\right)^{-1}</math>
 
where {{math|1=''Q''<sub>0</sub> {{=}} ''M''}}.
:<math>Q_{n + 1} = 2 M \left(Q_n^{-1} M + M^\mathrm{T} Q_n\right)^{-1}</math>
 
where {{math|''Q''<sub>0</sub> {{=}} ''M''}}.
 
These iterations are stable provided the [[condition number]] of {{mvar|M}} is less than three.<ref>[http://www.maths.manchester.ac.uk/~nareports/narep91.pdf "Newton's Method for the Matrix Square Root"] {{Webarchive|url=https://web.archive.org/web/20110929131330/http://www.maths.manchester.ac.uk/~nareports/narep91.pdf |date=2011-09-29 }}, Nicholas J. Higham, Mathematics of Computation, Volume 46, Number 174, 1986.</ref>
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Using a first-order approximation of the inverse and the same initialization results in the modified iteration:
 
:<math display="block">N_{n} = Q_n^\mathrm{T} Q_n</math>
:<math display="block">P_{n} = \frac 1 2 Q_n N_{n}</math>
:<math display="block">Q_{n + 1} = 2 Q_n + P_n N_n - 3 P_n</math>
 
==Spin and pin==
A subtle technical problem afflicts some uses of orthogonal matrices. Not only are the group components with determinant +1 and −1 not [[Connected space|connected]] to each other, even the +1 component, {{math|SO(''n'')}}, is not [[simply connected space|simply connected]] (except for SO(1), which is trivial). Thus it is sometimes advantageous, or even necessary, to work with a [[covering map|covering group]] of SO(''n''), the [[spinor group|spin group]], {{math|Spin(''n'')}}. Likewise, {{math|O(''n'')}} has covering groups, the [[pin group]]s, Pin(''n''). For {{nowrapmath|''n'' > 2}}, {{math|Spin(''n'')}} is simply connected and thus the universal covering group for {{math|SO(''n'')}}. By far the most famous example of a spin group is {{math|Spin(3)}}, which is nothing but {{math|SU(2)}}, or the group of unit [[quaternion]]s.
 
The Pin and Spin groups are found within [[Clifford algebra]]s, which themselves can be built from orthogonal matrices.