Editing Rotation matrix (section)

=== Polar decomposition ===
If the {{math|''n'' × ''n''}} matrix {{mvar|M}} is nonsingular, its columns are linearly independent vectors; thus the [[Gram–Schmidt process]] can adjust them to be an orthonormal basis. Stated in terms of [[numerical linear algebra]], we convert {{mvar|M}} to an orthogonal matrix, {{mvar|Q}}, using [[QR decomposition]]. However, we often prefer a {{mvar|Q}} closest to {{mvar|M}}, which this method does not accomplish. For that, the tool we want is the [[polar decomposition]] ({{Harvnb|Fan|Hoffman|1955}}; {{Harvnb|Higham|1989}}).

To measure closeness, we may use any [[matrix norm]] invariant under orthogonal transformations. A convenient choice is the [[Frobenius norm]], {{math|{{norm|''Q'' − ''M''}}<sub>F</sub>}}, squared, which is the sum of the squares of the element differences. Writing this in terms of the [[trace (linear algebra)|trace]], {{math|Tr}}, our goal is,
: Find {{mvar|Q}} minimizing {{math|Tr( (''Q'' − ''M'')<sup>T</sup>(''Q'' − ''M'') )}}, subject to {{math|''Q''<sup>T</sup>''Q'' {{=}} ''I''}}.

Though written in matrix terms, the [[objective function]] is just a quadratic polynomial. We can minimize it in the usual way, by finding where its derivative is zero. For a {{nowrap|3 × 3}} matrix, the orthogonality constraint implies six scalar equalities that the entries of {{mvar|Q}} must satisfy. To incorporate the constraint(s), we may employ a standard technique, [[Lagrange multipliers]], assembled as a symmetric matrix, {{mvar|Y}}. Thus our method is:
: Differentiate {{math|Tr( (''Q'' − ''M'')<sup>T</sup>(''Q'' − ''M'') + (''Q''<sup>T</sup>''Q'' − ''I'')''Y'' )}} with respect to (the entries of) {{mvar|Q}}, and equate to zero.

<div style="float:right; font-size:90%; border:1px solid black; padding:1em;">
Consider a {{nowrap|2 × 2}} example. Including constraints, we seek to minimize
:<math>\begin{align}
 &\left(Q_{xx} - M_{xx}\right)^2 + \left(Q_{xy} - M_{xy}\right)^2 + \left(Q_{yx} - M_{yx}\right)^2 + \left(Q_{yy} - M_{yy}\right)^2 \\
 &\quad {}+ \left(Q_{xx}^2 + Q_{yx}^2 - 1\right)Y_{xx} + \left(Q_{xy}^2 + Q_{yy}^2 - 1\right)Y_{yy} + 2\left(Q_{xx} Q_{xy} + Q_{yx} Q_{yy}\right)Y_{xy} .
\end{align}</math>

Taking the derivative with respect to {{mvar|Q<sub>xx</sub>}}, {{mvar|Q<sub>xy</sub>}}, {{mvar|Q<sub>yx</sub>}}, {{mvar|Q<sub>yy</sub>}} in turn, we assemble a matrix.
:<math>2\begin{bmatrix}
  Q_{xx} - M_{xx} + Q_{xx} Y_{xx} + Q_{xy} Y_{xy} & Q_{xy} - M_{xy} + Q_{xx} Y_{xy} + Q_{xy} Y_{yy} \\
  Q_{yx} - M_{yx} + Q_{yx} Y_{xx} + Q_{yy} Y_{xy} & Q_{yy} - M_{yy} + Q_{yx} Y_{xy} + Q_{yy} Y_{yy}
\end{bmatrix}</math>
</div>
In general, we obtain the equation
:<math> 0 = 2(Q - M) + 2QY , </math>
so that
:<math> M = Q(I + Y) = QS , </math>
where {{mvar|Q}} is orthogonal and {{mvar|S}} is symmetric. To ensure a minimum, the {{mvar|Y}} matrix (and hence {{mvar|S}}) must be positive definite. Linear algebra calls {{mvar|QS}} the [[polar decomposition]] of {{mvar|M}}, with {{mvar|S}} the positive square root of {{math|''S''<sup>2</sup> {{=}} ''M''<sup>T</sup>''M''}}.
:<math> S^2 = \left(Q^\mathsf{T} M\right)^\mathsf{T} \left(Q^\mathsf{T} M\right) = M^\mathsf{T} Q Q^\mathsf{T} M = M^\mathsf{T} M </math>

When {{mvar|M}} is [[non-singular matrix|non-singular]], the {{mvar|Q}} and {{mvar|S}} factors of the polar decomposition are uniquely determined. However, the determinant of {{mvar|S}} is positive because {{mvar|S}} is positive definite, so {{mvar|Q}} inherits the sign of the determinant of {{mvar|M}}. That is, {{mvar|Q}} is only guaranteed to be orthogonal, not a rotation matrix. This is unavoidable; an {{mvar|M}} with negative determinant has no uniquely defined closest rotation matrix.