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Polar decomposition

Polar decomposition in 2D

By the polar decomposition theorem, every nonsingular real square matrix M can be factored uniquely into the product of an orthogonal and a positive definite real matrix:
\begin{equation} \label{EqPolar.1} \mathbf{M} = {\bf V}\,{\bf P}, \qquad \mathbf{V}^T {\bf V} = {\bf I} \quad\& \quad \mathbf{P} = \mathbf{P}^T > 0 , \end{equation}
where VT and PT denote the transposes of V and P.
Theorem 1: Every nonsingular two-dimensional matrix M ∈ ℝ2×2 has the polar decomposition (1) where \[ \mathbf{P} = \left\vert \mathbf{M} + \vert \det\mathbf{M} \vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right\vert^{-1/2} \left( \mathbf{M}^{\mathrm T} \mathbf{M} + \vert \det \mathbf{M} \vert \mathbf{I} \right) \] and \[ \mathbf{V} = \left\vert \det \left( \mathbf{M} + \left\vert \det\mathbf{M} \right\vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right) \right\vert^{-1/2} \left( \mathbf{M} + \vert \det\mathbf{M} \vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right) . \]
The proof follows Frank Uhlig paper Explicit Polar Decomposition and a Near-Characteristic Polynomial: The 2 X 2 Case.

Instead of giving the lengthy and rather tedious arithmetic sug- gested by the constructive proof for the polar decomposition theorem here, we need only verify that V and P as given above are in fact the polar factors of any nonsingular \[ \mathbf{M} = \begin{bmatrix} a & b \\ c& d \end{bmatrix} \in \mathbb{R}^{2\times 2} . \] Clearly P is positive definite as the sum of positive definite matrices, and \[ \mathbf{V}^{\mathrm T} \mathbf{V} = \left\vert \mathbf{M} + \vert \det\mathbf{M} \vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right\vert^{-1} . \] \[ \times \left[ \mathbf{M}^{\mathrm T} \mathbf{M} + 2\left\vert \det\mathbf{M} \right\vert \mathbf{I} + \left( \det \mathbf{M}^2 \right) \left( \mathbf{M}^{\mathrm T} \mathbf{M} \right) \right] . \] Now \[ \mathbf{M}^{\mathrm T} \mathbf{M} = \begin{bmatrix} a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \end{bmatrix} , \] and by Cramer’s rule \[ \left( \mathbf{M}^{\mathrm T} \mathbf{M} \right)^{-1} = \frac{1}{\det\mathbf{M}^2} \,\begin{bmatrix} b^2 + d^2 & - \left( ab + cd \right) \\ - \left( ab + cd \right) & a^2 + c^2 \end{bmatrix} . \] This, with \( \displaystyle \quad \beta = \left\vert \det \left( \mathbf{M} + \left\vert \det\mathbf{M} \right\vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right) \right\vert^{1/2} , \quad \) we have \[ \mathbf{V}^{\mathrm T} \mathbf{V} = \frac{1}{\beta^2} \begin{bmatrix} a^2 + b^2 + c^2 + d^2 + 2 \left\vert \det \mathbf{M} \right\vert & 0 \\ 0 & a^2 + b^2 + c^2 + d^2 + 2 \left\vert \det \mathbf{M} \right\vert \end{bmatrix} . \] Since \[ \left( \mathbf{M}^{\mathrm T} \right)^{-1} = \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -c \\ -b & a \end{bmatrix} , \] we have \[ \left\vert \det\mathbf{M} \right\vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} = \pm \begin{bmatrix} d & -c \\ -b & a \end{bmatrix} , \] the sign depending on whether det M is positive or negative. If det(M) > 0, then \begin{align*} \det \left[ \det\mathbf{M} + \left( \det\mathbf{M} \right) \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right] &= \det \begin{bmatrix} a+d & b-c \\ c-b & d+a \end{bmatrix} \\ &= \left( a + d \right)^2 + \left( b-c \right)^2 \\ &= a^2 + b^2 + c^2 + d^2 + 2\,\det\mathbf{M} \\ &= \beta^2 > 0 . \end{align*} If det(M) < 0, then \begin{align*} \beta^2 &= \left\vert \det \left( \mathbf{M} + \left\vert \det\mathbf{M} \right\vert \left( \mathbf{M}^{\mathrm T} \right)^{-1} \right) \right\vert \\ &= \left\vert \det \begin{bmatrix} a-d & b+c \\ c+b & d-a \end{bmatrix} \right\vert \\ &= \left\vert - \left( a-d \right)^2 - \left( b+c \right)^2 \right\vert \\ &= a^2 + b^2 + c^2 + d^2 + 2 \left\vert \det\mathbf{M} \right\vert . \end{align*} Thus, in either case VTV = I.

It remains to show that M = V P. Let us take the case detM < 0 for example. Then \[ \mathbf{V} = \frac{1}{\beta} \begin{bmatrix} a-d & b+c \\ c+b & d-a \end{bmatrix} \] and \[ \mathbf{P} = \frac{1}{\beta} \begin{bmatrix} a^2 + c^2 - \left( ad - bc \right) & ab + cd \\ ab + cd & b^2 + d^2 - \left( ad - bc \right) \end{bmatrix} . \] So \begin{align*} {\bf V}\,{\bf P} = \frac{1}{\beta^2} \begin{bmatrix} a-d & b+c \\ c+b & d-a \end{bmatrix} \cdot \begin{bmatrix} a^2 + c^2 - \left( ad - bc \right) & ab + cd \\ ab + cd & b^2 + d^2 - \left( ad - bc \right) \end{bmatrix} \\ &= \frac{1}{\beta^2} \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{bmatrix} , \end{align*} where \begin{align*} a_{1,1} &= a \left[ a^2 + c^2 - \left( ad -bc \right) \right] - da^2 + ad^2 -bcd + a b^2 + bcd + abs + c^2 d \\ &= a \left[ a^2 + c^2 - \left( ad -bc \right) - ad + d^2 + b^2 + bc\right] \\ &= a \left[ a^2 + b^2 + c^2 + d^2 -2 \left( ad - bc \right) \right] = a\,\beta^2 . \end{align*} Similarly, 𝑎1,2 = bβ², 𝑎2,1 = cβ², and 𝑎2,2 = dβ². Therefore, \[ \mathbf{M} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = {\bf V}\,{\bf P} . \] The case detM > 0 can be verified analogously.

   

Example 1:    ■

End of Example 1
One can apply Theorem 1 to classify all integer 2 X 2 matrices whose polar decomposition can be achieved over the rationals:
Theorem 2 (rational polar decomposition): A nonsingular matrix \[ \mathbf{A} = \begin{bmatrix} a & b \\ c&d \end{bmatrix} \in \mathbb{Z}^{2\times 2} \] has a polar decomposition over ℚ if and only if

in case detA < 0: 𝑎 − d and b + c form a Pythagorean pair of integers, and

in case detA > 0: 𝑎 + d and b − c do.

In case detA > 0, we have \[ 0 < \det \left[ \mathbf{A} + \left( \det\mathbf{A} \right) \left( \mathbf{A}^{\mathrm T} \right)^{-1} \right] = \left( a + d \right)^2 + \left( b-c \right)^2 \] from the previous proof. Thus, in this case P ∈ ℚ2×2 and V ∈ ℚ2×2 if and only if \[ \beta^2 = ( a + d )^2 + ( b -c )^2 \] is an integer square. The proof in case detA < 0 is similar.
   

Example 2:    ■

End of Example 2
Theorem 3 (Matrix identity): Every nonsingular matrix A ∈ ℝ2×2 satisfies the identity \begin{align*} \tag{2} &\left\vert \det \left[ \mathbf{A} + \left\vert \det\mathbf{A} \right\vert \left( \mathbf{A}^{\mathrm T} \right)^{-1} \right] \right\vert \mathbf{A} \\ &= \mathbf{A} \,\mathbf{A}^{\mathrm T} \mathbf{A} + 2 \left\vert \det\mathbf{A} \right\vert \mathbf{A} + \left( \det\mathbf{A}^2 \right) \left( \mathbf{A}^{\mathrm T} \right)^{-1} . \notag \end{align*}
Formular (2) follows from the identity A = V P for any nonsingular real 2 × 2 matrix.
   

Example 3:    ■

End of Example 3
    If we multiply both sides of Eq.(2) by AT from the right and set X = A AT, then this equation can be written as
\[ \left\vert \det \left[ \mathbf{A} + \left\vert \det\mathbf{A} \right\vert \left( \mathbf{A}^{\mathrm T} \right)^{-1} \right] \right\vert \mathbf{X} = \mathbf{X}^2 + 2 \left\vert \det\mathbf{A} \right\vert \mathbf{X} + \left( \det\mathbf{X} \right) \mathbf{I} , \]
or
\[ \mathbf{X}^2 - \left\{ \left\vert \det \left[ \mathbf{A} + \left\vert \det\mathbf{A} \right\vert \left( \mathbf{A}^{\mathrm T} \right)^{-1} \right] \right\vert - 2 \left\vert \det\mathbf{A} \right\vert \right\} \mathbf{X} + \left( \det\mathbf{X} \right) \mathbf{I} = 0 . \]
This last equation must hold for all positive definite 2×2 real matrices X by Theorem 3.
Lemma 1: For any nonsingular matrix A ∈ ℝ2×2, we have \[ \mbox{trace} \left( \mathbf{A}\,\mathbf{A}^{\mathrm T} \right) = \left\vert \det \left[ \mathbf{A} + \left\vert \det\mathbf{A} \right\vert \left( \mathbf{A}^{\mathrm T} \right)^{-1} \right] \right\vert - 2 \left\vert \det\mathbf{A} \right\vert . \]
Without loss of generality assume \[ \mathbf{A} = \begin{bmatrix} a & b \\ c&d \end{bmatrix} \in \mathbb{Z}^{2\times 2} \qquad \mbox{with} \quad \det\mathbf{A} > 0 . \] Then tr(A AT) = 𝑎² + b² +c² + d² while \[ \left\vert \det \left[ \mathbf{A} + \left\vert \det\mathbf{A} \right\vert \left( \mathbf{A}^{\mathrm T} \right)^{-1} \right] \right\vert - 2 \left\vert \det\mathbf{A} \right\vert = ( a+b)^2 + (b-c)^2 - 2 \left( ad -bc \right) \] from the proof of Theorem 2.
   

Example 4:    ■

End of Example 4