I need help finding the polar

decomposition

of

$\displaystyle

\left( \begin{array}{cc} 11 & -5 \\ -2 & 10\end{array} \right)

$

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- Dec 12th 2009, 07:25 PMmancillaj3finding polar decomposition!
I need help finding the polar

decomposition

of

$\displaystyle

\left( \begin{array}{cc} 11 & -5 \\ -2 & 10\end{array} \right)

$ - Dec 14th 2009, 12:51 PMOpalg
If $\displaystyle A = \begin{bmatrix}11 & -5 \\ -2 & 10\end{bmatrix}$ then the polar decomposition of A is a factorisation $\displaystyle A = UR$, where U is unitary and R is positive. To find R use the fact that $\displaystyle R^2 = A^*A = \begin{bmatrix}125 & -75 \\ -75 & 125\end{bmatrix}$. So R is the positive square root of that matrix, which you can compute by diagonalising it. You should find that the eigenvalues of $\displaystyle R^2$ are 50 and 200, with corresponding normalised eigenvectors $\displaystyle \frac1{\sqrt2}\begin{bmatrix}1 \\ 1\end{bmatrix}$ and $\displaystyle \frac1{\sqrt2}\begin{bmatrix}1 \\ -1\end{bmatrix}$. So $\displaystyle R^2 = PDP^{-1}$, where $\displaystyle P = P^{-1} = \frac1{\sqrt2}\begin{bmatrix}1 &1 \\ 1 & -1\end{bmatrix}$ and $\displaystyle D = \begin{bmatrix}50 & 0 \\ 0 & 200\end{bmatrix}$. The square root is given by $\displaystyle R = PEP^{-1}$, where $\displaystyle E = D^{1/2} = \begin{bmatrix}5\sqrt2 & 0 \\ 0 & 10\sqrt2\end{bmatrix}$.

Having found R, you can then get U as $\displaystyle U = AR^{-1}$.