singular value decomposition

bamby · #1 $Old$ November 12th, 2008, 07:58 PM

Can you give me an adea how to compute the singular value decomposition of this

(1+i 1 0
1-i 0 1)

Thank you!

Opalg · #2 $Old$ November 13th, 2008, 12:50 PM

Quote:

Originally Posted by bamby $View Post$

Can you give me an idea how to compute the singular value decomposition of this

$T=\begin{bmatrix}1+i&1&0\\ 1-i&0&1\end{bmatrix}$

This is a somewhat long and laborious process. Here's an outline of it.

You have to start by finding the eigenvalues and eigenvectors of T*T and TT*. Take TT* first, because that is a 2×2 matrix. So, $TT^* =$ $\begin{bmatrix}1+i &1& 0\\1-i &0& 1\end{bmatrix}\begin{bmatrix}1-i &1+i\\1&0\\ 0&1\end{bmatrix} = \begin{bmatrix}3&2i\\-2i&3\end{bmatrix}$ . The eigenvalues of this matrix are 1 and 5, with corresponding normalised eigenvectors $\begin{bmatrix}1/\sqrt2\\i/\sqrt2\end{bmatrix}$ and $\begin{bmatrix}1/\sqrt2\\-i/\sqrt2\end{bmatrix}$ .

Next, $T^*T =$ $\begin{bmatrix}4&1-i&1+i\\1+i&1&0\\ 1-i&0&1\end{bmatrix}$ . You don't have to calculate the eigenvectors again, because the theory tell you that these will be the same as those of TT*, together with enough zeros added to make up the right number. Therefore they are 1, 5 and 0. You still have to find corresponding normalised eigenvectors, which are $\begin{bmatrix}0\\(1+i)/2\\(-1+i)/2\end{bmatrix}$ , $\begin{bmatrix}2/\sqrt5\\(1+i)/(2\sqrt5)\\ (1-i)/(2\sqrt5)\end{bmatrix}$ , $\begin{bmatrix}1/\sqrt5\\(-1-i)/(\sqrt5)\\ (-1+i)/(\sqrt5)\end{bmatrix}$ .

You can now write down the matrices for the singular value decomposition $T = U\Sigma V^*$ of T. These are constructed as follows (apart from a slight modification that will come at the end): U is the matrix whose columns are the normalised eigenvectors of T*T, V is the matrix whose columns are the normalised eigenvectors of TT*, and Σ is a matrix having the same shape as T, with zeros everywhere except down the main diagonal, where the entries are the singular values of T, namely the square roots of the eigenvalues. Thus $U = \begin{bmatrix}1/\sqrt2&1/\sqrt2\\ i/\sqrt2&-i/\sqrt2\end{bmatrix}$ , $V = \begin{bmatrix}0&2/\sqrt5&1/\sqrt5)\\ (1+i)/2&(1+i)/(2\sqrt5)&(-1-i)/(\sqrt5)\\ (-1+i)/2&(1-i)/(2\sqrt5)&(-1+i)/(\sqrt5)\end{bmatrix}$ , $\Sigma = \begin{bmatrix}1&0&0\\0&\sqrt5&0\end{bmatrix}$ .

There's one final catch. You should check your answer by writing down the adjoint V* of V and computing the product UΣV*, which should be equal to T. But when you do that, you find that $U\Sigma V^* =$ $\begin{bmatrix}2/\sqrt2&(1-i)/\sqrt2&0\\ -2i/\sqrt2&0&(1-i)/\sqrt2\end{bmatrix}$ . This looks very different from T. But in fact it's not so different. The process of finding U, Σ and V is only determined up to a scalar factor of modulus 1, which in this case is $(1-i)/\sqrt2$ . If you multiply each entry of UΣV* by $\sqrt2/(1-i)$ then you do indeed get T. So to get the final correct solution, you should multiply each entry of U by $\sqrt2/(1-i)$ . With that modified U it is true that T=UΣV*.