The matrix is
$$\pmatrix{1 &0 &-1\\ -1& 0& 1}$$
The $\Sigma$ matrix I get is
$$\pmatrix{2& 0& 0\\ 0& 0& 0}$$
which is not invertible, so I can't calculate $U$.
How do you calculate the SVD when $\Sigma$ is not invertible?
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$\begingroup$
linear-algebra
matrices
matrix-decomposition
svd
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0Why the downvote when the OP is showing the correct $S$? – 2017-01-27
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3We find the first column of $U$ as $u_1 =\begin{bmatrix} -\dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \end{bmatrix}$, but we can't find $u_2$ in the same way since $\sigma_2 = 0$. To find $u_2$, we just have to extend $u_1$ to an orthonormal basis of $\mathbb{R}^2$. Letting $u_2 = \begin{bmatrix} \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \end{bmatrix}$ will work and allow you to continue. Notice that $U = [u_1~~u_2]$ is a rotation matrix. – 2017-01-27
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0im still getting the wrong value for u1, can you explain a bit on how to actually get the value. Also i did not understand how to come to the conclusion of using u2 as the 1/(2**1/2) 1/(2**1/2). – 2017-01-27