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I want to calculate the SVD ($A = U\Sigma V^*$)of $$A = \begin{bmatrix} 0 & 2 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$$ but $$A^TA = \begin{bmatrix} 0 & 0 \\ 0 & 4 \end{bmatrix}$$

which has a zero eigenvalue. The problem with this is that the columns of $U$ are given by

$$u_i = \frac{Av_i}{\sigma_i}$$

where $\sigma_i = \sqrt{\lambda_i}$.

1 Answers1

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No, $Av_i=\sigma_iu_i$, which is perfectly well defined even when $\sigma_i=0$. The point is $U$ can be decomposed into vectors corresponding to $\sigma_1,\cdots,\sigma_k>0$ and, when $\sigma_i=0$, you pad $U$ with vectors spanning the cokernel (i.e. whatever the range of $A$ misses) of $A$. See the example calculation here:

https://en.wikipedia.org/wiki/Singular_value_decomposition#Example

Alex R.
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  • Is the number of zero eigenvalues of a matrix equal to the number of zero singular values of it ? – Fareed Abi Farraj Jul 15 '19 at 07:48
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    @Fareed AF: by definition a singular value satisfies $Au_i=\sigma_i v_i$. So zero singular values then there are as many as there are zero eigenvectors. – Alex R. Jul 15 '19 at 18:26
  • The example pointed to on Wiki does not seem to actually how the calculation steps (or am I missing things?)... – Confounded May 30 '23 at 06:55