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In a lot of texts I have seen involving the singulare value decomposition, it only says, that there are as many nonzero singular value as is the rank of the matrix $A$, which is to be decomposed.

Now I have looked at different examples throughout the net and everywhere these singular values are distinct. But as I understood the proof of the singular value decomposition, there are indeed $r$ singular values (if $rank(A)=r)$, but they may appear more than once: To be specific, they appear as often as the algebraic multiplicity of the eigenvalue, which corresponds to this singular value (since the sum of all algebraic multiplicities of all nonzero eigenvalues is $r$).

Is this correct ? Could anyone give me an example of a matrix $A$, such that this happens, i.e. such that $A^*A$ has eigenvalues with algebraic multiplicity $>1$ ?

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    Yes.$\ \ \ \ \ \ $2012-06-21

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You can build a matrix from its singular values by starting with a diagonal matrix with non-negative entries. Concretely, for any choice $\alpha_1,\ldots,\alpha_n\in[0,\infty)$, the matrix $ A=\begin{bmatrix}\alpha_1 \\ & \ddots \\ & & \alpha_n\end{bmatrix} $ has singular values $\alpha_1,\ldots,\alpha_n$. And of course you can choose them with as many repetitions as you want. They can even be all equal, as in Gerry's example: $\alpha_1=\cdots=\alpha_n=1$.