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Please have a look at the property of SVD - section "Singular values, singular vectors, and their relation to the SVD" from http://en.wikipedia.org/wiki/Singular_value_decomposition.

Could it be state that "and" between $Mv = \sigma\cdot u \quad\text{ and }\quad M^Tu = \sigma\cdot v$ is actually equivalence, and can be safely omitted?

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No. It is not an equivalence. We need both the properties to be satisfied.

Given any matrix $M$, and any vector $v$ and a scalar $\sigma$, we can get a $u$, if we just want to satisfy $Mv = \sigma u.$ This need not satisfy $M^*u = \sigma v$

For instance, let $M = \begin{bmatrix} 2 & 9\\ 3 & 6 \end{bmatrix}$, $v = \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $\sigma = 2$. This gives us $Mv = \begin{bmatrix} 2\\ 3 \end{bmatrix}$ and hence $u = \begin{bmatrix} 1 \\ 3/2 \end{bmatrix}$. However, $\dfrac1{\sigma} M^*u = \dfrac12 \begin{bmatrix} 2 & 3\\ 9 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 3/2 \end{bmatrix} = \dfrac12 \begin{bmatrix} 2 + 9/2 \\ 9 + 9 \end{bmatrix} = \dfrac12 \begin{bmatrix}11/2 \\ 18 \end{bmatrix} = \begin{bmatrix}11/4 \\ 9 \end{bmatrix} \neq v$

In fact, we see that we need to choose $v$ to be the eigenvector of $M^*M$ and choose $u$ to be the eigenvector of $MM^*$ to get what we want.