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For the SVD, $A_{m\times n}=U_m\Sigma_{m\times n}V^T_n$ where $U$ & $V$ are orthogonal matrices & $\Sigma$ is diagonal, I am trying to obtain a formula for $\Sigma$...

If $\Sigma\Sigma^T$ was found using $AA^T=U\Sigma\Sigma^TU^T$, then: $$\Sigma = \left\{\begin{array}{ll} (\Sigma\Sigma^T)^{1/2}\begin{bmatrix}I_n\\0\end{bmatrix}_{m\times n}, & m > n\\ (\Sigma\Sigma^T)^{1/2}, & m = n\\ (\Sigma\Sigma^T)^{1/2}\begin{bmatrix}I_m & 0\end{bmatrix}_{m\times n}, & m < n \end{array} \right.$$

Likewise, if $\Sigma^T\Sigma$ was found using $A^TA=V\Sigma^T\Sigma V^T$, then: $$\Sigma = \left\{\begin{array}{ll} \begin{bmatrix}I_n\\0\end{bmatrix}_{m\times n}(\Sigma^T\Sigma)^{1/2}, & m > n\\ (\Sigma^T\Sigma)^{1/2}, & m = n\\ \begin{bmatrix}I_m & 0\end{bmatrix}_{m\times n}(\Sigma^T\Sigma)^{1/2}, & m < n \end{array} \right.$$

Did I make any mistakes? If not, is there a more elegant/simpler formula?

Notes:

  • I knew that the square roots of $\Sigma\Sigma^T$ & $\Sigma^T\Sigma$ would have only real entries bc they are just the eigenvalue matrices of the positive semi-definite $AA^T$ & $A^TA$, resp. whose eigenvalues are always positive.
  • $\Sigma\Sigma^T$ & $\Sigma^T\Sigma$ are square diagonal matrices so finding the square roots of their diagonal elements is all that is required to find $(\Sigma\Sigma^T)^{1/2}$ & $(\Sigma^T\Sigma)^{1/2}$.
  • From $AV=U\Sigma$ I found that $\Sigma=U^TAV$
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