# A good formula for singular value matrix of SVD?

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 $$\Sigma=U^TAV$$