Knowing from theory that fro a matrix $A$ we $A = U\Sigma V^{T}$
I want to solve for $U$,$V$ and $\Sigma$. My effort is the following but I don't if I am correct. \begin{align*} A &= U\Sigma V^{T} \\ U^{T}A&= \Sigma V^{T}\\ U^{T}&= \Sigma V^{T}A^{-1}\\ U&= \Sigma^{-1}V A \end{align*} and
\begin{align*} A &= U\Sigma V^{T} \\ A\Sigma^{-1}&= UV^{T}\\ \Sigma^{-1}&= A^{-1} U V^{T}\\ \Sigma&= U^{T}A V \end{align*}
\begin{align*} A &= U\Sigma V^{T} \\ A V &= \Sigma U \\ V &=\Sigma U^{T}A ^{-1} \end{align*}
Along with $A^{T}=V\Sigma U^{T} $ \begin{align*} A&=USV^{T} \\ A^{T}A &= V\Sigma U^{T}U\Sigma V^{T} \\ A^{T}A &= V\Sigma^{2}V^{T} \\ A^{T}AV &= V\Sigma^{2} \end{align*}
I'd be really careful with $A^{-1}$, since the SVD decomposition exists for matrices with different dimensions. The third section is what tells em the morst information, mainly that $V$ contains the eigenvectors of $A{*} A$. If you try to simplify $A A{*}$, you'll find similar information about $U$.