SVD Transpose Equations $$Av_i=
\begin{cases}
\sigma_iu_i  & i = 1, \ldots , r \\
0 & i = r+1, \ldots , m
\end{cases}$$
$$A^Tu_i=
\begin{cases}
\sigma_iv_i  & i = 1, \ldots , r \\
0 & i = r+1, \ldots , m
\end{cases}$$
1.) Show that the top equation implies the matrix equation $AV = U\Sigma$.
2.) Show that the bottom equation implies the matrix equation $A^TU = V\Sigma^T$
3.) Show that either one implies the SVD $A=U\Sigma V^T$
I understand the implications of both equations, however, I'm not sure how to show it in a simplistic way.
 A: You have not defined the dimensions of the matrix. Since, the SVD is applicable to rectangular matrices, we assume that $A$ is an $m \times n$ matrix. Furthermore, we assume that $A$ is a tall matrix; that is $m \ge n$. If $m < n$, we work with the transpose of $A$. This tall matrix has $n$ singular values but some of them could be zero. You have assumed that there are only $r$ non-zero singular values.  It is not necessary to assume this restriction explicitly at this stage. 
Wit the new conditions, the top equation becomes
$$
Av_i = \sigma_i u_i, \;\; i=1,\ldots,n
$$
We stack all the columns given by the $n$ equations to get
$$
A[v_1,v_2,\ldots,v_n ] = [u_1,u_2,\ldots,u_n]
  \left[ \begin{array}{cccc}
   \sigma_1 & & & \\
             & \sigma_2 &   & \\
              &          &   & \\
              &          &    & \sigma_n 
           \end{array} \right]
$$
which can be recognised as
$$
A V = U\Sigma .
$$
The above is the first answer you required.
Post multiplying by $V^t$ we get
$$
AVV^t = U\Sigma V^t
$$
However, $VV^t = I$ (the identity matrix. Thus,
$ A = U\Sigma V^t$ which is the third answer you required. This is the "economical version" of the SVD which does not have the left null space.  
You can get the second answer in a similar way to the first answer.
