I was faced with the task of decomposing the following matrix via SVD, i.e. decomposing the matrix into the form $USV^T$.
\begin{align*} A = \begin{bmatrix}3&2&2\\2&3&-2\end{bmatrix} \end{align*}
To approach this question, I first attempted to find the SVD decomposition of $A^T$, i.e. finding $USV^T$, then transposing it into $VS^TU^T$. I have renamed these variables back to $USV^T$ in the end.
To find the matrix $U$ for $A^T$, I found $u_1$ and $u_2$, which are calculated from $\frac{1}{\sigma_1}A^Tv_1$ and $\frac{1}{\sigma_2}A^Tv_2$. Then I calculated the cross product of those two vectors to make $u_3$. However, there seems to be an alternate solution (second one below) which does not involve finding $u_3$ (which I was always told was a necessary step). Therefore, I am wondering why this is valid and whether it is always okay to do so, any help would be appreciated.
\begin{align*} U &= \begin{bmatrix}\frac{1}{\sqrt2}&\frac{1}{\sqrt2} \\ \frac{1}{\sqrt2}&\frac{-1}{\sqrt2}\end{bmatrix} & S &= \begin{bmatrix}5&0&0 \\ 0&3&0\end{bmatrix} & V^T &= \begin{bmatrix}\frac{1}{\sqrt2}&\frac{1}{3\sqrt2}&2 \\ \frac{1}{\sqrt2}&\frac{-1}{3\sqrt2}&-\sqrt2 \\ 0&\frac{4}{3\sqrt2}&\frac{-1}{3\sqrt2}\end{bmatrix} \end{align*}
\begin{align*} U &= \begin{bmatrix}\frac{1}{\sqrt2}&\frac{1}{\sqrt2} \\ \frac{1}{\sqrt2}&\frac{-1}{\sqrt2}\end{bmatrix} & S &= \begin{bmatrix}5&0 \\ 0&3\end{bmatrix} & V^T &= \begin{bmatrix}\frac{1}{\sqrt2}&\frac{1}{\sqrt2}&0 \\ \frac{1}{3\sqrt2}&\frac{-1}{3\sqrt2}&\frac{4}{3\sqrt2}\end{bmatrix} \end{align*}
