I do have the matrix $A=U\Sigma V^T$ with the SVD $$ U = \frac{1}{3} \begin{bmatrix}2 & 1&-2\\ 1&2&2\\ 2&-2&1\end{bmatrix}$$ $$ \Sigma = \begin{bmatrix} 3&0\\0&0.3\\0&0 \end{bmatrix}$$ $$V=\begin{bmatrix} 0.8&0.6\\-0.6&0.8 \end{bmatrix}$$ Now I would like to calculate the $\ker(AA^\top)$ & $\mathrm{range}(AA^\top)$.
Due to the SVD I know $AA^\top$ does only have $2$ eigenvalues, which results in $\mathrm{rank}(AA^\top)=2$.
With the SVD I know that it should be $\mathrm{range}(A)= \mathrm{Span}\{u_1, u_2\}$ & $\ker(A)=\mathrm{Span}\{0\}$. (as there is no $v_3$. Right?)
How can I proceed from here? Can I use properties of the SVD to get the kernel and range of $AA^\top$ quickly without calculating $A$?