Find a SVD of $A=u_1v_1^\intercal +2u_2v_2^\intercal +3u_3v_3^\intercal$ where $\{u_1,u_2,u_3\}\subset\mathbb R^M$ and $\{v_1,v_2,v_3\}\subset\mathbb R^N$ are orthonormal sets.
The SVD is $A=U\Sigma T^\intercal$, so I begin by finding $AA^\intercal=u_1u_1^\intercal+4u_2u_2^\intercal+9u_3u_3^\intercal\in\mathbb R^{M\times M}$. I now need to find the eigenvalues and eigenvectors, what could they be?
$(u_1u_1^\intercal+4u_2u_2^\intercal+9u_3u_3^\intercal)v=\lambda v\implies (u_1^\intercal v)u_1+4(u_2^\intercal v)u_2+9(u_3^\intercal v)u_3=\lambda v\implies...what?$
Likewise, $AA^\intercal=v_1v_1^\intercal+2v_2v_2^\intercal+3v_3v_3^\intercal\in\mathbb R^{N\times N}$, and what are the eigenvalues? Just need the eigenvalues for this one.
Any help is appreciated.
Since you already computed $AA^T=u_1u_1^T+4u_2u_2^T+9u_3u_3^T$, let us write its matrix representation $[AA^T]_\beta$ with respect to orthonormal basis $\beta=\{u_1,u_2,u_3\}$. The first column is $[AA^Tu_1]_\beta = [(u_1u_1^T+4u_2u_2^T+9u_3u_3^T)u_1]_\beta = [u_1]_\beta=\begin{pmatrix} 1 \\ 0\\ 0 \end{pmatrix}$; the second column is $[AA^Tu_2]_\beta = [4u_2]_\beta=\begin{pmatrix} 0 \\ 4\\ 0 \end{pmatrix}$; and the third column is $[AA^Tu_3]_\beta = [9u_3]_\beta=\begin{pmatrix} 0 \\ 0\\ 9 \end{pmatrix}$. This gives $[AA^T]_\beta=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 0\\ 0& 0 &9 \end{pmatrix}$, which I think you can see what its eigenvalues are.
Also, from here, can you see what the eigenvectors of $AA^T$ are? Now make a guess of what the eigenvalues and eigenvectors are for an expression like $a_1 u_1 u_1^T + a_2 u_2 u_2^T +\cdots + a_n u_n u_n^T$ where $\{u_1,\ldots,u_n\}$ forms an orthonormal basis.