If $ A = ab^T$, where $a$ and $b$ are n-dimentional vectors, find the SVD factorization of $A$.
Ideas:
I first notice that given the form of $A$ it only has two null spaces generated by $a$ and $b$ and I was trying to solve $$ A(\alpha a + \beta b) = \lambda (\alpha a + \beta b) $$ But I think this is about the eigenspace of $A$ not $A^T A$.
Since $A$ is a rank-one matrix, its null-space has dimension $n-1$.
You have that $$AA^\top a = \|a\|^2\|b\|^2a $$ so $a$ is an eigenvector of $AA^\top$, which is a left singular vector of $A$. Similarly
$$A^\top A b = \|a\|^2 \|b\|^2 b $$ so $b$ is an eigenvector of $A^\top A$, which is a right singular vector of $A$.
Both the matrix $A^\top A$ and $AA^\top $ have the same eigenvalues (in this case only one of them is different from zero and equal to $\lambda=\|a\|^2 \|b\|^2$ ), which are the squared singular values of $A$.
Thus the only singular value different from zero out of the $n$ is given by $\sigma= \sqrt\lambda= \|a\|\|b\|$.