moore-penrose inverse of a dyadic product

Question

Given $u,v \in \mathbb{K}^n$ and $||u||_2=||v||_2=1$, I've got to determine the pseudo inverse of the dyadic product $uv^*$, where $v^*$ is the conjugate transpose of $v$. How can I calculate the matrix $uv^*$? Anyone can give me a hint?

Answer 1

Hint: Recall that the SVD of a matrix $A = U\Sigma V^*$can be written in the form $$ A = \sum_{j=1}^n \sigma_j u_jv_j^* $$ Where the $u_j,v_j$ are the columns of $U,V$ respectively. The SVD of your matrix has the form $U\Sigma V^*$ in which the first column of $U$ is $u$ and the first column of $V$ is $v$.