Derive the Pseudo Inverse (Moore Penrose) of Rank 1 Matrix as a Scalar Multiple of Its Transpose

Question

Let $ A \in \mathbb{R}^{m \times n} $ be a matrix of rank 1.
Show that its pseudo inverse is given by:

$$ {c}^{-1} {A}^{T} $$

where $ c = {\rm trace} \left( {A}^{T} A \right) $.

I know that $A^{T}A$ is symmetric with rank 1, so it has exactly one non-zero eigenvalue. Since the eigenvalues of a symmetric matrix are its diagonal entries, this eigenvalue must be $c$. However, I am stuck from here. Thanks for the help.

Answer 1

Because $A^TA$ has rank one, it has nullity $n-1$, thus $A^TA$ is similar to the following $n\times n$ diagonal matrix $$\begin{pmatrix}c&&&\\&0&&\\&&\ddots&\\&&&0\end{pmatrix}.$$ So, if we write $A=U\Sigma V^T$ as the singular value decomposition, where $U$ and $V$ are $m\times m$ and $n\times n$ orthogonal matrices, respectively, and $\Sigma$ is the $m\times n$ matrix such that $$\Sigma=\begin{pmatrix}\sqrt{c}&\\&O_{(m-1)\times(n-1)}\end{pmatrix}.$$ Then it is clear that the pseudoinverse of $A$ is \begin{align} A^\dagger &=V\Sigma^\dagger U^T\\ &=V \begin{pmatrix}\frac{1}{\sqrt{c}}&\\&O_{(n-1)\times(m-1)}\end{pmatrix}U^T\\ &=c^{-1}V \begin{pmatrix}\sqrt{c}&\\&O_{(n-1)\times(m-1)}\end{pmatrix}U^T\\ &=c^{-1}A^T. \end{align}

Answer 2

Presumably $A$ is real. Let $A=uv^T\ne0$. Then $c=\|u\|^2\|v\|^2\ne0$. Now it's straightforward to verify that $B=c^{-1}vu^T$ satisfies the four defining properties of Moore-Penrose pseudoinverse: $ABA=A,\ BAB=B$, $AB$ is Hermitian, $BA$ is Hermitian.