Pseudoinverse of rank-$1$ matrix

Question

Prove that $$(xy^T)^{+} = (x^Tx)^{+}(y^Ty)^{+}yx^T$$ where $x,y \in \mathbb{R}^n$, and $A^{+}$ is defined to be such that $AA^{+}A = A$.

Answer 1

If $x=0$ or $y=0$, then $xy^\top$ is the zero matrix, so anything can be the pseudoinverse.

So now assume $x$ and $y$ are both nonzero. Then $(x^\top x)^+ = (x^\top x)^{-1}$ and $(y^\top y)^+ = (y^\top y)^{-1}$ are just scalars. \begin{align} (xy^\top) (xy^\top)^+ (xy^\top) &= (xy^\top) (x^\top x)^+ (y^\top y)^+ y x^\top (xy^\top) \\ &= (x^\top x)^{-1} (y^\top y)^{-1} \; xy^\top y x^\top x y^\top \\ &= xy^\top. \end{align}

Answer 2

There is a formula for the pseudoinverse of the product of two matrices. $$(AB)^+ = (A^+AB)^+(ABB^+)^+$$ The pseudoinverse of a vector has a simple formula. $$v^+ = \frac{v^T}{v^Tv} \;=\; (v^Tv)^+v^T$$ Although it looks strange, the expression on the far RHS has the advantage that it remains valid when $v=0$.

We also have the Penrose conditions: $\quad A^+AA^+\!= A^+,\quad AA^+A = A$

Apply the product formula and the Penrose conditions to evaluate $$\eqalign{ (v^+v)^+ &= ((v^+)^+v^+v)^+\;(v^+vv^+)^+ = (v)^+(v^+)^+ = v^+v \\ }$$ Now apply the above formulas to the outer product of two vectors. $$\eqalign{ (xy^T)^+ &= \Big(x^+xy^T\Big)^+\;\Big(xy^Ty^{T+}\Big)^+ \\ &= \Big((x^Tx)^+x^Txy^T\Big)^+\;\Big(xy^Ty(y^Ty)^+\Big)^+ \\ &= \Big(y^{T+}(x^Tx)^+(x^Tx)\Big)\;\Big((y^Ty)(y^Ty)^+x^+\Big) \\ &= y(y^Ty)^+(y^Ty)(y^Ty)^+\;\,(x^Tx)^+(x^Tx)(x^Tx)^+x^T \\ &= y\,(y^Ty)^+\;\,(x^Tx)^+x^T \\ &= (x^Tx)^+(y^Ty)^+\;yx^T \\ }$$