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I actually spent several days on this one. Really don't know how to prove. Can anyone help me? Thanks!

Show that for all $x,y\in\mathbb R^n$, $(xy^T)^+=(x^Tx)^+(y^Ty)^+yx^T$.

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    $\begingroup$ what you mean for + sign? $\endgroup$
    – user528935
    Commented Feb 28, 2018 at 3:33
  • $\begingroup$ Well, it means Moore-Penrose pseudoinverse. $\endgroup$
    – Sam
    Commented Mar 1, 2018 at 14:47

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This argument is valid in the case when $x$ and $y$ are both nonzero vectors, which is the nontrivial case (the result can easily be verified if $x=0$ or $y=0$).

If $U \Sigma V^*$ is the SVD of $A$, then $A^+ = V \Sigma^+ U^*$ and since $$ xy^\top = \frac{x}{||x||}(||x||\cdot||y||)\frac{y^\top}{||y||}, $$ it follows that \begin{align*} (xy^\top)^+ &= \frac{y}{||y||}(||x||\cdot||y||)^+\frac{x^\top}{||x||} \\ &= \frac{y}{||y||^2} \cdot\frac{x^\top}{||x||^2} \\ &= \frac{y}{y^\top y}\cdot \frac{x^\top}{x^\top x} \\ &= (x^\top x)^+ (y^\top y)^+y x^\top, \end{align*} which establishes the result when both vectors are nonzero.

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  • $\begingroup$ Thanks a lot! I see. $\endgroup$
    – Sam
    Commented Mar 2, 2018 at 3:08

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