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I'm trying to prove the following statement. It seems pretty interesting but I have no idea of how to prove it:

Let $X$ be a $n\times p$ matrix, rank deficient. We can continuously add row to $X$ which is linearly independent with the row space of $X$ until full rank. That it, we obtain $\begin{pmatrix} X\\ A\end{pmatrix}$, where $X$ and $A$ are linearly independent. Prove that: $$(X^TX+A^TA)^{-1}$$ is an generalized inverse of $X^TX$, i.e, $(X^TX)(X^TX+A^TA)^{-1}(X^TX)=X^TX$

Is it really correct? I try a lot of examples but can't find any counter-examples...

Yuyuan Wang
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Let $v$ be an arbitrary vector. We will show that $X^TX(X^TX + A^TA)^{-1}X^TXv = X^TXv$. It will then follow that $X^TX(X^TX + A^TA)^{-1}X^TX = X^TX$.

Labeling, $v' = X^TXv$ we have that $v'$ is in the span of the rows of $X$. This means that if $(X^TX + A^TA)^{-1}v = b$ we have $v' = X^TXb$ (take a moment to show this yourself, but it is in the spoiler).

Indeed, we have $$v' = (X^TX + A^TA)b$$ but the rows of $A$ and $X$ are linearly independent, which means that $A^TAb=0$

This shows that

$$X^TX(X^TX + A^TA)^{-1}(X^TX)v = X^TX(X^TX + A^TA)^{-1}v' = X^TXb = v' = X^TXv$$

  • Thanks a lot! The proof is clear and straightforward! I was thinking about SVD but couldn't find a way to utilize the condition that $X$ and $A$ are independent. Say the decomposition such as: $$\begin{pmatrix} X\ A\end{pmatrix}= \begin{pmatrix} P_1\ P_2\end{pmatrix}\begin{pmatrix} D \ 0\end{pmatrix} Q^T$$ won't work.. – Yuyuan Wang Oct 14 '21 at 15:39