If I have a matrix $P_{M\times N}$ where $M>N$, can I find a matrix $A_{M\times M}$ so that it satisfies this equation?
$P^{T}AP=I$
where $I$ is identity matrix.
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We have two options:
- $rank(P)<n$:
First of all, we use that $$rank(AB)\leq\min\{rank(A),rank(B)\}$$ As proved here. Then, we have that $$rank(P^TAP)\leq rank(P) < n = rank(I)$$ We have that $rank(P^TAP)\neq rank(I)$ for any choices over $A$, so there is no $A$ matrix that holds the equality.
- $rank(P)=n$:
If we consider the $QR$ decomposition of $P$ (not full QR), we can write $P=QR$ with $Q\in\Bbb R^{m\times n}$ such that $Q^TQ=I$ and $R\in\Bbb R^{n\times n}$ an upper triangular invertible matrix (because $P$ has full rank). Then, you can choose $A=QR^{-1^T}R^{-1}Q^T$, so: $$P^TAP=(R^TQ^T)(QR^{-1^T}R^{-1}Q^T)(QR) = R^T(Q^TQ)R^{-1^T}R^{-1}(Q^TQ)R $$ $$=(R^TR^{-1^T})(R^{-1}R)=I$$
So we conclude that you can find a matrix $A\in\Bbb R^{m\times m}$ such that $P^TAP=I$ if and only if $P$ is a full rank matrix.
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The identity matrix on the right is $n\times n$ identity. Thus, $rank I = n$. – Sungjin Kim Dec 03 '16 at 15:47
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Oh you're right.. I'll edit my answer in a few mins. The answer will be "yes" if and only if $rank(P)=n$ – GLay Dec 03 '16 at 15:50
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