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I am struggling with understanding one result in Buera and Nicolini (2004) published in JME. In page 538, they write $$A_t(h^t) b_{t-1}(h^{t-1}) = Z_t(h^{t-1})$$ where $A_t(h^t)$ is a $N \times J$ matrix, $b_{t-1}(h^{t-1})$ is a $J\times 1$ vector, and $Z_t(h^{t-1})$ is a $N\times 1$ vector. $N$ is the number of states and $J$ is the number of maturities. Then, in page 539, they say that

"a necessary condition for the Ramsey to be implementable is that the government can issue non-contingent bonds in at least as many maturities as possible realization of the stock, i.e., $J \ge N$.''

For an ease of notation, write the above equation as $A b = Z$. Suppose that $J > N$. Premultiplying a $J \times N$ matrix $D$, $$D Ab = D Z$$ If $D$ is an inverse of $A$, rank($DA$) = $J$. But, $\text{rank}(DA) \le \min(\text{rank}(D), \text{rank}(A))\le N$. Contradiction. I think I am missing something important. I would appreciate if someone helps me with this.

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  • $\begingroup$ There is something I can't understand. $A$ is a $n\times j$ matrix, if $n\neq j$ $A$ is a non-square matrix, and can't have an inverse, the inverse of a matrix is defined for square matrices only. $\endgroup$ Mar 26 at 12:11
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    $\begingroup$ @BakerStreet Perhaps it is a generalized inverse? The article is "an", as in "$D$ is an inverse of $A$". In that case rank of $DA$ would be $N \leq J$ though. $\endgroup$
    – Giskard
    Mar 26 at 16:41
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    $\begingroup$ @Giskard maybe, but nothing can be said without reading the paper, it sholuld be specified. $\endgroup$ Mar 26 at 17:01
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    $\begingroup$ Is this $D$ still in the paper though? I thought the papers part stops at the quotation marks at the middle. $\endgroup$
    – Giskard
    Mar 26 at 17:10
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    $\begingroup$ @BakerStreet Here is the relevant page, in case you are interested econ.ucla.edu/fjbuera/papers/matFINAL2.pdf#page=12 $\endgroup$
    – Giskard
    Mar 26 at 17:13

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