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Im aware of the question Derivation of IV estimator? on this site.

Im interested however in obtaining the way we derive it using linear algebra.

$$\beta^{IV}=(Z'X)^{-1}Z'Y$$

the reason why I ask is to obtain a clearer picture of how exactly we get this equation.

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  • $\begingroup$ Can you be more precise as to what you mean by a derivation "using linear algebra"? $\endgroup$ – Christoph Hanck Mar 7 at 8:45
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One possible answer to this question is that we know that OLS is the solution to $$ (y-X\beta)'(y-X\beta) $$ w.r.t. to $\beta$.

The 2SLS estimator $\beta^{2SLS}=(X'P_ZX)^{-1}X'P_Zy$ is the solution to $$ (y-X\beta)'P_Z(y-X\beta), $$ where $P_Z$ is the (symmetric) projection matrix $P_Z=Z(Z'Z)^{-1}Z'$.

The IV formula you quote results under exact identification, so that $X'Z$ is square and the expression for 2SLS simplifies accordingly, noting that $$ (X'Z(Z'Z)^{-1}Z'X)^{-1}=(Z'X)^{-1}(Z'Z)(X'Z)^{-1}. $$

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