# struggling on the properties of the idempotent matrix, namely for any n &#x00D7;<!-- × --> n

struggling on the properties of the idempotent matrix, namely for any $n×n$ matrix, ${A}^{2}=A.$. The projection matrix defined by $M={I}_{n}-A{\left({A}^{T}A\right)}^{-1}{A}^{T}$ is an idempotent matrix. The question is, for any given $n×m$ ($n>m$) matrix $B,$, do we have
$\begin{array}{rcl}M& =& {I}_{n}-A{\left({A}^{T}A\right)}^{-1}{A}^{T}\\ & =& {I}_{n}-AB{\left({B}^{T}{A}^{T}AB\right)}^{-1}{B}^{T}{A}^{T},\end{array}$
since $AB$ is basically the linear transformation of matrix $A.$
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Daniel Valdez
First, your projection matrix makes sense if $rank\phantom{\rule{thinmathspace}{0ex}}A=k. Otherwise, $M=0$. If so, $M$ maps to the $n-k$-dimensional linear space spanned by the vectors ${c}_{i}$ orthogonal to the columns of $A$.
Now, when you consider $\overline{M}={I}_{n}-AB{\left({B}^{T}{A}^{T}AB\right)}^{-1}{B}^{T}{A}^{T}$, this operator maps to the linear subspace orthogonal to the column space of $AB$. Here, you can have different options depending on the rank of $AB$.
Let, for instance, $rank\phantom{\rule{thinmathspace}{0ex}}AB=rank\phantom{\rule{thinmathspace}{0ex}}A$. In this case, the column space of $AB$ is the same as the column space of $A$ and $M=\overline{M}$. But this will not be the case if $rank\phantom{\rule{thinmathspace}{0ex}}AB.