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Let $A$ and $B$ be $n\times m$ matrices of full column rank such that $\mathrm{Range}(A)\cap\mathrm{Range}(B)^{\perp} = \{0\}$. Show that the projection on $\mathrm{Range}(A)$ along $\mathrm{Range}(B)^{\perp}$ in $\mathbb R^n$ is given by $P = A(B'A)^{-1}B'$ where $B'$ is $B$ transpose.

Can you tell me how to show this??

detnvvp
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ssandi
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1 Answers1

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Necessarily $n\geq m$. $A,B:K^m\rightarrow K^n$ are one to one. $B^T:K^m\rightarrow K^n$ is onto. $im(B)⊥=\ker(B^T)$ has dimension $n-m$ and $im(A)\oplus \ker(B^T)=K^n$. Note that $B^TA:K^m\rightarrow K^m$ is bijective because $B^TAx=0$ implies $Ax\in im(A)\cap \ker(B^T)=\{0\}$. Let $y=Ax\in im(A)$ ; then $Py=A(B^TA)^{-1}(B^TA)x=y$. Let $z\in \ker(B^T)$ ; then $Pz=0$ and we are done.