Do all projections matrices take this form?












4












$begingroup$


Do all projection matrices take the form $P = A{(A^TA)}^{-1}A^T$? If so, can you help me derive it and explain it intuitively?






linear-algebra matrices projective-geometry







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$endgroup$

















    4












    $begingroup$


    Do all projection matrices take the form $P = A{(A^TA)}^{-1}A^T$? If so, can you help me derive it and explain it intuitively?






    linear-algebra matrices projective-geometry







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      Do all projection matrices take the form $P = A{(A^TA)}^{-1}A^T$? If so, can you help me derive it and explain it intuitively?






      linear-algebra matrices projective-geometry







      share|cite|improve this question









      $endgroup$




      Do all projection matrices take the form $P = A{(A^TA)}^{-1}A^T$? If so, can you help me derive it and explain it intuitively?






      linear-algebra matrices projective-geometry




      linear-algebra matrices projective-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 1 at 11:58









      Kid CudiKid Cudi

      456




      456






















          2 Answers
          2






          active

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          9












          $begingroup$

          $A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $pmatrix{1&1\ 0&0}$. Thus the answer is clearly no.



          It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $Pin M_n(mathbb R)$ is an orthogonal projection, then $P|_U=operatorname{id}$ and $P|_{U^perp}=0$ for some subspace $Usubseteqmathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $vin U^perp$. Also, since the columns of $A$ span $U$, every vector $uin U$ can be written as $Ax$ for some $xinmathbb R^n$. Therefore
          $$
          left(A(A^TA)^{-1}A^Tright)u=left(A(A^TA)^{-1}A^Tright)(Ax)=left(A(A^TA)^{-1}A^TAright)x=Ax=u
          $$
          for every $u=Axin U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            $endgroup$
            – littleO
            Jan 1 at 15:24










          • $begingroup$
            You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
            $endgroup$
            – amd
            Jan 2 at 7:05










          • $begingroup$
            @amd I am. This is the usual setting when this kind of projections are encountered.
            $endgroup$
            – user1551
            Jan 2 at 9:36



















          3












          $begingroup$

          As pointed out by @user1551, this is only true for orthogonal projection matrices.



          Let $P$ be the orthogonal projection operator that projects a vector $b in mathbb R^n$ onto a subspace $S subset mathbb R^n$. Let $(a_1,ldots,a_m)$ be a basis for $S$, and let $A$ be the matrix whose $i$th column is $a_i$. Then $S ={Ax mid x in mathbb R^m}$, and projecting $b$ onto $S$ is equivalent to selecting $x$ so as to minimize the distance from $b$ to $Ax$. Equivalently, we want to minimize
          $$
          r(x) = | Ax - b |^2.
          $$
          This is a least squares problem.
          Setting the gradient equal to $0$, we find that $x$ satisfies
          $$
          tag{1} A^T(Ax-b) = 0
          $$
          or equivalently
          $$A^TA x = A^T b.$$
          (This system of equations is often called the "normal equations". Visually, equation (1) just says that the residual vector $b - Ax$ is orthogonal to the column space of $A$.)



          It follows that $x = (A^T A)^{-1} A^T b$. So the projection of $b$ onto $S$ is
          $$
          P(x) = Ax = A(A^T A)^{-1} A^T b.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So it isn't necessarily true that all projection matrices take that form?
            $endgroup$
            – Kid Cudi
            Jan 1 at 13:05











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          2 Answers
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          9












          $begingroup$

          $A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $pmatrix{1&1\ 0&0}$. Thus the answer is clearly no.



          It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $Pin M_n(mathbb R)$ is an orthogonal projection, then $P|_U=operatorname{id}$ and $P|_{U^perp}=0$ for some subspace $Usubseteqmathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $vin U^perp$. Also, since the columns of $A$ span $U$, every vector $uin U$ can be written as $Ax$ for some $xinmathbb R^n$. Therefore
          $$
          left(A(A^TA)^{-1}A^Tright)u=left(A(A^TA)^{-1}A^Tright)(Ax)=left(A(A^TA)^{-1}A^TAright)x=Ax=u
          $$
          for every $u=Axin U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            $endgroup$
            – littleO
            Jan 1 at 15:24










          • $begingroup$
            You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
            $endgroup$
            – amd
            Jan 2 at 7:05










          • $begingroup$
            @amd I am. This is the usual setting when this kind of projections are encountered.
            $endgroup$
            – user1551
            Jan 2 at 9:36
















          9












          $begingroup$

          $A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $pmatrix{1&1\ 0&0}$. Thus the answer is clearly no.



          It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $Pin M_n(mathbb R)$ is an orthogonal projection, then $P|_U=operatorname{id}$ and $P|_{U^perp}=0$ for some subspace $Usubseteqmathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $vin U^perp$. Also, since the columns of $A$ span $U$, every vector $uin U$ can be written as $Ax$ for some $xinmathbb R^n$. Therefore
          $$
          left(A(A^TA)^{-1}A^Tright)u=left(A(A^TA)^{-1}A^Tright)(Ax)=left(A(A^TA)^{-1}A^TAright)x=Ax=u
          $$
          for every $u=Axin U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            $endgroup$
            – littleO
            Jan 1 at 15:24










          • $begingroup$
            You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
            $endgroup$
            – amd
            Jan 2 at 7:05










          • $begingroup$
            @amd I am. This is the usual setting when this kind of projections are encountered.
            $endgroup$
            – user1551
            Jan 2 at 9:36














          9












          9








          9





          $begingroup$

          $A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $pmatrix{1&1\ 0&0}$. Thus the answer is clearly no.



          It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $Pin M_n(mathbb R)$ is an orthogonal projection, then $P|_U=operatorname{id}$ and $P|_{U^perp}=0$ for some subspace $Usubseteqmathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $vin U^perp$. Also, since the columns of $A$ span $U$, every vector $uin U$ can be written as $Ax$ for some $xinmathbb R^n$. Therefore
          $$
          left(A(A^TA)^{-1}A^Tright)u=left(A(A^TA)^{-1}A^Tright)(Ax)=left(A(A^TA)^{-1}A^TAright)x=Ax=u
          $$
          for every $u=Axin U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.






          share|cite|improve this answer











          $endgroup$



          $A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $pmatrix{1&1\ 0&0}$. Thus the answer is clearly no.



          It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $Pin M_n(mathbb R)$ is an orthogonal projection, then $P|_U=operatorname{id}$ and $P|_{U^perp}=0$ for some subspace $Usubseteqmathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $vin U^perp$. Also, since the columns of $A$ span $U$, every vector $uin U$ can be written as $Ax$ for some $xinmathbb R^n$. Therefore
          $$
          left(A(A^TA)^{-1}A^Tright)u=left(A(A^TA)^{-1}A^Tright)(Ax)=left(A(A^TA)^{-1}A^TAright)x=Ax=u
          $$
          for every $u=Axin U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 9:40

























          answered Jan 1 at 13:46









          user1551user1551

          72.9k566128




          72.9k566128








          • 1




            $begingroup$
            Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            $endgroup$
            – littleO
            Jan 1 at 15:24










          • $begingroup$
            You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
            $endgroup$
            – amd
            Jan 2 at 7:05










          • $begingroup$
            @amd I am. This is the usual setting when this kind of projections are encountered.
            $endgroup$
            – user1551
            Jan 2 at 9:36














          • 1




            $begingroup$
            Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            $endgroup$
            – littleO
            Jan 1 at 15:24










          • $begingroup$
            You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
            $endgroup$
            – amd
            Jan 2 at 7:05










          • $begingroup$
            @amd I am. This is the usual setting when this kind of projections are encountered.
            $endgroup$
            – user1551
            Jan 2 at 9:36








          1




          1




          $begingroup$
          Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
          $endgroup$
          – littleO
          Jan 1 at 15:24




          $begingroup$
          Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
          $endgroup$
          – littleO
          Jan 1 at 15:24












          $begingroup$
          You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
          $endgroup$
          – amd
          Jan 2 at 7:05




          $begingroup$
          You appear to be assuming the standard basis and Euclidean scalar product. In a different basis of the ambient space or with a different scalar product, the expression still yields the matrix of a projection ($P^2=P$ always), but it’s not likely to be orthogonal.
          $endgroup$
          – amd
          Jan 2 at 7:05












          $begingroup$
          @amd I am. This is the usual setting when this kind of projections are encountered.
          $endgroup$
          – user1551
          Jan 2 at 9:36




          $begingroup$
          @amd I am. This is the usual setting when this kind of projections are encountered.
          $endgroup$
          – user1551
          Jan 2 at 9:36











          3












          $begingroup$

          As pointed out by @user1551, this is only true for orthogonal projection matrices.



          Let $P$ be the orthogonal projection operator that projects a vector $b in mathbb R^n$ onto a subspace $S subset mathbb R^n$. Let $(a_1,ldots,a_m)$ be a basis for $S$, and let $A$ be the matrix whose $i$th column is $a_i$. Then $S ={Ax mid x in mathbb R^m}$, and projecting $b$ onto $S$ is equivalent to selecting $x$ so as to minimize the distance from $b$ to $Ax$. Equivalently, we want to minimize
          $$
          r(x) = | Ax - b |^2.
          $$
          This is a least squares problem.
          Setting the gradient equal to $0$, we find that $x$ satisfies
          $$
          tag{1} A^T(Ax-b) = 0
          $$
          or equivalently
          $$A^TA x = A^T b.$$
          (This system of equations is often called the "normal equations". Visually, equation (1) just says that the residual vector $b - Ax$ is orthogonal to the column space of $A$.)



          It follows that $x = (A^T A)^{-1} A^T b$. So the projection of $b$ onto $S$ is
          $$
          P(x) = Ax = A(A^T A)^{-1} A^T b.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So it isn't necessarily true that all projection matrices take that form?
            $endgroup$
            – Kid Cudi
            Jan 1 at 13:05
















          3












          $begingroup$

          As pointed out by @user1551, this is only true for orthogonal projection matrices.



          Let $P$ be the orthogonal projection operator that projects a vector $b in mathbb R^n$ onto a subspace $S subset mathbb R^n$. Let $(a_1,ldots,a_m)$ be a basis for $S$, and let $A$ be the matrix whose $i$th column is $a_i$. Then $S ={Ax mid x in mathbb R^m}$, and projecting $b$ onto $S$ is equivalent to selecting $x$ so as to minimize the distance from $b$ to $Ax$. Equivalently, we want to minimize
          $$
          r(x) = | Ax - b |^2.
          $$
          This is a least squares problem.
          Setting the gradient equal to $0$, we find that $x$ satisfies
          $$
          tag{1} A^T(Ax-b) = 0
          $$
          or equivalently
          $$A^TA x = A^T b.$$
          (This system of equations is often called the "normal equations". Visually, equation (1) just says that the residual vector $b - Ax$ is orthogonal to the column space of $A$.)



          It follows that $x = (A^T A)^{-1} A^T b$. So the projection of $b$ onto $S$ is
          $$
          P(x) = Ax = A(A^T A)^{-1} A^T b.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So it isn't necessarily true that all projection matrices take that form?
            $endgroup$
            – Kid Cudi
            Jan 1 at 13:05














          3












          3








          3





          $begingroup$

          As pointed out by @user1551, this is only true for orthogonal projection matrices.



          Let $P$ be the orthogonal projection operator that projects a vector $b in mathbb R^n$ onto a subspace $S subset mathbb R^n$. Let $(a_1,ldots,a_m)$ be a basis for $S$, and let $A$ be the matrix whose $i$th column is $a_i$. Then $S ={Ax mid x in mathbb R^m}$, and projecting $b$ onto $S$ is equivalent to selecting $x$ so as to minimize the distance from $b$ to $Ax$. Equivalently, we want to minimize
          $$
          r(x) = | Ax - b |^2.
          $$
          This is a least squares problem.
          Setting the gradient equal to $0$, we find that $x$ satisfies
          $$
          tag{1} A^T(Ax-b) = 0
          $$
          or equivalently
          $$A^TA x = A^T b.$$
          (This system of equations is often called the "normal equations". Visually, equation (1) just says that the residual vector $b - Ax$ is orthogonal to the column space of $A$.)



          It follows that $x = (A^T A)^{-1} A^T b$. So the projection of $b$ onto $S$ is
          $$
          P(x) = Ax = A(A^T A)^{-1} A^T b.
          $$






          share|cite|improve this answer











          $endgroup$



          As pointed out by @user1551, this is only true for orthogonal projection matrices.



          Let $P$ be the orthogonal projection operator that projects a vector $b in mathbb R^n$ onto a subspace $S subset mathbb R^n$. Let $(a_1,ldots,a_m)$ be a basis for $S$, and let $A$ be the matrix whose $i$th column is $a_i$. Then $S ={Ax mid x in mathbb R^m}$, and projecting $b$ onto $S$ is equivalent to selecting $x$ so as to minimize the distance from $b$ to $Ax$. Equivalently, we want to minimize
          $$
          r(x) = | Ax - b |^2.
          $$
          This is a least squares problem.
          Setting the gradient equal to $0$, we find that $x$ satisfies
          $$
          tag{1} A^T(Ax-b) = 0
          $$
          or equivalently
          $$A^TA x = A^T b.$$
          (This system of equations is often called the "normal equations". Visually, equation (1) just says that the residual vector $b - Ax$ is orthogonal to the column space of $A$.)



          It follows that $x = (A^T A)^{-1} A^T b$. So the projection of $b$ onto $S$ is
          $$
          P(x) = Ax = A(A^T A)^{-1} A^T b.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 15:22

























          answered Jan 1 at 12:33









          littleOlittleO

          29.9k646109




          29.9k646109












          • $begingroup$
            So it isn't necessarily true that all projection matrices take that form?
            $endgroup$
            – Kid Cudi
            Jan 1 at 13:05


















          • $begingroup$
            So it isn't necessarily true that all projection matrices take that form?
            $endgroup$
            – Kid Cudi
            Jan 1 at 13:05
















          $begingroup$
          So it isn't necessarily true that all projection matrices take that form?
          $endgroup$
          – Kid Cudi
          Jan 1 at 13:05




          $begingroup$
          So it isn't necessarily true that all projection matrices take that form?
          $endgroup$
          – Kid Cudi
          Jan 1 at 13:05


















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          Olmecas

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          Monumento 1, uma das quatro cabeças colossais encontradas em La Venta, com altura aproximada de 3 metros Olmecas é a designação do povo e da civilização que estiveram na origem da antiga cultura pré-colombiana da Mesoamérica e que se desenvolveram nas regiões tropicais do centro-sul do atual México durante o pré-clássico, próximo de onde hoje estão localizados os estados mexicanos de Veracruz e Tabasco, no Istmo de Tehuantepec, numa zona designada área nuclear olmeca. A cultura olmeca floresceu nesta região aproximadamente entre 1500 e 400 a.C., [ 1 ] e crê-se que tenha sido a civilização-mãe de todas as civilizações mesoamericanas que se desenvolveram posteriormente. [ 2 ] No entanto, desconhece-se a sua exacta filiação étnica, ainda que existam numerosas hipóteses colocadas para tentar resolver esta questão. O etnónimo olmeca foi cunhado pelos arqueólogos do século XX, e não devem confundir-se com os muito posteriores olmecas-xicalancas que ocuparam vários locais do México
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