Do all projections matrices take this form?












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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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    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

























      2












      2








      2







      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













      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






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      asked 4 hours ago









      Kid Cudi

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          $A(A^TA)^{-1}A^T$ 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 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























          • Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            – littleO
            38 mins ago



















          1














          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























          • So it isn't necessarily true that all projection matrices take that form?
            – Kid Cudi
            2 hours ago











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          2 Answers
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          oldest

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          2 Answers
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          active

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          4














          $A(A^TA)^{-1}A^T$ 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 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























          • Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            – littleO
            38 mins ago
















          4














          $A(A^TA)^{-1}A^T$ 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 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























          • Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            – littleO
            38 mins ago














          4












          4








          4






          $A(A^TA)^{-1}A^T$ 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 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














          $A(A^TA)^{-1}A^T$ 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 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 9 mins ago

























          answered 2 hours ago









          user1551

          71.4k566125




          71.4k566125












          • Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            – littleO
            38 mins ago


















          • Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
            – littleO
            38 mins ago
















          Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
          – littleO
          38 mins ago




          Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
          – littleO
          38 mins ago











          1














          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























          • So it isn't necessarily true that all projection matrices take that form?
            – Kid Cudi
            2 hours ago
















          1














          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























          • So it isn't necessarily true that all projection matrices take that form?
            – Kid Cudi
            2 hours ago














          1












          1








          1






          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














          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 40 mins ago

























          answered 3 hours ago









          littleO

          29.1k644108




          29.1k644108












          • So it isn't necessarily true that all projection matrices take that form?
            – Kid Cudi
            2 hours ago


















          • So it isn't necessarily true that all projection matrices take that form?
            – Kid Cudi
            2 hours ago
















          So it isn't necessarily true that all projection matrices take that form?
          – Kid Cudi
          2 hours ago




          So it isn't necessarily true that all projection matrices take that form?
          – Kid Cudi
          2 hours ago


















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