Do all projections matrices take this form?
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
asked Jan 1 at 11:58
Kid Cudi
456
add a comment |
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
asked Jan 1 at 11:58
Kid Cudi
456
add a comment |
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
asked Jan 1 at 11:58
Kid Cudi
456
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
asked Jan 1 at 11:58
Kid Cudi
456
asked Jan 1 at 11:58
Kid Cudi
456
asked Jan 1 at 11:58
Kid Cudi
456
asked Jan 1 at 11:58
Kid Cudi
456
asked Jan 1 at 11:58
Kid Cudi
456
456
add a comment |
add a comment |
2 Answers
2
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oldest
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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 (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. $P=A(A^TA)^{-1}A^T$.
answered Jan 1 at 13:46
user1551
71.6k566125
1
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
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.
– amd
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
add a comment |
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
$$
P(x) = Ax = A(A^T A)^{-1} A^T b.
$$
answered Jan 1 at 12:33
littleO
29.2k644109
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
add a comment |
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2 Answers
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2 Answers
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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 (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. $P=A(A^TA)^{-1}A^T$.
answered Jan 1 at 13:46
user1551
71.6k566125
1
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
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.
– amd
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
add a comment |
$A(A^TA)^{-1}A^T$ 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. $P=A(A^TA)^{-1}A^T$.
answered Jan 1 at 13:46
user1551
71.6k566125
1
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
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.
– amd
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
add a comment |
$A(A^TA)^{-1}A^T$ 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. $P=A(A^TA)^{-1}A^T$.
answered Jan 1 at 13:46
user1551
71.6k566125
$A(A^TA)^{-1}A^T$ 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. $P=A(A^TA)^{-1}A^T$.
answered Jan 1 at 13:46
user1551
71.6k566125
edited 2 days ago
answered Jan 1 at 13:46
user1551
71.6k566125
answered Jan 1 at 13:46
user1551
71.6k566125
answered Jan 1 at 13:46
user1551
71.6k566125
71.6k566125
1
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
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.
– amd
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
add a comment |
1
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
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.
– amd
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
1
1
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
Thanks, I learned something from your answer -- I had forgotten that a projection matrix does not have to be an orthogonal projection matrix.
– littleO
Jan 1 at 15:24
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.
– amd
2 days ago
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.
– amd
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
@amd I am. This is the usual setting when this kind of projections are encountered.
– user1551
2 days ago
add a comment |
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
$$
P(x) = Ax = A(A^T A)^{-1} A^T b.
$$
answered Jan 1 at 12:33
littleO
29.2k644109
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
add a comment |
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
$$
P(x) = Ax = A(A^T A)^{-1} A^T b.
$$
answered Jan 1 at 12:33
littleO
29.2k644109
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
add a comment |
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
$$
P(x) = Ax = A(A^T A)^{-1} A^T b.
$$
answered Jan 1 at 12:33
littleO
29.2k644109
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
$$
P(x) = Ax = A(A^T A)^{-1} A^T b.
$$
answered Jan 1 at 12:33
littleO
29.2k644109
edited Jan 1 at 15:22
answered Jan 1 at 12:33
littleO
29.2k644109
answered Jan 1 at 12:33
littleO
29.2k644109
answered Jan 1 at 12:33
littleO
29.2k644109
29.2k644109
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
add a comment |
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
So it isn't necessarily true that all projection matrices take that form?
– Kid Cudi
Jan 1 at 13:05
add a comment |
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