Projection of a vector into the nullspace of a matrix
$begingroup$
I need a clarification about the correct way to compute the projection of a vector into the nullspace of a matrix.
For sake of clarity, let's call $A$ the matrix, $N(A)$ it's kernel and $A^sharp$ it's pseudoinverse.
Let's call $v$ the vector.
It was told me that $(I-A^sharp A)cdot v$ corresponds to the projection of $v$ into the kernel of $A$.
However, given the matrix $A$, I can find a basis for it's kernel. Let's denote with $N$ the matrix which has as columns the vectors of this basis.
Why I can't easily multiply $Ncdot v$ to get the projection of $v$ into the kernel of $A$?
Thanks for the explanation
matrices pseudoinverse projection-matrices
asked Jan 15 at 19:44
sunrisesunrise
6191618
$endgroup$
add a comment |
$begingroup$
I need a clarification about the correct way to compute the projection of a vector into the nullspace of a matrix.
For sake of clarity, let's call $A$ the matrix, $N(A)$ it's kernel and $A^sharp$ it's pseudoinverse.
Let's call $v$ the vector.
It was told me that $(I-A^sharp A)cdot v$ corresponds to the projection of $v$ into the kernel of $A$.
However, given the matrix $A$, I can find a basis for it's kernel. Let's denote with $N$ the matrix which has as columns the vectors of this basis.
Why I can't easily multiply $Ncdot v$ to get the projection of $v$ into the kernel of $A$?
Thanks for the explanation
matrices pseudoinverse projection-matrices
asked Jan 15 at 19:44
sunrisesunrise
6191618
$endgroup$
3
$begingroup$
Ordinarily, projection here should mean orthogonal projection. If you start with $v$ orthogonal to $N(A)$, do you get $0$? More important, every projection $P$ should satisfy $P^2=P$ (so that you stay where you are if you start in the subspace). Is that true with what you're trying to do?
$endgroup$
– Ted Shifrin
Jan 15 at 19:49
$begingroup$
To do something like that you need an orthonormal basis for $N(A)$. See math.stackexchange.com/q/2035114/265466 for an explanation. Also, the basis vectors should be rows of the matrix.
$endgroup$
– amd
Jan 15 at 20:29
$begingroup$
@TedShifrin you're right :) So the idea is to have a matrix in which the columns are the vectors of the 'starting space' basis projected onto the kernel of the matrix... right? In other words, the columns should be the images of the vectors composing the space basis through the map ending in the kernel space.. right? Please, tell me if my question is not clear. A lot of thanks!!
$endgroup$
– sunrise
Jan 21 at 12:17
add a comment |
$begingroup$
I need a clarification about the correct way to compute the projection of a vector into the nullspace of a matrix.
For sake of clarity, let's call $A$ the matrix, $N(A)$ it's kernel and $A^sharp$ it's pseudoinverse.
Let's call $v$ the vector.
It was told me that $(I-A^sharp A)cdot v$ corresponds to the projection of $v$ into the kernel of $A$.
However, given the matrix $A$, I can find a basis for it's kernel. Let's denote with $N$ the matrix which has as columns the vectors of this basis.
Why I can't easily multiply $Ncdot v$ to get the projection of $v$ into the kernel of $A$?
Thanks for the explanation
matrices pseudoinverse projection-matrices
asked Jan 15 at 19:44
sunrisesunrise
6191618
$endgroup$
I need a clarification about the correct way to compute the projection of a vector into the nullspace of a matrix.
For sake of clarity, let's call $A$ the matrix, $N(A)$ it's kernel and $A^sharp$ it's pseudoinverse.
Let's call $v$ the vector.
It was told me that $(I-A^sharp A)cdot v$ corresponds to the projection of $v$ into the kernel of $A$.
However, given the matrix $A$, I can find a basis for it's kernel. Let's denote with $N$ the matrix which has as columns the vectors of this basis.
Why I can't easily multiply $Ncdot v$ to get the projection of $v$ into the kernel of $A$?
Thanks for the explanation
matrices pseudoinverse projection-matrices
matrices pseudoinverse projection-matrices
asked Jan 15 at 19:44
sunrisesunrise
6191618
asked Jan 15 at 19:44
sunrisesunrise
6191618
asked Jan 15 at 19:44
sunrisesunrise
6191618
asked Jan 15 at 19:44
sunrisesunrise
6191618
asked Jan 15 at 19:44
sunrisesunrise
6191618
6191618
3
$begingroup$
Ordinarily, projection here should mean orthogonal projection. If you start with $v$ orthogonal to $N(A)$, do you get $0$? More important, every projection $P$ should satisfy $P^2=P$ (so that you stay where you are if you start in the subspace). Is that true with what you're trying to do?
$endgroup$
– Ted Shifrin
Jan 15 at 19:49
$begingroup$
To do something like that you need an orthonormal basis for $N(A)$. See math.stackexchange.com/q/2035114/265466 for an explanation. Also, the basis vectors should be rows of the matrix.
$endgroup$
– amd
Jan 15 at 20:29
$begingroup$
@TedShifrin you're right :) So the idea is to have a matrix in which the columns are the vectors of the 'starting space' basis projected onto the kernel of the matrix... right? In other words, the columns should be the images of the vectors composing the space basis through the map ending in the kernel space.. right? Please, tell me if my question is not clear. A lot of thanks!!
$endgroup$
– sunrise
Jan 21 at 12:17
add a comment |
3
$begingroup$
Ordinarily, projection here should mean orthogonal projection. If you start with $v$ orthogonal to $N(A)$, do you get $0$? More important, every projection $P$ should satisfy $P^2=P$ (so that you stay where you are if you start in the subspace). Is that true with what you're trying to do?
$endgroup$
– Ted Shifrin
Jan 15 at 19:49
$begingroup$
To do something like that you need an orthonormal basis for $N(A)$. See math.stackexchange.com/q/2035114/265466 for an explanation. Also, the basis vectors should be rows of the matrix.
$endgroup$
– amd
Jan 15 at 20:29
$begingroup$
@TedShifrin you're right :) So the idea is to have a matrix in which the columns are the vectors of the 'starting space' basis projected onto the kernel of the matrix... right? In other words, the columns should be the images of the vectors composing the space basis through the map ending in the kernel space.. right? Please, tell me if my question is not clear. A lot of thanks!!
$endgroup$
– sunrise
Jan 21 at 12:17
3
3
$begingroup$
Ordinarily, projection here should mean orthogonal projection. If you start with $v$ orthogonal to $N(A)$, do you get $0$? More important, every projection $P$ should satisfy $P^2=P$ (so that you stay where you are if you start in the subspace). Is that true with what you're trying to do?
$endgroup$
– Ted Shifrin
Jan 15 at 19:49
$begingroup$
Ordinarily, projection here should mean orthogonal projection. If you start with $v$ orthogonal to $N(A)$, do you get $0$? More important, every projection $P$ should satisfy $P^2=P$ (so that you stay where you are if you start in the subspace). Is that true with what you're trying to do?
$endgroup$
– Ted Shifrin
Jan 15 at 19:49
$begingroup$
To do something like that you need an orthonormal basis for $N(A)$. See math.stackexchange.com/q/2035114/265466 for an explanation. Also, the basis vectors should be rows of the matrix.
$endgroup$
– amd
Jan 15 at 20:29
$begingroup$
To do something like that you need an orthonormal basis for $N(A)$. See math.stackexchange.com/q/2035114/265466 for an explanation. Also, the basis vectors should be rows of the matrix.
$endgroup$
– amd
Jan 15 at 20:29
$begingroup$
@TedShifrin you're right :) So the idea is to have a matrix in which the columns are the vectors of the 'starting space' basis projected onto the kernel of the matrix... right? In other words, the columns should be the images of the vectors composing the space basis through the map ending in the kernel space.. right? Please, tell me if my question is not clear. A lot of thanks!!
$endgroup$
– sunrise
Jan 21 at 12:17
$begingroup$
@TedShifrin you're right :) So the idea is to have a matrix in which the columns are the vectors of the 'starting space' basis projected onto the kernel of the matrix... right? In other words, the columns should be the images of the vectors composing the space basis through the map ending in the kernel space.. right? Please, tell me if my question is not clear. A lot of thanks!!
$endgroup$
– sunrise
Jan 21 at 12:17
add a comment |
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3
$begingroup$
Ordinarily, projection here should mean orthogonal projection. If you start with $v$ orthogonal to $N(A)$, do you get $0$? More important, every projection $P$ should satisfy $P^2=P$ (so that you stay where you are if you start in the subspace). Is that true with what you're trying to do?
$endgroup$
– Ted Shifrin
Jan 15 at 19:49
$begingroup$
To do something like that you need an orthonormal basis for $N(A)$. See math.stackexchange.com/q/2035114/265466 for an explanation. Also, the basis vectors should be rows of the matrix.
$endgroup$
– amd
Jan 15 at 20:29
$begingroup$
@TedShifrin you're right :) So the idea is to have a matrix in which the columns are the vectors of the 'starting space' basis projected onto the kernel of the matrix... right? In other words, the columns should be the images of the vectors composing the space basis through the map ending in the kernel space.. right? Please, tell me if my question is not clear. A lot of thanks!!
$endgroup$
– sunrise
Jan 21 at 12:17