Why there are n-r dimensional set of vectors x for rank-deficient least square problem
I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem
Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$
What's the reason behind the "$n-r$" theorem?
linear-algebra optimization numerical-linear-algebra least-squares
asked Nov 22 '18 at 3:04
WeiShan Ng
617
add a comment |
I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem
Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$
What's the reason behind the "$n-r$" theorem?
linear-algebra optimization numerical-linear-algebra least-squares
asked Nov 22 '18 at 3:04
WeiShan Ng
617
$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47
@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50
Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55
add a comment |
I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem
Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$
What's the reason behind the "$n-r$" theorem?
linear-algebra optimization numerical-linear-algebra least-squares
asked Nov 22 '18 at 3:04
WeiShan Ng
617
I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem
Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$
What's the reason behind the "$n-r$" theorem?
linear-algebra optimization numerical-linear-algebra least-squares
linear-algebra optimization numerical-linear-algebra least-squares
asked Nov 22 '18 at 3:04
WeiShan Ng
617
asked Nov 22 '18 at 3:04
WeiShan Ng
617
asked Nov 22 '18 at 3:04
WeiShan Ng
617
asked Nov 22 '18 at 3:04
WeiShan Ng
617
asked Nov 22 '18 at 3:04
WeiShan Ng
617
617
$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47
@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50
Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55
add a comment |
$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47
@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50
Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55
$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47
$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47
@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50
@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50
Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55
Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55
add a comment |
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I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
$$A^T A vec{x} = A^T vec{b}$$
And we also know that
$R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$
So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$
Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$
And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
$$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$
which is of $n-r$ dimension
answered Nov 24 '18 at 5:13
WeiShan Ng
617
add a comment |
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1 Answer
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I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
$$A^T A vec{x} = A^T vec{b}$$
And we also know that
$R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$
So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$
Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$
And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
$$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$
which is of $n-r$ dimension
answered Nov 24 '18 at 5:13
WeiShan Ng
617
add a comment |
I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
$$A^T A vec{x} = A^T vec{b}$$
And we also know that
$R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$
So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$
Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$
And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
$$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$
which is of $n-r$ dimension
answered Nov 24 '18 at 5:13
WeiShan Ng
617
add a comment |
I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
$$A^T A vec{x} = A^T vec{b}$$
And we also know that
$R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$
So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$
Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$
And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
$$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$
which is of $n-r$ dimension
answered Nov 24 '18 at 5:13
WeiShan Ng
617
I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
$$A^T A vec{x} = A^T vec{b}$$
And we also know that
$R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$
So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$
Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$
And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
$$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$
which is of $n-r$ dimension
answered Nov 24 '18 at 5:13
WeiShan Ng
617
answered Nov 24 '18 at 5:13
WeiShan Ng
617
answered Nov 24 '18 at 5:13
WeiShan Ng
617
answered Nov 24 '18 at 5:13
WeiShan Ng
617
617
add a comment |
add a comment |
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$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47
@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50
Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55