Least Squares Solution of Minimal Norm when $A^{*}b = 0$
$begingroup$
Suppose, given a matrix $textbf{A} in mathbb{C}^{m times n}$ and a vector $textbf{b} in mathbb{C}^{n}$, I want to find the minimal norm solution of
$$min_{textbf{x}}|textbf{A}textbf{x} - textbf{b} |_{2} $$
with the condition that $textbf{A}^{*}textbf{b} = textbf{0}_{n}$.
If $rank(textbf{A}) =n$, then the solution is
$$textbf{y} = (textbf{A}^{*}textbf{A})^{-1}textbf{A}^{*}textbf{b}.$$
This would give $textbf{y} = textbf{0}$ with the stated condition. However, for $textbf{A}$ of arbitrary rank, the the minimal norm solution for this problem is
$$textbf{y} = textbf{A}^{dagger}textbf{b}$$
Where $textbf{A}^{dagger}$ is the pseudoinverse.
Can we say anything about $textbf{y}$ with stated condition in the general case, or is it arbitrary
linear-algebra matrices numerical-linear-algebra least-squares constraints
asked Nov 28 '18 at 12:16
SimonPSimonP
434
$endgroup$
add a comment |
$begingroup$
Suppose, given a matrix $textbf{A} in mathbb{C}^{m times n}$ and a vector $textbf{b} in mathbb{C}^{n}$, I want to find the minimal norm solution of
$$min_{textbf{x}}|textbf{A}textbf{x} - textbf{b} |_{2} $$
with the condition that $textbf{A}^{*}textbf{b} = textbf{0}_{n}$.
If $rank(textbf{A}) =n$, then the solution is
$$textbf{y} = (textbf{A}^{*}textbf{A})^{-1}textbf{A}^{*}textbf{b}.$$
This would give $textbf{y} = textbf{0}$ with the stated condition. However, for $textbf{A}$ of arbitrary rank, the the minimal norm solution for this problem is
$$textbf{y} = textbf{A}^{dagger}textbf{b}$$
Where $textbf{A}^{dagger}$ is the pseudoinverse.
Can we say anything about $textbf{y}$ with stated condition in the general case, or is it arbitrary
linear-algebra matrices numerical-linear-algebra least-squares constraints
asked Nov 28 '18 at 12:16
SimonPSimonP
434
$endgroup$
add a comment |
$begingroup$
Suppose, given a matrix $textbf{A} in mathbb{C}^{m times n}$ and a vector $textbf{b} in mathbb{C}^{n}$, I want to find the minimal norm solution of
$$min_{textbf{x}}|textbf{A}textbf{x} - textbf{b} |_{2} $$
with the condition that $textbf{A}^{*}textbf{b} = textbf{0}_{n}$.
If $rank(textbf{A}) =n$, then the solution is
$$textbf{y} = (textbf{A}^{*}textbf{A})^{-1}textbf{A}^{*}textbf{b}.$$
This would give $textbf{y} = textbf{0}$ with the stated condition. However, for $textbf{A}$ of arbitrary rank, the the minimal norm solution for this problem is
$$textbf{y} = textbf{A}^{dagger}textbf{b}$$
Where $textbf{A}^{dagger}$ is the pseudoinverse.
Can we say anything about $textbf{y}$ with stated condition in the general case, or is it arbitrary
linear-algebra matrices numerical-linear-algebra least-squares constraints
asked Nov 28 '18 at 12:16
SimonPSimonP
434
$endgroup$
Suppose, given a matrix $textbf{A} in mathbb{C}^{m times n}$ and a vector $textbf{b} in mathbb{C}^{n}$, I want to find the minimal norm solution of
$$min_{textbf{x}}|textbf{A}textbf{x} - textbf{b} |_{2} $$
with the condition that $textbf{A}^{*}textbf{b} = textbf{0}_{n}$.
If $rank(textbf{A}) =n$, then the solution is
$$textbf{y} = (textbf{A}^{*}textbf{A})^{-1}textbf{A}^{*}textbf{b}.$$
This would give $textbf{y} = textbf{0}$ with the stated condition. However, for $textbf{A}$ of arbitrary rank, the the minimal norm solution for this problem is
$$textbf{y} = textbf{A}^{dagger}textbf{b}$$
Where $textbf{A}^{dagger}$ is the pseudoinverse.
Can we say anything about $textbf{y}$ with stated condition in the general case, or is it arbitrary
linear-algebra matrices numerical-linear-algebra least-squares constraints
linear-algebra matrices numerical-linear-algebra least-squares constraints
asked Nov 28 '18 at 12:16
SimonPSimonP
434
asked Nov 28 '18 at 12:16
SimonPSimonP
434
asked Nov 28 '18 at 12:16
SimonPSimonP
434
asked Nov 28 '18 at 12:16
SimonPSimonP
434
asked Nov 28 '18 at 12:16
SimonPSimonP
434
434
add a comment |
add a comment |
1 Answer
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$begingroup$
It is known that a vector $hat {mathbf x}$ is a least squares solution of the system
$$
A {mathbf x}= {mathbf b}
$$ iff $$A^*Ahat {mathbf x}=A^* {mathbf b}.$$
Thus, if the condition $A^* {mathbf b}={mathbf 0}$ is satisfied, then $hat {mathbf x}= {mathbf 0}$ is a least squares solution. Obviously, it has the minimum possible norm (zero).
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
It is known that a vector $hat {mathbf x}$ is a least squares solution of the system
$$
A {mathbf x}= {mathbf b}
$$ iff $$A^*Ahat {mathbf x}=A^* {mathbf b}.$$
Thus, if the condition $A^* {mathbf b}={mathbf 0}$ is satisfied, then $hat {mathbf x}= {mathbf 0}$ is a least squares solution. Obviously, it has the minimum possible norm (zero).
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
$endgroup$
add a comment |
$begingroup$
It is known that a vector $hat {mathbf x}$ is a least squares solution of the system
$$
A {mathbf x}= {mathbf b}
$$ iff $$A^*Ahat {mathbf x}=A^* {mathbf b}.$$
Thus, if the condition $A^* {mathbf b}={mathbf 0}$ is satisfied, then $hat {mathbf x}= {mathbf 0}$ is a least squares solution. Obviously, it has the minimum possible norm (zero).
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
$endgroup$
add a comment |
$begingroup$
It is known that a vector $hat {mathbf x}$ is a least squares solution of the system
$$
A {mathbf x}= {mathbf b}
$$ iff $$A^*Ahat {mathbf x}=A^* {mathbf b}.$$
Thus, if the condition $A^* {mathbf b}={mathbf 0}$ is satisfied, then $hat {mathbf x}= {mathbf 0}$ is a least squares solution. Obviously, it has the minimum possible norm (zero).
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
$endgroup$
It is known that a vector $hat {mathbf x}$ is a least squares solution of the system
$$
A {mathbf x}= {mathbf b}
$$ iff $$A^*Ahat {mathbf x}=A^* {mathbf b}.$$
Thus, if the condition $A^* {mathbf b}={mathbf 0}$ is satisfied, then $hat {mathbf x}= {mathbf 0}$ is a least squares solution. Obviously, it has the minimum possible norm (zero).
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
answered Nov 28 '18 at 15:14
AVKAVK
2,0961517
2,0961517
add a comment |
add a comment |
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