Show that the solution space of $Ax=b$ varies continously with $b$, $A$ is $n times (n+1)$
$begingroup$
Suppose $A$ is a $n times (n+1)$ matrix of rank $n$.
(a) Show that the one dimensional solution space of $A x =b$ varies continuously with $b in mathbb{R}^n$.
(b) Generalize
My attempt:
(a) The nullspace of $A$ has dimension 1.
If I think of the RREF of $A$ obtained by premultiplication with an $n times n$ matrix $E$, then the solution to $Ax= b$ is
$x = begin{pmatrix} Eb \ 0 \ end{pmatrix} + alpha begin{pmatrix} 0 \ 1 end{pmatrix}$
where the second term on the right is a general vector in the nullspace of $A$.
Thus for RHS $b_1, b_2$, (now denoting the nullspace basis vector by $e_n$
$||x_1 - x_2|| leq ||E|| cdot||b_1-b_2|| + |gamma| cdot ||e_n||, forall ; gamma ;in mathbb{R} $.
This would prove the claim if $gamma=0$, but as it stands, I do not see if this is correct. How do I show the claim ?
(b) I can think of
Rank of A = n, but dimension of nullspace > 1.
Rank of A < n. In this case it is not clear to me how to even begin, since there may be no solution for a given $b in mathbb{R}^n$
So I am puzzled by the point of this question.
linear-algebra vector-spaces linear-transformations systems-of-equations
asked Jan 12 at 23:14
me10240me10240
450212
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a $n times (n+1)$ matrix of rank $n$.
(a) Show that the one dimensional solution space of $A x =b$ varies continuously with $b in mathbb{R}^n$.
(b) Generalize
My attempt:
(a) The nullspace of $A$ has dimension 1.
If I think of the RREF of $A$ obtained by premultiplication with an $n times n$ matrix $E$, then the solution to $Ax= b$ is
$x = begin{pmatrix} Eb \ 0 \ end{pmatrix} + alpha begin{pmatrix} 0 \ 1 end{pmatrix}$
where the second term on the right is a general vector in the nullspace of $A$.
Thus for RHS $b_1, b_2$, (now denoting the nullspace basis vector by $e_n$
$||x_1 - x_2|| leq ||E|| cdot||b_1-b_2|| + |gamma| cdot ||e_n||, forall ; gamma ;in mathbb{R} $.
This would prove the claim if $gamma=0$, but as it stands, I do not see if this is correct. How do I show the claim ?
(b) I can think of
Rank of A = n, but dimension of nullspace > 1.
Rank of A < n. In this case it is not clear to me how to even begin, since there may be no solution for a given $b in mathbb{R}^n$
So I am puzzled by the point of this question.
linear-algebra vector-spaces linear-transformations systems-of-equations
asked Jan 12 at 23:14
me10240me10240
450212
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a $n times (n+1)$ matrix of rank $n$.
(a) Show that the one dimensional solution space of $A x =b$ varies continuously with $b in mathbb{R}^n$.
(b) Generalize
My attempt:
(a) The nullspace of $A$ has dimension 1.
If I think of the RREF of $A$ obtained by premultiplication with an $n times n$ matrix $E$, then the solution to $Ax= b$ is
$x = begin{pmatrix} Eb \ 0 \ end{pmatrix} + alpha begin{pmatrix} 0 \ 1 end{pmatrix}$
where the second term on the right is a general vector in the nullspace of $A$.
Thus for RHS $b_1, b_2$, (now denoting the nullspace basis vector by $e_n$
$||x_1 - x_2|| leq ||E|| cdot||b_1-b_2|| + |gamma| cdot ||e_n||, forall ; gamma ;in mathbb{R} $.
This would prove the claim if $gamma=0$, but as it stands, I do not see if this is correct. How do I show the claim ?
(b) I can think of
Rank of A = n, but dimension of nullspace > 1.
Rank of A < n. In this case it is not clear to me how to even begin, since there may be no solution for a given $b in mathbb{R}^n$
So I am puzzled by the point of this question.
linear-algebra vector-spaces linear-transformations systems-of-equations
asked Jan 12 at 23:14
me10240me10240
450212
$endgroup$
Suppose $A$ is a $n times (n+1)$ matrix of rank $n$.
(a) Show that the one dimensional solution space of $A x =b$ varies continuously with $b in mathbb{R}^n$.
(b) Generalize
My attempt:
(a) The nullspace of $A$ has dimension 1.
If I think of the RREF of $A$ obtained by premultiplication with an $n times n$ matrix $E$, then the solution to $Ax= b$ is
$x = begin{pmatrix} Eb \ 0 \ end{pmatrix} + alpha begin{pmatrix} 0 \ 1 end{pmatrix}$
where the second term on the right is a general vector in the nullspace of $A$.
Thus for RHS $b_1, b_2$, (now denoting the nullspace basis vector by $e_n$
$||x_1 - x_2|| leq ||E|| cdot||b_1-b_2|| + |gamma| cdot ||e_n||, forall ; gamma ;in mathbb{R} $.
This would prove the claim if $gamma=0$, but as it stands, I do not see if this is correct. How do I show the claim ?
(b) I can think of
Rank of A = n, but dimension of nullspace > 1.
Rank of A < n. In this case it is not clear to me how to even begin, since there may be no solution for a given $b in mathbb{R}^n$
So I am puzzled by the point of this question.
linear-algebra vector-spaces linear-transformations systems-of-equations
linear-algebra vector-spaces linear-transformations systems-of-equations
asked Jan 12 at 23:14
me10240me10240
450212
asked Jan 12 at 23:14
me10240me10240
450212
asked Jan 12 at 23:14
me10240me10240
450212
asked Jan 12 at 23:14
me10240me10240
450212
asked Jan 12 at 23:14
me10240me10240
450212
450212
add a comment |
add a comment |
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