Let $A$ be an $8 times 5$ matrix of rank 3, and let $b$ be a nonzero vector in $N(A^T)$. Show $Ax=b$ must be...
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Here's the entire question: Let $A$ be an 8 $times$ 5 matrix of rank 3, and let $b$ be a nonzero vector in $N(A^T)$.
a) Show that the system $Ax = b$ must be inconsistent.
Gonna take a wild stab at this one... If the rank is 3, that means the dimension of the column space is 3. But $A$ has 5 columns, so they are not all linearly independent and therefore $Ax = b$ is inconsistent.
b) How many least squares solutions will the system $Ax = b$ have? Explain.
On previous problems, I found the best least squares linear fit, where the approximation of $x$ was a vector that contained sometimes regular numbers, and sometimes variables. Does this mean that there must be either 1 linear solution or infinite (because you can always find an approximation)? In the example that apparently had an infinite number of least squares solutions, it appeared that one row of $A^TA$ was a constant multiple of another row, leading to a row of zeros in reduced row echelon form. From this problem I know that $A^TA$ is a 5x5 matrix, but I don't think I can prove that any rows are a scalar multiple of other rows, so I'm guessing I have to use some other means of figuring this out.
Sorry if I sound like I have no idea what I'm talking about. Just wanted to try out the problem to my best ability before asking about it.
linear-algebra matrices matrix-rank least-squares transpose
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
asked Nov 11 '15 at 6:40
Chris
3521213
|
show 2 more comments
up vote
2
down vote
favorite
Here's the entire question: Let $A$ be an 8 $times$ 5 matrix of rank 3, and let $b$ be a nonzero vector in $N(A^T)$.
a) Show that the system $Ax = b$ must be inconsistent.
Gonna take a wild stab at this one... If the rank is 3, that means the dimension of the column space is 3. But $A$ has 5 columns, so they are not all linearly independent and therefore $Ax = b$ is inconsistent.
b) How many least squares solutions will the system $Ax = b$ have? Explain.
On previous problems, I found the best least squares linear fit, where the approximation of $x$ was a vector that contained sometimes regular numbers, and sometimes variables. Does this mean that there must be either 1 linear solution or infinite (because you can always find an approximation)? In the example that apparently had an infinite number of least squares solutions, it appeared that one row of $A^TA$ was a constant multiple of another row, leading to a row of zeros in reduced row echelon form. From this problem I know that $A^TA$ is a 5x5 matrix, but I don't think I can prove that any rows are a scalar multiple of other rows, so I'm guessing I have to use some other means of figuring this out.
Sorry if I sound like I have no idea what I'm talking about. Just wanted to try out the problem to my best ability before asking about it.
linear-algebra matrices matrix-rank least-squares transpose
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
asked Nov 11 '15 at 6:40
Chris
3521213
For (a), that's not good enough, because your argument doesn't mention what $b$ is. You need to use the fact that the column space of $A$ is the set of all vectors $c$ such that $Ax=c$ is consistent. ... The vector $b$ being in the null space of $A^top$ means $b$ is perpendicular to every vector in the column space (why?); hence, $b$ can't be in the column space, unless $b$ is zero (why?).
– Christopher Carl Heckman
Nov 11 '15 at 7:11
For (b), use the fact that $b$ is in the null space of $A^top$ means $A^top b = 0$, and use the fact that the normal equation is $(A^top A)x = (A^top b)$.
– Christopher Carl Heckman
Nov 11 '15 at 7:13
So $R(A)$ = $N(A^T)^⊥$ (just found that formula in my book), and because b is in the latter, that means it must be perpendicular to R(A). Though I can't find anything about b having to be 0 because of this. I'll keep looking, though.
– Chris
Nov 11 '15 at 7:37
The book says that $Ax = b$ is consistent if and only if $b$ is an element of $R(A)$. But is that not possible because $b$ doesn't contain the 0 vector? I remember doing that when checking to see if something is a subspace.
– Chris
Nov 11 '15 at 7:49
And for the part $b$, I said $A^Tb$ = $0$, meaning $(A^TA)x$ = $0$, and because the rank of $A$ is equal to the rank of $A^TA$, that means $A^TA$ has a rank of 3 and therefore has a nullity of 5, which means there are 5 free variables in $A^Tb$ = $0$ and that means $Ax=b$ has infinitely many least squares solutions. Is that right?
– Chris
Nov 11 '15 at 8:06
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Here's the entire question: Let $A$ be an 8 $times$ 5 matrix of rank 3, and let $b$ be a nonzero vector in $N(A^T)$.
a) Show that the system $Ax = b$ must be inconsistent.
Gonna take a wild stab at this one... If the rank is 3, that means the dimension of the column space is 3. But $A$ has 5 columns, so they are not all linearly independent and therefore $Ax = b$ is inconsistent.
b) How many least squares solutions will the system $Ax = b$ have? Explain.
On previous problems, I found the best least squares linear fit, where the approximation of $x$ was a vector that contained sometimes regular numbers, and sometimes variables. Does this mean that there must be either 1 linear solution or infinite (because you can always find an approximation)? In the example that apparently had an infinite number of least squares solutions, it appeared that one row of $A^TA$ was a constant multiple of another row, leading to a row of zeros in reduced row echelon form. From this problem I know that $A^TA$ is a 5x5 matrix, but I don't think I can prove that any rows are a scalar multiple of other rows, so I'm guessing I have to use some other means of figuring this out.
Sorry if I sound like I have no idea what I'm talking about. Just wanted to try out the problem to my best ability before asking about it.
linear-algebra matrices matrix-rank least-squares transpose
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
asked Nov 11 '15 at 6:40
Chris
3521213
Here's the entire question: Let $A$ be an 8 $times$ 5 matrix of rank 3, and let $b$ be a nonzero vector in $N(A^T)$.
a) Show that the system $Ax = b$ must be inconsistent.
Gonna take a wild stab at this one... If the rank is 3, that means the dimension of the column space is 3. But $A$ has 5 columns, so they are not all linearly independent and therefore $Ax = b$ is inconsistent.
b) How many least squares solutions will the system $Ax = b$ have? Explain.
On previous problems, I found the best least squares linear fit, where the approximation of $x$ was a vector that contained sometimes regular numbers, and sometimes variables. Does this mean that there must be either 1 linear solution or infinite (because you can always find an approximation)? In the example that apparently had an infinite number of least squares solutions, it appeared that one row of $A^TA$ was a constant multiple of another row, leading to a row of zeros in reduced row echelon form. From this problem I know that $A^TA$ is a 5x5 matrix, but I don't think I can prove that any rows are a scalar multiple of other rows, so I'm guessing I have to use some other means of figuring this out.
Sorry if I sound like I have no idea what I'm talking about. Just wanted to try out the problem to my best ability before asking about it.
linear-algebra matrices matrix-rank least-squares transpose
linear-algebra matrices matrix-rank least-squares transpose
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
asked Nov 11 '15 at 6:40
Chris
3521213
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
asked Nov 11 '15 at 6:40
Chris
3521213
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
edited Nov 22 at 18:04
Lorenzo B.
1,8322520
1,8322520
asked Nov 11 '15 at 6:40
Chris
3521213
asked Nov 11 '15 at 6:40
Chris
3521213
asked Nov 11 '15 at 6:40
Chris
3521213
3521213
For (a), that's not good enough, because your argument doesn't mention what $b$ is. You need to use the fact that the column space of $A$ is the set of all vectors $c$ such that $Ax=c$ is consistent. ... The vector $b$ being in the null space of $A^top$ means $b$ is perpendicular to every vector in the column space (why?); hence, $b$ can't be in the column space, unless $b$ is zero (why?).
– Christopher Carl Heckman
Nov 11 '15 at 7:11
For (b), use the fact that $b$ is in the null space of $A^top$ means $A^top b = 0$, and use the fact that the normal equation is $(A^top A)x = (A^top b)$.
– Christopher Carl Heckman
Nov 11 '15 at 7:13
So $R(A)$ = $N(A^T)^⊥$ (just found that formula in my book), and because b is in the latter, that means it must be perpendicular to R(A). Though I can't find anything about b having to be 0 because of this. I'll keep looking, though.
– Chris
Nov 11 '15 at 7:37
The book says that $Ax = b$ is consistent if and only if $b$ is an element of $R(A)$. But is that not possible because $b$ doesn't contain the 0 vector? I remember doing that when checking to see if something is a subspace.
– Chris
Nov 11 '15 at 7:49
And for the part $b$, I said $A^Tb$ = $0$, meaning $(A^TA)x$ = $0$, and because the rank of $A$ is equal to the rank of $A^TA$, that means $A^TA$ has a rank of 3 and therefore has a nullity of 5, which means there are 5 free variables in $A^Tb$ = $0$ and that means $Ax=b$ has infinitely many least squares solutions. Is that right?
– Chris
Nov 11 '15 at 8:06
|
show 2 more comments
For (a), that's not good enough, because your argument doesn't mention what $b$ is. You need to use the fact that the column space of $A$ is the set of all vectors $c$ such that $Ax=c$ is consistent. ... The vector $b$ being in the null space of $A^top$ means $b$ is perpendicular to every vector in the column space (why?); hence, $b$ can't be in the column space, unless $b$ is zero (why?).
– Christopher Carl Heckman
Nov 11 '15 at 7:11
For (b), use the fact that $b$ is in the null space of $A^top$ means $A^top b = 0$, and use the fact that the normal equation is $(A^top A)x = (A^top b)$.
– Christopher Carl Heckman
Nov 11 '15 at 7:13
So $R(A)$ = $N(A^T)^⊥$ (just found that formula in my book), and because b is in the latter, that means it must be perpendicular to R(A). Though I can't find anything about b having to be 0 because of this. I'll keep looking, though.
– Chris
Nov 11 '15 at 7:37
The book says that $Ax = b$ is consistent if and only if $b$ is an element of $R(A)$. But is that not possible because $b$ doesn't contain the 0 vector? I remember doing that when checking to see if something is a subspace.
– Chris
Nov 11 '15 at 7:49
And for the part $b$, I said $A^Tb$ = $0$, meaning $(A^TA)x$ = $0$, and because the rank of $A$ is equal to the rank of $A^TA$, that means $A^TA$ has a rank of 3 and therefore has a nullity of 5, which means there are 5 free variables in $A^Tb$ = $0$ and that means $Ax=b$ has infinitely many least squares solutions. Is that right?
– Chris
Nov 11 '15 at 8:06
For (a), that's not good enough, because your argument doesn't mention what $b$ is. You need to use the fact that the column space of $A$ is the set of all vectors $c$ such that $Ax=c$ is consistent. ... The vector $b$ being in the null space of $A^top$ means $b$ is perpendicular to every vector in the column space (why?); hence, $b$ can't be in the column space, unless $b$ is zero (why?).
– Christopher Carl Heckman
Nov 11 '15 at 7:11
For (a), that's not good enough, because your argument doesn't mention what $b$ is. You need to use the fact that the column space of $A$ is the set of all vectors $c$ such that $Ax=c$ is consistent. ... The vector $b$ being in the null space of $A^top$ means $b$ is perpendicular to every vector in the column space (why?); hence, $b$ can't be in the column space, unless $b$ is zero (why?).
– Christopher Carl Heckman
Nov 11 '15 at 7:11
For (b), use the fact that $b$ is in the null space of $A^top$ means $A^top b = 0$, and use the fact that the normal equation is $(A^top A)x = (A^top b)$.
– Christopher Carl Heckman
Nov 11 '15 at 7:13
For (b), use the fact that $b$ is in the null space of $A^top$ means $A^top b = 0$, and use the fact that the normal equation is $(A^top A)x = (A^top b)$.
– Christopher Carl Heckman
Nov 11 '15 at 7:13
So $R(A)$ = $N(A^T)^⊥$ (just found that formula in my book), and because b is in the latter, that means it must be perpendicular to R(A). Though I can't find anything about b having to be 0 because of this. I'll keep looking, though.
– Chris
Nov 11 '15 at 7:37
So $R(A)$ = $N(A^T)^⊥$ (just found that formula in my book), and because b is in the latter, that means it must be perpendicular to R(A). Though I can't find anything about b having to be 0 because of this. I'll keep looking, though.
– Chris
Nov 11 '15 at 7:37
The book says that $Ax = b$ is consistent if and only if $b$ is an element of $R(A)$. But is that not possible because $b$ doesn't contain the 0 vector? I remember doing that when checking to see if something is a subspace.
– Chris
Nov 11 '15 at 7:49
The book says that $Ax = b$ is consistent if and only if $b$ is an element of $R(A)$. But is that not possible because $b$ doesn't contain the 0 vector? I remember doing that when checking to see if something is a subspace.
– Chris
Nov 11 '15 at 7:49
And for the part $b$, I said $A^Tb$ = $0$, meaning $(A^TA)x$ = $0$, and because the rank of $A$ is equal to the rank of $A^TA$, that means $A^TA$ has a rank of 3 and therefore has a nullity of 5, which means there are 5 free variables in $A^Tb$ = $0$ and that means $Ax=b$ has infinitely many least squares solutions. Is that right?
– Chris
Nov 11 '15 at 8:06
And for the part $b$, I said $A^Tb$ = $0$, meaning $(A^TA)x$ = $0$, and because the rank of $A$ is equal to the rank of $A^TA$, that means $A^TA$ has a rank of 3 and therefore has a nullity of 5, which means there are 5 free variables in $A^Tb$ = $0$ and that means $Ax=b$ has infinitely many least squares solutions. Is that right?
– Chris
Nov 11 '15 at 8:06
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
a
The system is inconsistent. Example:
$$
left[
begin{array}{cc}
1 & 0 \ 0 & 0
end{array}
right]
%
left[
begin{array}{cc}
x_{1} \ x_{2}
end{array}
right]
=
left[
begin{array}{c}
0 \ 1
end{array}
right]
$$
b
No, the system is not unique. The solution is an affine space: infinite solutions. In linear algebra the number of solution is either 0 (no existence), 1 (existence and uniqueness), or infinite (existence, no uniqueness).
A pencil and paper exercise: Unique least square solutions
Theoretical treatment: Is a least squares solution to Ax=bAx=b necessarily unique
To see a diagram Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?
How the SVD produces the Moore-Penrose pseudoinverse: How does the SVD solve the least squares problem?
answered Mar 10 '17 at 23:39
dantopa
6,38132042
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
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down vote
a
The system is inconsistent. Example:
$$
left[
begin{array}{cc}
1 & 0 \ 0 & 0
end{array}
right]
%
left[
begin{array}{cc}
x_{1} \ x_{2}
end{array}
right]
=
left[
begin{array}{c}
0 \ 1
end{array}
right]
$$
b
No, the system is not unique. The solution is an affine space: infinite solutions. In linear algebra the number of solution is either 0 (no existence), 1 (existence and uniqueness), or infinite (existence, no uniqueness).
A pencil and paper exercise: Unique least square solutions
Theoretical treatment: Is a least squares solution to Ax=bAx=b necessarily unique
To see a diagram Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?
How the SVD produces the Moore-Penrose pseudoinverse: How does the SVD solve the least squares problem?
answered Mar 10 '17 at 23:39
dantopa
6,38132042
add a comment |
up vote
0
down vote
a
The system is inconsistent. Example:
$$
left[
begin{array}{cc}
1 & 0 \ 0 & 0
end{array}
right]
%
left[
begin{array}{cc}
x_{1} \ x_{2}
end{array}
right]
=
left[
begin{array}{c}
0 \ 1
end{array}
right]
$$
b
No, the system is not unique. The solution is an affine space: infinite solutions. In linear algebra the number of solution is either 0 (no existence), 1 (existence and uniqueness), or infinite (existence, no uniqueness).
A pencil and paper exercise: Unique least square solutions
Theoretical treatment: Is a least squares solution to Ax=bAx=b necessarily unique
To see a diagram Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?
How the SVD produces the Moore-Penrose pseudoinverse: How does the SVD solve the least squares problem?
answered Mar 10 '17 at 23:39
dantopa
6,38132042
add a comment |
up vote
0
down vote
up vote
0
down vote
a
The system is inconsistent. Example:
$$
left[
begin{array}{cc}
1 & 0 \ 0 & 0
end{array}
right]
%
left[
begin{array}{cc}
x_{1} \ x_{2}
end{array}
right]
=
left[
begin{array}{c}
0 \ 1
end{array}
right]
$$
b
No, the system is not unique. The solution is an affine space: infinite solutions. In linear algebra the number of solution is either 0 (no existence), 1 (existence and uniqueness), or infinite (existence, no uniqueness).
A pencil and paper exercise: Unique least square solutions
Theoretical treatment: Is a least squares solution to Ax=bAx=b necessarily unique
To see a diagram Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?
How the SVD produces the Moore-Penrose pseudoinverse: How does the SVD solve the least squares problem?
answered Mar 10 '17 at 23:39
dantopa
6,38132042
a
The system is inconsistent. Example:
$$
left[
begin{array}{cc}
1 & 0 \ 0 & 0
end{array}
right]
%
left[
begin{array}{cc}
x_{1} \ x_{2}
end{array}
right]
=
left[
begin{array}{c}
0 \ 1
end{array}
right]
$$
b
No, the system is not unique. The solution is an affine space: infinite solutions. In linear algebra the number of solution is either 0 (no existence), 1 (existence and uniqueness), or infinite (existence, no uniqueness).
A pencil and paper exercise: Unique least square solutions
Theoretical treatment: Is a least squares solution to Ax=bAx=b necessarily unique
To see a diagram Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?
How the SVD produces the Moore-Penrose pseudoinverse: How does the SVD solve the least squares problem?
answered Mar 10 '17 at 23:39
dantopa
6,38132042
answered Mar 10 '17 at 23:39
dantopa
6,38132042
answered Mar 10 '17 at 23:39
dantopa
6,38132042
answered Mar 10 '17 at 23:39
dantopa
6,38132042
6,38132042
add a comment |
add a comment |
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For (a), that's not good enough, because your argument doesn't mention what $b$ is. You need to use the fact that the column space of $A$ is the set of all vectors $c$ such that $Ax=c$ is consistent. ... The vector $b$ being in the null space of $A^top$ means $b$ is perpendicular to every vector in the column space (why?); hence, $b$ can't be in the column space, unless $b$ is zero (why?).
– Christopher Carl Heckman
Nov 11 '15 at 7:11
For (b), use the fact that $b$ is in the null space of $A^top$ means $A^top b = 0$, and use the fact that the normal equation is $(A^top A)x = (A^top b)$.
– Christopher Carl Heckman
Nov 11 '15 at 7:13
So $R(A)$ = $N(A^T)^⊥$ (just found that formula in my book), and because b is in the latter, that means it must be perpendicular to R(A). Though I can't find anything about b having to be 0 because of this. I'll keep looking, though.
– Chris
Nov 11 '15 at 7:37
The book says that $Ax = b$ is consistent if and only if $b$ is an element of $R(A)$. But is that not possible because $b$ doesn't contain the 0 vector? I remember doing that when checking to see if something is a subspace.
– Chris
Nov 11 '15 at 7:49
And for the part $b$, I said $A^Tb$ = $0$, meaning $(A^TA)x$ = $0$, and because the rank of $A$ is equal to the rank of $A^TA$, that means $A^TA$ has a rank of 3 and therefore has a nullity of 5, which means there are 5 free variables in $A^Tb$ = $0$ and that means $Ax=b$ has infinitely many least squares solutions. Is that right?
– Chris
Nov 11 '15 at 8:06