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.










share|cite|improve this question
























  • 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

















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.










share|cite|improve this question
























  • 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















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.










share|cite|improve this question















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






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edited Nov 22 at 18:04









Lorenzo B.

1,8322520




1,8322520










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




















  • 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












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?






share|cite|improve this answer





















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    1 Answer
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    1 Answer
    1






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    active

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    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?






    share|cite|improve this answer

























      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?






      share|cite|improve this answer























        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?






        share|cite|improve this answer












        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?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 10 '17 at 23:39









        dantopa

        6,38132042




        6,38132042






























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