Rank, nullity and consistency for two matrices












1












$begingroup$


I have this question here which says the following.



Let $A$,$B$ be $3 times 6$ matrices with the following properties.



$(i)$ For every b$epsilon mathbb{R}^3$, rank$(A)$ $=$ rank$([A|b])$



$(ii)$ dim$($Row$(B)$$)$$=2$.



Answer the following questions.



$(a)$ What is rank$(A)$?



My reasoning: I know that rank$(A)leq min(m,n)$ but I am not sure if that justifies in me saying that rank$(A)=3$. I want to say yes, but I am not certain.



$(b)$ What is nullity$(B)$?



My reasoning: rank$(B)+{}$nullity$(B)=$ # of columns.



Therefore, $2$+nullity$(B)$ $=6$



nullity$(B)$ $=6-2$



nullity$(B)$ $=4$



$(c)$ Is there a vector b$epsilon mathbb{R}^3$ such that $A$x=b is consistent?



My reasoning: Yes there is. Since rank$(A)$ $=$ rank$([A|b])$, there must be a consistent solution.



$(d)$ Is there a vector b$epsilon mathbb{R}^3$ such that $B$x=b is consistent?



My reasoning: Yes there is. Since rank$(B)$ $=2$, this means that rank$(B)$ $<6$ so the system is not only consistent, but has infinitely many solutions.



$(e)$ Is there a non-zero vector x $epsilon mathbb{R}^6$ such that Ax=$0$ and Bx=$0$ simultaneously?



I'm not sure about this one... Any guidance would be much appreciated!



Are my other answers reasonable? Thanks!






linear-algebra matrices systems-of-equations matrix-rank







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have this question here which says the following.



    Let $A$,$B$ be $3 times 6$ matrices with the following properties.



    $(i)$ For every b$epsilon mathbb{R}^3$, rank$(A)$ $=$ rank$([A|b])$



    $(ii)$ dim$($Row$(B)$$)$$=2$.



    Answer the following questions.



    $(a)$ What is rank$(A)$?



    My reasoning: I know that rank$(A)leq min(m,n)$ but I am not sure if that justifies in me saying that rank$(A)=3$. I want to say yes, but I am not certain.



    $(b)$ What is nullity$(B)$?



    My reasoning: rank$(B)+{}$nullity$(B)=$ # of columns.



    Therefore, $2$+nullity$(B)$ $=6$



    nullity$(B)$ $=6-2$



    nullity$(B)$ $=4$



    $(c)$ Is there a vector b$epsilon mathbb{R}^3$ such that $A$x=b is consistent?



    My reasoning: Yes there is. Since rank$(A)$ $=$ rank$([A|b])$, there must be a consistent solution.



    $(d)$ Is there a vector b$epsilon mathbb{R}^3$ such that $B$x=b is consistent?



    My reasoning: Yes there is. Since rank$(B)$ $=2$, this means that rank$(B)$ $<6$ so the system is not only consistent, but has infinitely many solutions.



    $(e)$ Is there a non-zero vector x $epsilon mathbb{R}^6$ such that Ax=$0$ and Bx=$0$ simultaneously?



    I'm not sure about this one... Any guidance would be much appreciated!



    Are my other answers reasonable? Thanks!






    linear-algebra matrices systems-of-equations matrix-rank







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have this question here which says the following.



      Let $A$,$B$ be $3 times 6$ matrices with the following properties.



      $(i)$ For every b$epsilon mathbb{R}^3$, rank$(A)$ $=$ rank$([A|b])$



      $(ii)$ dim$($Row$(B)$$)$$=2$.



      Answer the following questions.



      $(a)$ What is rank$(A)$?



      My reasoning: I know that rank$(A)leq min(m,n)$ but I am not sure if that justifies in me saying that rank$(A)=3$. I want to say yes, but I am not certain.



      $(b)$ What is nullity$(B)$?



      My reasoning: rank$(B)+{}$nullity$(B)=$ # of columns.



      Therefore, $2$+nullity$(B)$ $=6$



      nullity$(B)$ $=6-2$



      nullity$(B)$ $=4$



      $(c)$ Is there a vector b$epsilon mathbb{R}^3$ such that $A$x=b is consistent?



      My reasoning: Yes there is. Since rank$(A)$ $=$ rank$([A|b])$, there must be a consistent solution.



      $(d)$ Is there a vector b$epsilon mathbb{R}^3$ such that $B$x=b is consistent?



      My reasoning: Yes there is. Since rank$(B)$ $=2$, this means that rank$(B)$ $<6$ so the system is not only consistent, but has infinitely many solutions.



      $(e)$ Is there a non-zero vector x $epsilon mathbb{R}^6$ such that Ax=$0$ and Bx=$0$ simultaneously?



      I'm not sure about this one... Any guidance would be much appreciated!



      Are my other answers reasonable? Thanks!






      linear-algebra matrices systems-of-equations matrix-rank







      share|cite|improve this question











      $endgroup$




      I have this question here which says the following.



      Let $A$,$B$ be $3 times 6$ matrices with the following properties.



      $(i)$ For every b$epsilon mathbb{R}^3$, rank$(A)$ $=$ rank$([A|b])$



      $(ii)$ dim$($Row$(B)$$)$$=2$.



      Answer the following questions.



      $(a)$ What is rank$(A)$?



      My reasoning: I know that rank$(A)leq min(m,n)$ but I am not sure if that justifies in me saying that rank$(A)=3$. I want to say yes, but I am not certain.



      $(b)$ What is nullity$(B)$?



      My reasoning: rank$(B)+{}$nullity$(B)=$ # of columns.



      Therefore, $2$+nullity$(B)$ $=6$



      nullity$(B)$ $=6-2$



      nullity$(B)$ $=4$



      $(c)$ Is there a vector b$epsilon mathbb{R}^3$ such that $A$x=b is consistent?



      My reasoning: Yes there is. Since rank$(A)$ $=$ rank$([A|b])$, there must be a consistent solution.



      $(d)$ Is there a vector b$epsilon mathbb{R}^3$ such that $B$x=b is consistent?



      My reasoning: Yes there is. Since rank$(B)$ $=2$, this means that rank$(B)$ $<6$ so the system is not only consistent, but has infinitely many solutions.



      $(e)$ Is there a non-zero vector x $epsilon mathbb{R}^6$ such that Ax=$0$ and Bx=$0$ simultaneously?



      I'm not sure about this one... Any guidance would be much appreciated!



      Are my other answers reasonable? Thanks!






      linear-algebra matrices systems-of-equations matrix-rank




      linear-algebra matrices systems-of-equations matrix-rank






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 0:08









      Bernard

      120k740113




      120k740113










      asked Dec 12 '18 at 0:02









      Future Math personFuture Math person

      972817




      972817






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          (a) The answer is correct but the argument is not quite there. From the given information we can say that every $bin mathbb{R}^3$ is in the column space of $mathbb{R}^3$. Why we can conclude that the rank of $A$ is 3 from this fact?



          (b) It is ok.



          (c) The argument needs to be more precise.



          (d) You need to connect the relationship between the rank and the consistency of system of equations.



          (e) Use the fact that $text{dim}(U+V)=text{dim}(U)+text{dim}(V)-text{dim}(Ucap V)$.




          For part (e) try this way. From the previous part we know that $text{nullity}(A)=3$ and $text{nullity}(B)=4$. Let $X={x_1,x_2,x_3}$ and $Y={y_1,y_2,y_3,y_4}$ be respectively be the basis of nullspace of $A$ and $b$.



          We want to show that $text{null}(A)cap text{null}{B}neq emptyset$. Assume otherwise and show that the assumption leads to the conclusion that $ Xcup Y$ is linearly independent, which is impossible (why?)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
            $endgroup$
            – Future Math person
            Dec 12 '18 at 1:17










          • $begingroup$
            Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:21










          • $begingroup$
            I add something for part (e) above.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:30










          • $begingroup$
            I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 2:49










          • $begingroup$
            $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
            $endgroup$
            – user9077
            Dec 12 '18 at 2:53













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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          (a) The answer is correct but the argument is not quite there. From the given information we can say that every $bin mathbb{R}^3$ is in the column space of $mathbb{R}^3$. Why we can conclude that the rank of $A$ is 3 from this fact?



          (b) It is ok.



          (c) The argument needs to be more precise.



          (d) You need to connect the relationship between the rank and the consistency of system of equations.



          (e) Use the fact that $text{dim}(U+V)=text{dim}(U)+text{dim}(V)-text{dim}(Ucap V)$.




          For part (e) try this way. From the previous part we know that $text{nullity}(A)=3$ and $text{nullity}(B)=4$. Let $X={x_1,x_2,x_3}$ and $Y={y_1,y_2,y_3,y_4}$ be respectively be the basis of nullspace of $A$ and $b$.



          We want to show that $text{null}(A)cap text{null}{B}neq emptyset$. Assume otherwise and show that the assumption leads to the conclusion that $ Xcup Y$ is linearly independent, which is impossible (why?)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
            $endgroup$
            – Future Math person
            Dec 12 '18 at 1:17










          • $begingroup$
            Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:21










          • $begingroup$
            I add something for part (e) above.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:30










          • $begingroup$
            I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 2:49










          • $begingroup$
            $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
            $endgroup$
            – user9077
            Dec 12 '18 at 2:53


















          1












          $begingroup$

          (a) The answer is correct but the argument is not quite there. From the given information we can say that every $bin mathbb{R}^3$ is in the column space of $mathbb{R}^3$. Why we can conclude that the rank of $A$ is 3 from this fact?



          (b) It is ok.



          (c) The argument needs to be more precise.



          (d) You need to connect the relationship between the rank and the consistency of system of equations.



          (e) Use the fact that $text{dim}(U+V)=text{dim}(U)+text{dim}(V)-text{dim}(Ucap V)$.




          For part (e) try this way. From the previous part we know that $text{nullity}(A)=3$ and $text{nullity}(B)=4$. Let $X={x_1,x_2,x_3}$ and $Y={y_1,y_2,y_3,y_4}$ be respectively be the basis of nullspace of $A$ and $b$.



          We want to show that $text{null}(A)cap text{null}{B}neq emptyset$. Assume otherwise and show that the assumption leads to the conclusion that $ Xcup Y$ is linearly independent, which is impossible (why?)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
            $endgroup$
            – Future Math person
            Dec 12 '18 at 1:17










          • $begingroup$
            Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:21










          • $begingroup$
            I add something for part (e) above.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:30










          • $begingroup$
            I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 2:49










          • $begingroup$
            $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
            $endgroup$
            – user9077
            Dec 12 '18 at 2:53
















          1












          1








          1





          $begingroup$

          (a) The answer is correct but the argument is not quite there. From the given information we can say that every $bin mathbb{R}^3$ is in the column space of $mathbb{R}^3$. Why we can conclude that the rank of $A$ is 3 from this fact?



          (b) It is ok.



          (c) The argument needs to be more precise.



          (d) You need to connect the relationship between the rank and the consistency of system of equations.



          (e) Use the fact that $text{dim}(U+V)=text{dim}(U)+text{dim}(V)-text{dim}(Ucap V)$.




          For part (e) try this way. From the previous part we know that $text{nullity}(A)=3$ and $text{nullity}(B)=4$. Let $X={x_1,x_2,x_3}$ and $Y={y_1,y_2,y_3,y_4}$ be respectively be the basis of nullspace of $A$ and $b$.



          We want to show that $text{null}(A)cap text{null}{B}neq emptyset$. Assume otherwise and show that the assumption leads to the conclusion that $ Xcup Y$ is linearly independent, which is impossible (why?)






          share|cite|improve this answer











          $endgroup$



          (a) The answer is correct but the argument is not quite there. From the given information we can say that every $bin mathbb{R}^3$ is in the column space of $mathbb{R}^3$. Why we can conclude that the rank of $A$ is 3 from this fact?



          (b) It is ok.



          (c) The argument needs to be more precise.



          (d) You need to connect the relationship between the rank and the consistency of system of equations.



          (e) Use the fact that $text{dim}(U+V)=text{dim}(U)+text{dim}(V)-text{dim}(Ucap V)$.




          For part (e) try this way. From the previous part we know that $text{nullity}(A)=3$ and $text{nullity}(B)=4$. Let $X={x_1,x_2,x_3}$ and $Y={y_1,y_2,y_3,y_4}$ be respectively be the basis of nullspace of $A$ and $b$.



          We want to show that $text{null}(A)cap text{null}{B}neq emptyset$. Assume otherwise and show that the assumption leads to the conclusion that $ Xcup Y$ is linearly independent, which is impossible (why?)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 1:29

























          answered Dec 12 '18 at 0:36









          user9077user9077

          1,239612




          1,239612












          • $begingroup$
            Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
            $endgroup$
            – Future Math person
            Dec 12 '18 at 1:17










          • $begingroup$
            Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:21










          • $begingroup$
            I add something for part (e) above.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:30










          • $begingroup$
            I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 2:49










          • $begingroup$
            $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
            $endgroup$
            – user9077
            Dec 12 '18 at 2:53




















          • $begingroup$
            Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
            $endgroup$
            – Future Math person
            Dec 12 '18 at 1:17










          • $begingroup$
            Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:21










          • $begingroup$
            I add something for part (e) above.
            $endgroup$
            – user9077
            Dec 12 '18 at 1:30










          • $begingroup$
            I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
            $endgroup$
            – Future Math person
            Dec 12 '18 at 2:49










          • $begingroup$
            $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
            $endgroup$
            – user9077
            Dec 12 '18 at 2:53


















          $begingroup$
          Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
          $endgroup$
          – Future Math person
          Dec 12 '18 at 1:17




          $begingroup$
          Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before.
          $endgroup$
          – Future Math person
          Dec 12 '18 at 1:17












          $begingroup$
          Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
          $endgroup$
          – user9077
          Dec 12 '18 at 1:21




          $begingroup$
          Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem.
          $endgroup$
          – user9077
          Dec 12 '18 at 1:21












          $begingroup$
          I add something for part (e) above.
          $endgroup$
          – user9077
          Dec 12 '18 at 1:30




          $begingroup$
          I add something for part (e) above.
          $endgroup$
          – user9077
          Dec 12 '18 at 1:30












          $begingroup$
          I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
          $endgroup$
          – Future Math person
          Dec 12 '18 at 2:49




          $begingroup$
          I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed?
          $endgroup$
          – Future Math person
          Dec 12 '18 at 2:49












          $begingroup$
          $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
          $endgroup$
          – user9077
          Dec 12 '18 at 2:53






          $begingroup$
          $Xcup Y$ has 7 elements. So they must be linearly dependent. Hence there are $alpha_i,beta_iin mathbb{R}$ not all zero such that $$alpha_1x_1+cdots +alpha_3x_3+beta_1y_1+cdots beta_3y_3=0.$$ So $$alpha_1x_1+cdots+alpha_3x_3=-(beta_1y_1+cdots+beta_4y_4)in text{null}(A)cap text{null}(B)$$
          $endgroup$
          – user9077
          Dec 12 '18 at 2:53




















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