homogeneous only trivial or infinite











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This question has three parts. There are similar questions on stack exchange, but if you read all of the questions, then you'll see that this is not a duplicate [at least not one that I could find.]



From the Oregon math website:




"A $ntimes n$ homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions."





  1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?


  2. If not, then what values of the determinant for $A$ imply that there is a non-trivial, unique solution, for a homogeneous equation?


  3. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?











share|cite|improve this question




























    up vote
    -2
    down vote

    favorite












    This question has three parts. There are similar questions on stack exchange, but if you read all of the questions, then you'll see that this is not a duplicate [at least not one that I could find.]



    From the Oregon math website:




    "A $ntimes n$ homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions."





    1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?


    2. If not, then what values of the determinant for $A$ imply that there is a non-trivial, unique solution, for a homogeneous equation?


    3. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?











    share|cite|improve this question


























      up vote
      -2
      down vote

      favorite









      up vote
      -2
      down vote

      favorite











      This question has three parts. There are similar questions on stack exchange, but if you read all of the questions, then you'll see that this is not a duplicate [at least not one that I could find.]



      From the Oregon math website:




      "A $ntimes n$ homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions."





      1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?


      2. If not, then what values of the determinant for $A$ imply that there is a non-trivial, unique solution, for a homogeneous equation?


      3. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?











      share|cite|improve this question















      This question has three parts. There are similar questions on stack exchange, but if you read all of the questions, then you'll see that this is not a duplicate [at least not one that I could find.]



      From the Oregon math website:




      "A $ntimes n$ homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions."





      1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?


      2. If not, then what values of the determinant for $A$ imply that there is a non-trivial, unique solution, for a homogeneous equation?


      3. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?








      linear-algebra systems-of-equations determinant






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 18:22









      Martin Sleziak

      44.6k7115269




      44.6k7115269










      asked Nov 22 at 18:16









      Bertram

      41




      41






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote













          It is good to notice that a homogeneous system $Ax=0$ always has at least zero solution $x=0$. (So for homogeneous system you cannot have no solutions at all.)



          If you have any solution $x$, than any scalar multiple $cx$ is also a solution of the same homogeneous system. So once you have at least one non-zero solution, you immediately get many other solutions. (If you are working with real numbers or some other infinite field, you get infinitely many solutions.)



          Things are different if you look at systems which are not homogeneous - then there is also a possibility that there is no solution.






          share|cite|improve this answer





















          • If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
            – Bertram
            Nov 22 at 18:44


















          up vote
          0
          down vote













          From the given theorem we have that





          1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?





          • recall that $det A=0$ or $det A neq 0$




          1. If not, then what values of the determinant for A imply that there is a non-trivial, unique solution, for a homogeneous equation?





          • what about $det A neq 0$?




          1. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?





          • it is not weird, it is a theorem






          share|cite|improve this answer





















          • Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
            – Bertram
            Nov 22 at 18:39










          • @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
            – gimusi
            Nov 22 at 18:42










          • Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
            – Bertram
            Nov 23 at 17:12











          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

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          active

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          up vote
          1
          down vote













          It is good to notice that a homogeneous system $Ax=0$ always has at least zero solution $x=0$. (So for homogeneous system you cannot have no solutions at all.)



          If you have any solution $x$, than any scalar multiple $cx$ is also a solution of the same homogeneous system. So once you have at least one non-zero solution, you immediately get many other solutions. (If you are working with real numbers or some other infinite field, you get infinitely many solutions.)



          Things are different if you look at systems which are not homogeneous - then there is also a possibility that there is no solution.






          share|cite|improve this answer





















          • If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
            – Bertram
            Nov 22 at 18:44















          up vote
          1
          down vote













          It is good to notice that a homogeneous system $Ax=0$ always has at least zero solution $x=0$. (So for homogeneous system you cannot have no solutions at all.)



          If you have any solution $x$, than any scalar multiple $cx$ is also a solution of the same homogeneous system. So once you have at least one non-zero solution, you immediately get many other solutions. (If you are working with real numbers or some other infinite field, you get infinitely many solutions.)



          Things are different if you look at systems which are not homogeneous - then there is also a possibility that there is no solution.






          share|cite|improve this answer





















          • If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
            – Bertram
            Nov 22 at 18:44













          up vote
          1
          down vote










          up vote
          1
          down vote









          It is good to notice that a homogeneous system $Ax=0$ always has at least zero solution $x=0$. (So for homogeneous system you cannot have no solutions at all.)



          If you have any solution $x$, than any scalar multiple $cx$ is also a solution of the same homogeneous system. So once you have at least one non-zero solution, you immediately get many other solutions. (If you are working with real numbers or some other infinite field, you get infinitely many solutions.)



          Things are different if you look at systems which are not homogeneous - then there is also a possibility that there is no solution.






          share|cite|improve this answer












          It is good to notice that a homogeneous system $Ax=0$ always has at least zero solution $x=0$. (So for homogeneous system you cannot have no solutions at all.)



          If you have any solution $x$, than any scalar multiple $cx$ is also a solution of the same homogeneous system. So once you have at least one non-zero solution, you immediately get many other solutions. (If you are working with real numbers or some other infinite field, you get infinitely many solutions.)



          Things are different if you look at systems which are not homogeneous - then there is also a possibility that there is no solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 18:26









          Martin Sleziak

          44.6k7115269




          44.6k7115269












          • If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
            – Bertram
            Nov 22 at 18:44


















          • If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
            – Bertram
            Nov 22 at 18:44
















          If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
          – Bertram
          Nov 22 at 18:44




          If the solution is only x=0, then cx=0 is still 0. I do see your point that the theorem points out that the homogeneous system always at least has the solution x=0 whereas the inhomogeneous system may not have any solution at all.
          – Bertram
          Nov 22 at 18:44










          up vote
          0
          down vote













          From the given theorem we have that





          1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?





          • recall that $det A=0$ or $det A neq 0$




          1. If not, then what values of the determinant for A imply that there is a non-trivial, unique solution, for a homogeneous equation?





          • what about $det A neq 0$?




          1. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?





          • it is not weird, it is a theorem






          share|cite|improve this answer





















          • Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
            – Bertram
            Nov 22 at 18:39










          • @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
            – gimusi
            Nov 22 at 18:42










          • Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
            – Bertram
            Nov 23 at 17:12















          up vote
          0
          down vote













          From the given theorem we have that





          1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?





          • recall that $det A=0$ or $det A neq 0$




          1. If not, then what values of the determinant for A imply that there is a non-trivial, unique solution, for a homogeneous equation?





          • what about $det A neq 0$?




          1. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?





          • it is not weird, it is a theorem






          share|cite|improve this answer





















          • Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
            – Bertram
            Nov 22 at 18:39










          • @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
            – gimusi
            Nov 22 at 18:42










          • Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
            – Bertram
            Nov 23 at 17:12













          up vote
          0
          down vote










          up vote
          0
          down vote









          From the given theorem we have that





          1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?





          • recall that $det A=0$ or $det A neq 0$




          1. If not, then what values of the determinant for A imply that there is a non-trivial, unique solution, for a homogeneous equation?





          • what about $det A neq 0$?




          1. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?





          • it is not weird, it is a theorem






          share|cite|improve this answer












          From the given theorem we have that





          1. Does this imply that a homogeneous system $Ax=0$ has only a trivial solution or infinite number of solutions?





          • recall that $det A=0$ or $det A neq 0$




          1. If not, then what values of the determinant for A imply that there is a non-trivial, unique solution, for a homogeneous equation?





          • what about $det A neq 0$?




          1. If it does imply that $Ax=0$ only has a trivial or infinite (free parameter) solution, isn't that kind of weird?





          • it is not weird, it is a theorem







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 18:25









          gimusi

          92.7k94495




          92.7k94495












          • Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
            – Bertram
            Nov 22 at 18:39










          • @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
            – gimusi
            Nov 22 at 18:42










          • Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
            – Bertram
            Nov 23 at 17:12


















          • Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
            – Bertram
            Nov 22 at 18:39










          • @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
            – gimusi
            Nov 22 at 18:42










          • Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
            – Bertram
            Nov 23 at 17:12
















          Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
          – Bertram
          Nov 22 at 18:39




          Thank you for answering. The case of "what about det A not = 0" is addressed in the quote as leading to a trivial solution for a homogeneous system.
          – Bertram
          Nov 22 at 18:39












          @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
          – gimusi
          Nov 22 at 18:42




          @Bertram The key fact is that if $det A neq 0$ and the columns $c_1,c_2,c_3$ of $A$ are linearly independent then $$Ax=x_1c_1+x_2c_2+x_3c_3=0 iff x=0$$
          – gimusi
          Nov 22 at 18:42












          Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
          – Bertram
          Nov 23 at 17:12




          Yes thank you gimusi. I see now that the definition of the linear independence of the columns directly leads to the requirement that x=0 if it does solve the system of equations. Thank you.
          – Bertram
          Nov 23 at 17:12


















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