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edited Mar 17, 2011 at 15:32

It is not entirely true what you wrote about linear equations. If your equations are contradictory, then no matter how many additional variables you add, you will not find a solution. For instance $x+y=0$, $x+2y=1$, $2x+y=1$ has no solution regardless possible additional variables.
If you mean homogenous equations, that is, the constant terms are zero, then your claim is right. Well, it is obviously right since taking all variables to be zero gives you a solution. However, it is true that in this case if the number of variables is larger than the number of equations, then you have a non-trivial solution.
For homogenous systems of equations over $\mathbb C$ the same bound holds. Here is why: since it is homogenous, again, taking all variables to be zero gives a solution. Therefore the hypersurfaces determined by the equations have a common point and by the affine dimension theoremaffine dimension theorem (which is essentially saying that one equation can cut down the dimension of the solution set by at most one) the solution set has positive dimension.
Over $\mathbb R$ you run into the problem of not having solutions no matter what. If your equations contain something like $x^2+y^2+...=0$ involving all the variables, then you are out of luck.

It is not entirely true what you wrote about linear equations. If your equations are contradictory, then no matter how many additional variables you add, you will not find a solution. For instance $x+y=0$, $x+2y=1$, $2x+y=1$ has no solution regardless possible additional variables.
If you mean homogenous equations, that is, the constant terms are zero, then your claim is right. Well, it is obviously right since taking all variables to be zero gives you a solution. However, it is true that in this case if the number of variables is larger than the number of equations, then you have a non-trivial solution.
For homogenous systems of equations over $\mathbb C$ the same bound holds. Here is why: since it is homogenous, again, taking all variables to be zero gives a solution. Therefore the hypersurfaces determined by the equations have a common point and by the affine dimension theorem (which is essentially saying that one equation can cut down the dimension of the solution set by at most one) the solution set has positive dimension.
Over $\mathbb R$ you run into the problem of not having solutions no matter what. If your equations contain something like $x^2+y^2+...=0$ involving all the variables, then you are out of luck.

It is not entirely true what you wrote about linear equations. If your equations are contradictory, then no matter how many additional variables you add, you will not find a solution. For instance $x+y=0$, $x+2y=1$, $2x+y=1$ has no solution regardless possible additional variables.
If you mean homogenous equations, that is, the constant terms are zero, then your claim is right. Well, it is obviously right since taking all variables to be zero gives you a solution. However, it is true that in this case if the number of variables is larger than the number of equations, then you have a non-trivial solution.
For homogenous systems of equations over $\mathbb C$ the same bound holds. Here is why: since it is homogenous, again, taking all variables to be zero gives a solution. Therefore the hypersurfaces determined by the equations have a common point and by the affine dimension theorem (which is essentially saying that one equation can cut down the dimension of the solution set by at most one) the solution set has positive dimension.
Over $\mathbb R$ you run into the problem of not having solutions no matter what. If your equations contain something like $x^2+y^2+...=0$ involving all the variables, then you are out of luck.

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created Mar 17, 2011 at 7:41

It is not entirely true what you wrote about linear equations. If your equations are contradictory, then no matter how many additional variables you add, you will not find a solution. For instance $x+y=0$, $x+2y=1$, $2x+y=1$ has no solution regardless possible additional variables.
If you mean homogenous equations, that is, the constant terms are zero, then your claim is right. Well, it is obviously right since taking all variables to be zero gives you a solution. However, it is true that in this case if the number of variables is larger than the number of equations, then you have a non-trivial solution.
For homogenous systems of equations over $\mathbb C$ the same bound holds. Here is why: since it is homogenous, again, taking all variables to be zero gives a solution. Therefore the hypersurfaces determined by the equations have a common point and by the affine dimension theorem (which is essentially saying that one equation can cut down the dimension of the solution set by at most one) the solution set has positive dimension.
Over $\mathbb R$ you run into the problem of not having solutions no matter what. If your equations contain something like $x^2+y^2+...=0$ involving all the variables, then you are out of luck.