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I have a homework at linear algebra and we have this system of linear equations:

$ x+y+z+w=0 $
$ x+2*y+9*z+13*w=0 $
$4*x+41*y+6*z+656*w = 0 $

And we add this equation:

$ x^3 + y^4 + 8* z^5 + 8* w^6 = 0 $

What can be said about the solution set of this system?

I tried wolfram alpha and got this:

Real solutions:

  • w~~0, x~~0, y~~0, z~~0
  • w~~-0.0417376, x~~-0.665814, y~~0.73708, z~~-0.0295286

Doesn't this mean that the solution set is a non empty set?
And also that it is a finite set (with 2 elements? the (0,0,0,0) and (-0.0417376, -0.665814, 0.73708, -0.0295286) ?

Also, how can I find out if the solution set of the above system is a vector space?

Thank you!

  • 1
    Over $\mathbb Q$, the $3$ linear equations give you a matrix of rank $3$. So the nullspace is of dimension $1$, with basis vector $b = (\frac{335}{21}, \frac{-2596}{147}, \frac{-2596}{147}, 1),$ i.e., the solution space is all scalar multiples of $b$. All solutions to the system however, should be zeros of the polynomial. I'm not sure though how to use the polynomial to pick a particular solution.2012-03-12

3 Answers 3

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As for whether or not it's a vector space: one of the axioms of vector spaces says that if $\mathbb v$ is a vector in the space, then so is $k \mathbb v$ for any $k \in \mathbb R$. This means that $k (-0.0417376, -0.665814, 0.73708, -0.0295286)$ would have to be a solution for any $k$.

  • 1
    But it is not: the $x^3 + y^4 + 8 z^5 + 8 w^6 = 0$ prevents it, except for $k=0,1$ or the two complex values.2012-03-12
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You can solve for any three of x, y, z, and w in terms of the other by using the first three equations. For example, from the first two, you get $y+z+w = 2∗y+9∗z+13∗w$, or $y = -8z -12 w$.

From the first and third equations, you get (multiplying the first by 4), $4y+4z+4w = 41y+6z+656w$, or $37y = -2z-652w$.

This lets you get z as a multiple of w, then y as another multiple of w, and, finally, x as another multiple of w.

Substituting these into the last equation, you will get a sextic (sixth degree) equation in w. This will have 6 roots, and since the coefficients will be real, they will be real or in complex conjugate pairs.

Looking at the equation, there will be three zero roots. Since Alpha has found a non-zero real root, the other two roots are real or complex conjugate.

Since the resulting equation, after removing the zero roots is a cubic, you can solve it of divide out the root found by Alpha to get a quadratic.

I see that a solution has been entered while I was typing.

Let's look.

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Your three linear equations can be satisfied by $w = 147 t, x = 2345 t, y = -2596 t, z = 104 t$ for any $t$.

Substitute that into your fourth equation and you have a sextic equation for $t$, $80722386956232 t^6+97332232192 t^5+45417032294656 t^4+12895213625 t^3=0$ but you can take out a factor of $t^3$ so you have in effect three roots at $t=0$ (and so $w=y=z=0$) and a cubic: $80722386956232 t^3+97332232192 t^2+45417032294656 t+12895213625=0.$ This particular cubic has one real root and two complex roots, which you can translate back to values for $w,x,y,z$.

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    Yes - it is not a vector space2012-03-13