Proving R is Infinite-Dimensional & Solving 3 Variable System

[seju79]

(1) Show that the space R over the field of rational numbers Q with the usual operations is infinite dimensional.

(2) Find the values of lambda and μ so that the system
2x1 + 3x2 + 5x3 = 9
7x1 + 3x2 − 2x3 = 8
2x1 + 3x2 + lamdax3 = μ
has (i) no solution, (ii) a unique solution, and (iii) an infinite number of
solutions.

Hints:::(2) (i) lamda= 5, μ $$\neq$$ 9 (ii) lamda $$\neq$$ 5, μ arbitrary (iii) lamda = 5, μ = 9.

 

[HallsofIvy]

1) If it the space were finite dimensional, there would exist a finite basis: a finite number of real numbers such that every real number can be written as a sum of rational numbers times those real numbers. Since the set of rational numbers is countable, that would imply that the set of real numbers is countable.

2) Row reduce the augmented matrix to a triangular form (you don't need to get 0s above the diagonal). The last row will be all 0s except for functions of $\mu$ and $\lambda$ in the last two places, corresponding to the equation $f(\mu,\lambda)x_3= g(\mu,\lambda)$. If $f(\mu,\lambda)$ is not 0, then you can divide both sides by $f(\mu,\lambda)$ to get a single value for $x_3$. You will then have to look at the row above to see if the coefficient of $x_2$ is also non-zero. Values of $\mu$ and $\lambda$ that make both of those coefficients non-zero give a unique solution.

If $f(\mu,\lambda)= 0$, then if $g(\mu,\lambda)$ not 0, there is no solution. Values of $\mu$ and $\lambda$ that make f= 0, but g not equal to 0 give no solution.

If $f(\mu,\lambda)$ and $g(\mu,\lambda)$ are both 0, you again need to look at the second row. Values of $\mu$ and $\lambda$ that make the entire row 0 give an infinite number of solutions, values that make the coefficient of [math]x_2[/math] 0 while the last number in the row non-zero give no solutions.