What are the non-trivial solutions of this system?

Question

Let $a, b, c \in R$. How could I show that the homogeneous system of linear equations has only the trivial solution $\left(\begin{array}{c}0\\0\\0\end{array}\right)$? \begin{align} (a^2 + 1)x_1 + (ab)x_2 + (ac)x_3 = 0\\ (ab)x_1 + (b^2 + 1)x_2 + (bc)x_3 = 0\\ (ac)x_1 + (bc)x_2 + (c^2 + 1)x_3= 0 \end{align}

Answer 1

Hint: rewrite your system of linear equations in the form of $(I+vv^T)x=0$ and left-multiply it by $x^T$ on both sides.

Edit. If you are not familiar with basic linear algebra, here is a high school version: multiply both sides of the first, second and third equations in your question by $x_1,x_2$ and $x_3$ respectively. Sum them up, you get $x_1^2+x_2^2+x_3^2 + (ax_1+bx_2+cx_3)^2=0$ and you may continue from here.

Answer 2

If you are not familiar with determinants then you could use echelon row reduction on the matrix \begin{pmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ac&bc&c^2+1 \end{pmatrix} and should eventually end up with the identity matrix hence telling you that $(0,0,0)$ is the only solution. But note that this may be quite long and be careful when multiplying rows by terms in $a,b,c$ since you don't want to multiply or divide by zero. It's a bit tedious but it can be done.

Answer 3

As user 1551 writes, rewriting as $({\bf I}+{\bf vv}^T)$ is useful. Then you can just use knowledge of how a rank 1 perturbation affects the eigensystem. In this case adding 1 to the eigenvalue in the direction of $\hat{\bf v}$. So the eigenvalues will be all 1 except in the $\bf \hat{v}$ direction which would be 2. Since no eigenvalue is 0, the null space is empty and the only solution could be the zero vector.