How can I prove that a 3x3 symmetric matrix has eigenvalue = 1

Question

I need to prove that a non-zero Symmetric Matrix

$$M=\begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \\ \end{bmatrix}$$ has an eigenvalue = 1.

So for any non-zero vector $$V=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}$$

The following equation is true:

$$\begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}$$

Any help is very much appreciated.

Answer 1

This is not true in general. The number $1$ is an eigenvalue of $M$ if and only if $a^2+b^2+c^2=1$, since the characteristic polynomial of $M$ is $(a^2+b^2+c^2)x^2-x^3$. Besides, asserting that $1$ is an eigenvalue of $M$ means that there is some non-zero vector $v$ such that $M.v=v$.