I would like to make sure if I made any mistakes.
For a linear system $Ax = x$, if there are more than one eigenvalue of $A$ equal to 1, is that means there is not an unique x solves this linear system?
How should I claim this?
P.S. I forgot to add a restriction that $x$ is an unit vector. I am sorry about this.
The number $1$ is an eigenvalue of $A$ if and only if the equation $Ax=x$ has a non-zeror solution $x_0$. But for each such solution, $-x_0$ will be another solution and, if $\|x_0\|=1$, then $\|-x_0\|=1$ too.
Therefore, if $1$ is an eigenvalue of $A$, there will always be at least $2$ solutions.