A be a normal operator and has all its eigenvalues collinear in the complex plane iff it is of the form aH + bI A be a normal operator on a finite dimensional Hilbert space H of dimension n . and has all its eigenvalues collinear in the complex plane iff it is of the form aH + bI , where H is Hermitian and I is the identity and a,b are complex numbers.
$\mathbf Idea$: 'only if' part : the eigen values of the matrix A lies in a line . Then we can shift the line passing through the origin and then rotate it . Then we can get the eigen values of the new matrix lies in the real axis . so the new matrix is Hermitian .
'if' part: if A = aH + bI then A is normal and eigen values of H are $a_1,a_2, ... ,a_n$ where $a_i$ is real then if we can find eigen values of A then we will see that those are lies in a line .
My idea is this but I can not argue this mathematically . Can anybody help me to prove it formally .
 A: The parametric equation of a line in the complex plane is $at + b$ where $t \in \mathbb{R}$ and $a,b \in \mathbb{C}$ with $a \neq 0$.
In one direction, if $A = aH + bI$ where $H$ is Hermitian then $H$ has real eigenvalues $\lambda_1, \dots, \lambda_n$ (possibly including multiplicities) and so (verify!) the eigenvalues of $A$ are $a \lambda_1 + b, \dots, a \lambda_n + b$. If $a = 0$ then $A = bI$ so the only eigenvalue of $A$ is $b$ (of multiplicity $n$) and in particular all the eigenvalues of $A$ are collinear (and lie on infinitely many lines). If $a \neq 0$ then they all lie on the line $at + b$.
For the other direction, let $A$ be a normal operator whose eigenvalues lie on the line $at + b$ where $a \neq 0$. Denote the eigenvalues of $A$ by $a t_1 + b, \dots, a t_n + b$ where $t \in \mathbb{R}$ and set $H = \frac{1}{a}(A - bI)$. Then $H$ is normal with real eigenvalues $t_1,\dots,t_n$, hence Hermitian and so $A = aH + bI$.
A: If A = aH + bI
Then A is normal .
Now the eigen values of H are $\alpha_1, ... , \alpha_n$ where $\alpha_i$ is real.
the eigen values of A are $a\alpha_1 + b ,...,a\alpha_n+b $
there exists real number $\lambda_1,... ,\lambda_{n-1}$ such that
take$\alpha_1$ = $\alpha$
the eigen values of A are $a\alpha+b , ... ,a\lambda_{n-1}\alpha+b$
the eigen values are colinier .
