Matrix A has eigenvalue $λ$ , Prove the eigenvalues of Matrix $(A+kI)$ is (λ + k) 
The matrix A has an eigenvalue $λ$ with corresponding eigenvector $e$. 
Prove that the matrix $(A + kI)$, where $k$ is a real constant and I is the identity matrix, has an eigenvalue $(λ + k)$


My Attempt:
$$(A + kI)e$$ 
$$= Ae + kIe = λe + ke = (λ + k)e$$

Yes I proved it, but I'm not happy with the proof and I don't think its a good proof. Reasons:


*

*I started out assuming this : $(A + kI)e$ , But It should be :


$$(A + kI)x$$

And I don't know how to prove this way ^^^

Even though it might seem obvious to some of you (not for me) and after the proof it's obvious that $x=e$ , it wasn't right for me  to start my proof with it (since its not mentioned that x=e.
So How do I prove this? 

The point is you need to find a non zero vector $v$ such that  $(A+KI) v = \beta v$ and $ \beta $ is said to be the eigenvalue of $(A+KI)$. 
So consider $ x \in \mathbb{R}^n$ such that $Ax=\lambda x$ then : 
$(A+KI) x = Ax + Kx = \lambda x + Kx = (\lambda + K) x $ . 
This implies that  $\lambda+ K $ is an eigenvalue. 
