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? 
 A: If you want to start with $x$, then this means you want to find $x$ such that 
$$(A+kI)x = (\lambda + k)x$$
since $(A+kI)x = Ax + kx$, this reduces to find $x$ such that 
$$Ax + kx = (\lambda + k)x$$
i.e. $$Ax = \lambda x$$
then one will naturally think of taking $x = e$
A: So what you don't want is an "ad hoc" proof that doesn't make it clear where it comes from, is that it?
Then you could use the characteristic polynomial of A :
Let P be that polynomial : 
P(X) = det(A - XI) = det( A+kI - (X+k)I) = Q(X+k) , where Q is the characteristic polynomial of A+kI
P(X) = 0 <=> X= λ, with λ an eigenvalue of A. 
An eigenvalue λ' of A+kI is defined as such : λ' eigenvalue of A+kI <=> Q(λ') =0
But : Q(λ')=0 = P(λ' -k ) <=> λ'-k is an eigenvalue of A, since it verifies P(λ'-k) = 0 
So you can say that it exists a λ, eigenvalue of A, such as : λ' -k = λ
So you get the result you were looking for, and you can even affirm that the set of eigenvalues of A+k*I is {λ +k, λ eigenvalue of A} since the proof is made with equivalency all along the way
A: 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. 
