Prove the Linear Algebra about Matrices

Question

I have to do a proof about a linear algebra matrix but I'm stuck on how to go about doing the proof.

Here's the proof:

Let $A$ be an $n \times n$ matrix, and $λ$ an eigenvalue of $A$. Then $λ+µ$ is an eigenvalue of the matrix $M=A+µI$, where $I$ is the $n \times n$ matrix.

How do I go about doing this proof?

I can see that if we do the $det(A-λI)$ to find the eigenvalues we can get the eigenvalue of the matrix $M$ to be $λ+µ$, but how do I prove this, and I also don't really understand what I'm trying to prove in the above statement. Any help would be really appreciated.

Thanks

Answer 1

Consider the eigenvector $v$ then

$$Mv=(A+µI)v=Av+µv=(λ+µ)v$$

Answer 2

\begin{align}\lambda+\mu\text{ is an eigenvalue of }A+\mu\operatorname{Id}&\iff\det\bigl(A+\mu\operatorname{Id}-(\lambda+\mu)\operatorname{Id}\bigr)=0\\&\iff\det(A-\lambda\operatorname{Id})=0,\end{align}which is true, since $\lambda$ is an eigenvalue of $A$.

Answer 3

Let A be an n×n matrix, and λ an eigenvalue of A. Then λ+µ is an eigenvalue of the matrix M=A+µI, where I is the n×n matrix.

$$AV=\lambda V for V\ne 0 $$

$$ (A+\mu I-(\lambda +\mu))V= AV+ \mu V-\lambda V-\mu V =AV-\lambda V=0$$

That is $ \lambda +\mu$ is an eigenvalue for A+\mu I$