I just started studying Spectral Theory, and I'm trying to show that $\lambda$ is an eigenvalue of a matrix $A$ if and only if $\lambda + \alpha$ is an eigenvalue of $A + \alpha \cdot I$.
Here's my work so far:
- Forward direction: Say $\lambda$ is an eigenvalue of $A$. Then $\lambda$ is a root of A's characteristic polynomial. Then, I'd want to somehow show that the characteristic polynomial of $A + \alpha \cdot I$ has the root $\lambda + \alpha$. My current idea is to consider the factoring of $A$'s characteristic polynomial as some $c(x - \lambda_1)\ldots(x - \lambda_n)$, and to prove that the characteristic polynomial of $A + \alpha \cdot I$ factors into some $d(x - (\lambda_1 + \alpha))\ldots(x - (\lambda_n + \alpha))$. However, I'm not quite sure how to do this; my only idea is to use the determinant formula, but it was really messy.
- Backward direction: Say $\lambda + \alpha$ is an eigenvalue of $A + \lambda \cdot I$. Then, similar to my idea for the forward direction, I'd want to show that the roots of the characteristic polynomial of $A$ are just the roots of the characteristic polynomial of $A + \lambda \cdot I$, but subtracted by $\alpha$.
I've also thought about using the definition of eigenvalues ($\lambda$ is an eigenvalue if it is a scalar such that the equation $Av = \lambda v$ has some nontrivial solution for $v$). I haven't gotten very far this way either.
Can somehow show me how to solve this problem using concepts of Spectral Theory?
It follows directly from definition
Let's say $\lambda$ is eigenvalue for $A$
$$\Longleftrightarrow A\vec{v}=\lambda\vec{v}$$ $$\Longleftrightarrow A\vec{v}+\alpha I\vec{v}=\lambda\vec{v}+\alpha \vec{v}$$ $$\Longleftrightarrow (A+\alpha I)\vec{v}=(\lambda+\alpha)\vec{v}$$
$\Longleftrightarrow$ $\lambda + \alpha$ is eigenvalue of $A+\alpha I$