TITLE: Finding eigenvalues of an unknown matrix subtracted by the identity matrix
QUESTION [8 upvotes]: The question:

If the eigenvalues of $A$ are 0, 1, and 3, find the eigenvalues of $A-I$. Explain how you obtained them.

My intuition is telling me that I just subtract one from each of the eigenvalues since they are related to the diagonal, and we are subtracting one from each of the members on the diagonal.

REPLY [6 votes]: For any matrix $B$, we have that $\lambda$ is an eigenvalue of $B$ with eigenvector $v$ if and only if, for all scalars $\mu$, $\lambda + \mu$ is an eigenvalue of $B + \mu I$ with the same eigenvector, since
$Bv = \lambda v \tag{1}$
is clearly equivalent to
$(B + \mu I)v = Bv + \mu I v = (\lambda + \mu)v. \tag{2}$
Applying this equivalence to the case at hand shows that the eigenvalues of $A$ are indeed $-1$, $0$, and $2$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!