How to find the eigenvalues of $A+I$ given the eigenvalues of $A$? Say the $3 \times 3$ matrix $A$ has eigenvalues $a$, $b$ and $c$. Given $I$ is the $3 \times 3$ identity matrix, how to find the eigenvalues of $A+I$?
I think it could be $a+1$, $b+1$ and $c+1$ but I don't have any basis at all. Thanks in advance.
 A: $$ AV=\lambda V \implies (A+I)V= \lambda V +V = (\lambda +1)V$$
Thus $\lambda +1$ does the trick.
A: Hint:
$$
\det(\lambda I-A)=\det[(\lambda+1)I-(A+I)]
$$
A: The eigenvalues of $A + I$ are indeed $a+ 1$, $b + 1$, $c + 1$; this follows from the following
Fact:
$A \vec v = \mu \vec v \Longleftrightarrow (A + \alpha I) \vec v = (\mu + \alpha ) \vec v; \tag 2$
Proof of Fact:
if
$A\vec v = \mu \vec v, \tag 3$
then
$(A + \alpha I)\vec v = A \vec v + \alpha I \vec v = \mu \vec v + \alpha \vec v = (\mu + \alpha) \vec v; \tag 4$
furthermore, if
$(A + \alpha I)\vec v = (\mu + \alpha) \vec v, \tag 5$
then
$A \vec v + \alpha \vec v = A \vec v + \alpha I\vec v = (A + \alpha I)\vec v = (\mu + \alpha) \vec v = \mu \vec v+ \alpha\vec v , \tag 5$
whence,
$A\vec v = \mu \vec v. \tag 6$
End:  Proof of Fact.
We now simply apply this Fact to the problem at hand to see the our opening affirmation holds.  Furthermore, the proof of our little Fact also shows that the eigenvectors corresponding to $a$, $b$, $c$ are those associated with $a + 1$, $b + 1$, $c + 1$, respectively.
A: Actually, a much stronger statement is true. Namely, if $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of the $n\times n$ matrix $A$, then for any polynomial $p$ the eigenvalues of $p(A)$ are $p(\lambda_1),\ldots,p(\lambda_n)$. 
Easiest way to see it is probably via the Jordan form, but there are several other arguments. 
A: It is indeed $a+1$, $b+1$ and $c+1$, and the basis you have for $A$ works for $A+I$. Make sure you can see for yourself why this is.
