Eigenvalues of sum of matrices where one is I have a matrix $A$ with known eigenvalues.  ($A$ is a real tridiagonal symmetric matrix and is positive definite, if those properties end up being useful).  I am wondering if there are any special theorems or properties that would allow me to know/calculate the eigenvalues of $I-A$ from the eigenvalues of $A$.
If not, is there any way a bound on the eigenvalues of $I-A$ can be set?
 A: In fact, the properties of $A$ don't really matter here.  For any (square) matrix $A$, $\mu$ will be an eigenvalue of $I - A$ if and only if $\mu = 1 - \lambda$ for some eigenvalue $\lambda$ of $A$.
Here's a quick proof: we say that the eigenvalues of $A$ are the values of $t$ such that
$$
\det(A - tI) = 0
$$
We note that
$$
\det((I-A) - \mu I) = 0 \iff\\
\det(-A + (1 - \mu)I) = 0 \iff\\
\det(A - (1 -\mu)I) = 0
$$
A: Here's a way to see it from first principles, without using determinants or characteristic polynomials, and which works no matter what the dimension of the underlying vector space, on which $A$ operates, may be.
Proposition:  Let $A$ be a linear operator on a vector space $V$ over an arbitrary field $\Bbb F$.  Then $\lambda \in \Bbb F$ is an eigenvalue of $A$ if and only if $\lambda + \mu$ is an eigenvalue if $A + \mu I$ for any $\mu \in \Bbb F$.
Proof:  If
$Ax = \lambda x\tag{1}$
for some vector $x \ne 0$, then
$(A + \mu I)x = Ax + \mu I x = \lambda x + \mu x = (\lambda + \mu)x, \tag{2}$
showing $\lambda + \mu$ to be an eigenvalue of $A + \mu I$ (in fact, with the same eigenvector, $x$).  It is easy to further see that (2) implies (1) by subtracting $\mu I x = \mu x$.  QED.
Now simply recall that $\lambda$ is an eigenvalue of $A$ iff $-\lambda$ is an eigenvalue of $-A$ and use the proposition to see that $1 - \lambda$ is an eigenvalue of $I - A$ iff $\lambda$ is same for $A$.
A: If $\lambda$ be an eigenvalue of $A$, then $\det[A-\lambda I] =0$, so $\det[(A-I)-(\lambda-1) I] =0$. Thus $\lambda-1$ is an eigenvalue of $A-I$ and then $1-\lambda$ is an eigenvalue of $I-A$. 
Similarly, one can show that if $1-\lambda$ be an eigenvalue of $I-A$, then $\lambda$ is an eigenvalue of $A$,
