Assume $A$ is a $n \times n$ symmetric and positive semidefinite matrix with eigenvalues $\lambda_1 \leq \ldots \leq \lambda_n$.
I know that the eigenvalues of $A + cI$ are $\lambda_i + c$, but is there an expression for the eigenvalues of $A + c$ (matrix plus a scalar), where $c$ is a scalar? I'm also interested in the upper bound for the largest eigenvalue of $A + c$. Note that $A + c$ can be written as matrix sum $A + cJ$, where $J$ is a $n \times n$ matrix of all $1$'s.
No, not in general. That is, the eigenvalues of $A + cJ$ are not a function of the eigenvalues of $A$ and the eigenvalues of $J$ (even when $A$ is positive-semidefinite).
For example, take $$A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad A_2 = \begin{pmatrix} \frac{3}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{3}{2} \end{pmatrix},$$ which are two orthogonally similar, positive definite matrices, and add $J$ to each of them. The resulting matrices have different eigenvalues. $A_2 + J$ has eigenvalues $1$ and $4$, while $A_1 + J$ has eigenvalues $\frac{1}{2}(5 \pm \sqrt{5})$.
There is a somewhat notable special exception. If your positive-semidefinite matrix (or even just symmetric matrix) has constant row/column sums, then this constant sum will be an eigenvalue. Adding $cJ$ will increase this eigenvalue by $cn$, where $A$ and $J$ are $n \times n$, while leaving the other eigenvalues untouched.
This is because a matrix $A$ having a constant row sum $\lambda$ is equivalent to having $(1, 1, \ldots, 1)^\top$ be an eigenvector with corresponding eigenvalue $\lambda$. If $A$ is symmetric, we can extend this to an orthogonal basis of eigenvectors.
Since $cJ$ also has a constant row sum of $cn$, and the entire orthogonal complement of $\operatorname{span}(1, 1, \ldots, 1)^\top$ is the eigenspace of $cJ$ corresponding to $0$ (i.e. the nullspace), the orthogonal basis of eigenvectors of $A$ will also be an orthogonal basis of eigenvectors for $J$, so the eigenvalues sum as described.