I am looking for an efficient solution to the following problem $$ A+\lambda I = b \tag{1} $$ where $A\in\mathbf{S}^{n}$ is a symmetric matrix with nonzero eigenvalues, $b\in\mathbf{R}^n$ is fixed, $I$ is the $n$ dimensional identity matrix, and $\lambda$ is variable. I will need to solve the above problem for multiple instances of $\lambda$ so an efficient factorization is desirable. I have considered using the Woodbury matrix identity for $(A+\lambda I)^{-1}$, which gives us
$$(A+\lambda I)^{-1} = \frac{1}{\lambda}I-\frac{1}{\lambda}\left(\lambda A^{-1} - I\right)^{-1}$$
but I am not sure if this helps. Another thing I considered is simultaneous diagonalization of A and I, that is writing $A=Q^T D Q$, where $Q\in\mathbf{R}^{n\times n}$ is a unitary matrix, and $D\in\mathbf{R}^{n\times n}$ is a diagonal matrix. This would give us
$$(A+\lambda I)^{-1}b = Q^T(D +\lambda I)^{-1}\left(Qb\right)$$
which would simplify our solution by only requiring a single diagonalization, a single matrix multiplication, $Q b$, and a matrix multiplication for each instance of $\lambda$ that I solve (1). However, really, I am looking to take advantage of LU decomposition if possible. Does anyone know if there is a way to use LU decomposition here? Thanks!