Show that λ +c is an eigenvalue of B

Question

$A$ is a matrix with eigenvalue $λ$.

$B=A+cI$ where $c \in R$ and I is the identity matrix.

How do I prove that $λ+c$ is an eigenvalue of $B$? I don't know how to do this, so I don't have what to share regarding my work.

Second, we suppose (1) $X_A(λ)=λ(λ-1)(λ^2+1)$ and c=1. I have to compute $X_B(λ)$ and determine $det(B)$. Regarding the computation of $X_B(λ)$, I am thinking of replacing $λ$ with $λ+1$ (since $λ+c$ is the eigenvalue of B) from the formula number 1. And for the determinant, I am thinking of simply solving the equation I will get in the end. Is this correct?

Thank you

Answer 1

Notice that $$B-(\lambda+c)I=A+cI-(\lambda+c)I=A-\lambda I$$ hence $$\det(B-(\lambda+c)I)=\det(A-\lambda I)=0$$

For your second question, your claims are correct.