Given $R$ and its eigenvalues, find the eigenvalues of $R + 2I$:

Question

I have a problem solving an exercice, that I expose in the following.

Let $\textbf{R}$ be a $3\times3$ matrix with eigenvalues $\lambda = \{-4,-2,\ 2\}$. What are the eigenvalues of $\textbf{R} + 2\textbf{I}$ with $\textbf{I}$ the identity matrix?

My answer is that, given that $$det(\textbf{R} -\lambda\textbf{I})$$

and that $$det(\textbf{R} + 2\textbf{I}-\lambda\textbf{I}) = det(\textbf{R} -\textbf{I}(\lambda-2))$$ I assume that for every eigenvalue $\lambda$ of $\textbf{R}$, there is an eigen value $\lambda '$ of $\textbf{R} + 2\textbf{I}$ such that $$\lambda ' = \lambda - 2$$ and therefore $$\lambda' = \{-6,-4,\ 0\}$$

which I'm said it's not correct. Why?

Answer 1

Let $v$ be an eigenvector of $\mathbf{R}$ with respect to the eigenvalue $\lambda$, that is, $\mathbf{R} v = \lambda v$. Then $$ (\mathbf{R} + 2 \mathbf{I}) v = \mathbf{R} v + 2 \mathbf{I} v = \lambda v + 2 v = (\lambda + 2) v, $$ which tells you that $v$ is an eigenvector of $\mathbf{R} + 2 \mathbf{I}$ with respect to the eigenvalue...

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The correct calculation you were doing is the following. If $\lambda$ is an eigenvalue of $\mathbf{R}$, then $$ 0 = \det(\mathbf{R} - \lambda \mathbf{I}) = \det(\mathbf{R} + 2 \mathbf{I} - 2 \mathbf{I} - \lambda \mathbf{I}) = \det(\mathbf{R} + 2 \mathbf{I} - (\lambda + 2) \mathbf{I}), $$ so that $\lambda + 2$ is an eigenvalue of $\mathbf{R} + 2 \mathbf{I}$.

Answer 2

Let $\lambda$ be an Eigenvalue of $R$,

$$\text{det}(R-\lambda I)=0$$

and $\mu$ an Eigenvalue of the modified matrix,

$$\text{det}(R+2I-\mu I)=\text{det}(R-(\mu-2) I)=0.$$

Then, $\mu=\lambda+2$.