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

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?

• If $p$ is a polynomial and $\lambda_j$ are the eigenvalues of $R$, then $p(\lambda_j)$ are the eigenvalues of $p(R)$. To see this: just consider the definition $Rv=\lambda_j v$. – Dave Apr 22 at 16:16
• Just a sign error. – Yves Daoust Apr 22 at 16:34

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...

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 $$\mathbf{R} + 2 \mathbf{I}$$.

• Very clarifying. What a silly mistake I made employing twice $\lambda$ to talk about different sets of eigenvalues. – torito verdejo Apr 22 at 17:04

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

• Let me tell you I find your answer as helpful as the one I marked as the answer. Thank you. – torito verdejo Apr 22 at 17:00