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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?

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  • $\begingroup$ 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 $Rv=\lambda_j v$. $\endgroup$ – Dave Apr 22 at 16:16
  • $\begingroup$ Just a sign error. $\endgroup$ – Yves Daoust Apr 22 at 16:34
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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 $\mathbf{R} + 2 \mathbf{I}$.

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  • $\begingroup$ Very clarifying. What a silly mistake I made employing twice $\lambda$ to talk about different sets of eigenvalues. $\endgroup$ – torito verdejo Apr 22 at 17:04
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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, $\mu=\lambda+2$.

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

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