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Let's consider a matrix $A\in\mathbb{R}^{n\times n}$, with eigenvalues $\lambda_i$. We assume $A$ is positive definite, so $\lambda_i>0$. Now lets consider the matrix

$$ B =\lambda I - A,\qquad \lambda \in \lambda_i.$$ We know that $B$ is singular, even better we know $\lambda_i$ are intended to be found by computing for which $\lambda\ \det(B)=0$. Now let us denote $\mu_i$ to be the eigenvalues of $B$. We know that $\mu_1=0$, but I was wondering if we can say anyting about the rest of the eigenvalues of B.

I considered the following example:

$$A = \begin{pmatrix}14 & 38 & 26\\ 38 & 110 &94\\ 26 & 94 & 145\end{pmatrix}$$ With eigenvalues $\lambda = [0.1879,\ 36.6743,\ 232.1378] $. Then if we choos $B = \lambda_1I-A$, we get that $$ \mu = [0,\ -36.6743,\ -232.1378] = \lambda_1-\lambda. $$

Now my question is if A) the statement $\mu = \lambda_i -\lambda$ holds for any matrix $A$ and all eigenvalues $\lambda_i$ and B) if this only holds for eigenvalues $\lambda_i$, or holds for all matrices structured as $B = cI-A$, where $c$ can be any scalar value and this is actually a known property.

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Suppose that $\mu$ is an eigenvalue of $B$ and that $B=c\operatorname{Id}-A$ for some matrix $A$ and some number $c$. Then there is a non-zero vector $v$ such that $B.v=\mu v$. But\begin{align}B.v=\mu v&\iff(c\operatorname{Id}-A).v=\mu v\\&\iff cv-A.v=\mu v\\&\iff A.v=cv-\mu v\\&\iff A.v=(c-\mu)v.\end{align}So, whenever $\mu$ is an eigenvalue of $B$, $c-\mu$ is an eigenvalue of $A$. By the same argument, if $\lambda$ is an eigenvalue of $A$, $c-\lambda$ is an eigenvalue of $B$. Note that the only thing I need for this to work is that $B$ is a square matrix. Being symmetric is not relevant.

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  • $\begingroup$ Thanks! Putting it like this I feel a bit stupid that I did not notice this myself... $\endgroup$ – User123456789 Dec 21 '18 at 12:17
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    $\begingroup$ Believe me, I know that feeling. I'm glad I could help. $\endgroup$ – José Carlos Santos Dec 21 '18 at 12:19

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