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I'm studying for an exam on Tuesday and have been stumped on this question for a little while. Any hints or help at all would be appreciate!

I'm given 2 matrices, $A$ and $R$ as shown below. I'm also given that the eigenvalues for matrix $A$ are: 4, 3, -3, & -2. I am then told to determine the eigenvalues of $C(\alpha, \beta) = \alpha A + \beta R$, and thus determine when the limit $\lim \limits_{n \to \infty} C(\alpha, \beta)^n$ exists, given that $\alpha$ and $\beta$ are both greater than 0.

For the second part (determining when the limit exists), I think I have to use the rule that $\lim \limits_{n \to \infty} C(\alpha, \beta)^n$ will only exist if the eigenvalues $\lvert \lambda\rvert < 1$, so I imagine I'll have to set up some inequalities to do so. Any help is appreciated, since I've been stuck on this for over a day now! I've attempted to use the fact that the rank of matrix $R$ is 1, but I'm unsure how.

$A$:

$$ \begin{pmatrix} 2 & 0 & 2 & 0 \\ 2 & -1 & 3 & 0 \\ 2 & -1 & 2 & 1 \\ -16 & 8 & 13 & -1 \\ \end{pmatrix} $$

$R$:

$$ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{pmatrix} $$

EDIT: I realise that given the eigenvalues of A, the eigenvalues of $\alpha A$ will simply be the eigenvalues of A multiplied by $\alpha$. Is this property true of all matrices? Or is it just because they have the same row sums?

EDIT #2: I also realise that the eigenvalues of $A + R$ are 8, 3, -3, 2 i.e. the highest eigenvalue was multiplied by 2 while the rest remained the same. Similarly, eigenvalues of $A + 2R$ are 12, 3, -3, 2 and so on. I again fail to see why this is the case - would appreciate any pointers.

EDIT #3: Based on the previous 2 edits, the eigenvalues for $C(\alpha, \beta) = \alpha A + \beta R$ will be: $4\alpha + 4\beta, 3\alpha, -3\alpha, -2\alpha$. I figured this out by manually calculating the eigenvalues for the first couple of $\alpha's$ and $\beta's$, however is there a trick I'm missing to simplifying this problem, since on the exam I don't think I'll be asked to compute the eigenvalues of a 4X4 matrix.

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    $\begingroup$ I guess you meant the limit exists if $\;|\lambda|<1\;$ ... $\endgroup$ – DonAntonio May 11 at 17:55
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    $\begingroup$ @DonAntonio Yes, you are correct. I will fix it now. $\endgroup$ – user3424575 May 11 at 17:57
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    $\begingroup$ What is $B$ in $C(\alpha,\beta)=\alpha A+\beta B$? Do you mean $R$ instead? Start by noting $A$ has constant row sum, hence... $\endgroup$ – user10354138 May 11 at 18:25
  • $\begingroup$ @user10354138 yes you are correct, I'll fix that as well. Sorry for the typos. $\endgroup$ – user3424575 May 11 at 18:28
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    $\begingroup$ Do you mean the eigenvalues of $\alpha A + \beta R$? Otherwise, what is $B$? $\endgroup$ – Robert Lewis May 11 at 18:55
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Hint: If $\ e_\lambda\ $ is an eigenvector of $\ A\ $ with eigenvalue $\ \lambda\ $, $\ \mathbb{1}^\top e_\lambda=1\ $, and $\ k = \frac{\alpha\lambda-4\alpha-4\beta}{\beta}\ $, then, since $\ R=\mathbb{1}\mathbb{1}^\top\ $, \begin{eqnarray} (\alpha A + \beta R)(k e_\lambda + \mathbb{1})&=&\alpha\lambda ke_\lambda +4\alpha \mathbb{1} + \beta k\mathbb{1}\mathbb{1}^\top e_\lambda + \beta \mathbb{1}\mathbb{1}^\top \mathbb{1}\\ &=& \alpha\lambda k e_\lambda + \left(4\alpha+\beta k+4\beta\right)\mathbb{1}\\ &=& \alpha\lambda\left(ke_\lambda+\mathbb{1}\right)\ , \end{eqnarray} so $\ ke_\lambda +\mathbb{1}\ $ is an eigenvector of $\ \alpha A + \beta R\ $ with eigenvalue $\ \alpha\lambda\ $, except for the eigenvector $\ e_4 =\frac {1}{4}\mathbb{1}\ $, for which $\ ke_4 + \mathbb{1}= 0\ $. But $\ \mathbb{1}\ $ is an eigenvector of $\ \alpha A + \beta R\ $ with eigenvalue $\ 4\alpha + 4\beta\ $.

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