# Determining eigenvalues of sum of 2 matrices, and then evaluating whether the limit exists

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.

• I guess you meant the limit exists if $\;|\lambda|<1\;$ ... – DonAntonio May 11 at 17:55
• @DonAntonio Yes, you are correct. I will fix it now. – user3424575 May 11 at 17:57
• 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... – user10354138 May 11 at 18:25
• @user10354138 yes you are correct, I'll fix that as well. Sorry for the typos. – user3424575 May 11 at 18:28
• Do you mean the eigenvalues of $\alpha A + \beta R$? Otherwise, what is $B$? – Robert Lewis May 11 at 18:55

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\$$.