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I am to find all the eigenvalues of $$ A= \left( \begin{matrix} \alpha & \beta & \beta & \beta \\ \beta & \alpha & \beta & \beta \\ \beta & \beta & \alpha & \beta \\ \beta & \beta & \beta & \alpha \\ \end{matrix} \right) $$

My thoughts:

Sum of the elements of each row and column is $\alpha+3\beta$. Hence $\alpha+3\beta$ is an eigenvalue of $A$. How can I find the other eigenvalues?

Edit: Following @Gae. S. answer I have reached till here.

The characteristic polynomial of the matrix is given by $$ \begin{vmatrix} \alpha-x & \beta & \beta & \beta \\ \beta & \alpha-x & \beta & \beta \\ \beta & \beta & \alpha-x & \beta \\ \beta & \beta & \beta & \alpha-x \\ \end{vmatrix} $$ which has 2 linear factors $(x-(\alpha+3\beta))$ and $(x-(\alpha-\beta))$

Saikat
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1 Answers1

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$(1,-1,1,-1)^\top$ is a second eigenvector, and now that you have two roots of $\det(A-xI)$, computing the remaining two is trivial.

  • For finding the other two roots, should I compute by brute force method the polynomial and then factor it by two linear polynomial corresponding to the known roots? and then proceed to compute the other root by solving the remaining quadratic equation?

    Or is there some other small technique?

     – Saikat Jul 06 '21 at 07:45
  • @Saikat That's what I was saying. There is a completely different way to obtain the result, but I'll leave it to someone else to figure out. –  Jul 06 '21 at 07:47
  • Can you please give me some hints? I will try thinking that way then. – Saikat Jul 06 '21 at 07:49
  • @Saikat It revolves around the fact that if $v_i=1$ for all $i$, then $A=(\alpha-\beta)I+\beta vv^\top$. –  Jul 06 '21 at 07:52
  • If possible, can you please make me understand what is $v_i$? – Saikat Jul 06 '21 at 07:54
  • I suppose that Gae. S. refers to the componentes of the vector $v$, I mean, $v = (1,\dots,1)^T$ – Mario G Jul 06 '21 at 10:01