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I would like to calculate the eigenvalues of the following matrix $A$, but the factorization of the characteristic polynomial does not seem to be easy to compute.

$A=\pmatrix{ a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \\ },\ a\neq1,\ a\neq-2$

$f(\lambda)$ = Char$(A,\lambda)$ = $(a-\lambda)^3-3(a-\lambda)+2 = -\lambda^3 + 3a\lambda^2 + 3\lambda(1-3a^2) + (a-1)^2(a+2)$

I have thought about using the Rational-Root Theorem (RRT), so possible roots of $f(\lambda)$ are $(a-1),\ (-a+1),\ (a+2),\ (-a-2)$, and much more, as for example in the case $a=2$ we should also test whether $f(\pm2)=0$ or not, am I wrong?

The eigenvalues of $A$ are $a-1$ and $a+2$ (computed with Wolfram Alpha). This result can be obtained using RRT, computing $f(a-1)$ and $f(a+2)$ and realizing that both are equal to zero but, is there an easier and 'elegant' way to find these eigenvalues?

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  • $\begingroup$ Can you do it for $a=1$? $\endgroup$ – Lord Shark the Unknown Apr 18 at 16:09
  • $\begingroup$ Well, but that's a particular case. Can I assume from that case that eigenvalues are $a-1$ and $a+2$ ($0$ and $-2$ if $a=1$)? $\endgroup$ – Gibbs Apr 18 at 16:11
  • $\begingroup$ Can you go from one particular case to the general case? $\endgroup$ – Lord Shark the Unknown Apr 18 at 16:21
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  • $\begingroup$ Sometimes it’s easier to find eigenvectors first. Try simple linear combinations of the columns, such as summing all of them or taking two at a time. You can also take advantage of the symmetry of $A$ to narrow the search. $\endgroup$ – amd Apr 18 at 17:19
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Basically, you need to solve $(a-\lambda)^3-3(a-\lambda)+2 =0$ for $\lambda$. Don't expand the brackets, instead denote: $t=a-\lambda$. Then: $$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\ t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\ t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$

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  • $\begingroup$ What an easier and practical way to do it! Thanks a lot! $\endgroup$ – Gibbs Apr 18 at 17:29
  • $\begingroup$ You are welcome. $\endgroup$ – farruhota Apr 18 at 17:31
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One way to see $a+2$ is an eigenvalue is that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}a+2\\a+2\\a+2\end{bmatrix}.$$ Then you can use the fact that $x-(a+2)$ divides the characteristic polynomial.


More generally: if all the rows of $A$ add up to $\lambda$, then $\lambda$ is an eigenvalue.

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Hint: if $I$ denotes the identity matrix, then the eigenvalues of $A+cI$ are easily obtained from the eigenvalues of $A$: $$ (A+cI)v=\lambda v \iff Av=(\lambda-c)v $$ What if you take $c=1-a$?

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  • $\begingroup$ If $A=I$, $\lambda_1-c=0$ and $\lambda_2-c=3$, then $(A+(a-1))v=\lambda v$, so $\lambda_1 - (a-1) = 0 \to \lambda_1=a-1$, and $\lambda_2 - (a-1) = 3 \to \lambda_2 = a+2$. Good, thanks for your answer!! $\endgroup$ – Gibbs Apr 18 at 16:42
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i think so. the matrix in question is called the rank one perturbation of the identity matrix. that is $A = (a-1)I +uu^\top$ where $u$ is called the unit vector with all entries one. it is know that $uu^t$ has eigenvalues $uu^\top$ and zero with multiplicity dimension of $u$ - 1 and the associated eigenvectors $u$ and $u^\perp.$ the eigenvalues of $(a-1)I + uu^\top$ are $(a-1+u^top u = a+1$ and two fold $a-1$ the determinant is the product of these eigenvalues. that is $(a-1)^2(a+2).$

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