# Finding the eigenvalues of $A=\left(\begin{smallmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \\ \end{smallmatrix}\right)$

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?

• Can you do it for $a=1$? – Lord Shark the Unknown Apr 18 at 16:09
• 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 $a=1$)? – Gibbs Apr 18 at 16:11
• Can you go from one particular case to the general case? – Lord Shark the Unknown Apr 18 at 16:21
• – StubbornAtom Apr 18 at 16:45
• 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. – amd Apr 18 at 17:19

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: $$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\ t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\ t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$

• What an easier and practical way to do it! Thanks a lot! – Gibbs Apr 18 at 17:29
• You are welcome. – farruhota Apr 18 at 17:31

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.

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 $$c=1-a$$?

• 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!! – Gibbs Apr 18 at 16:42

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 $$(a-1)^2(a+2).$$